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Unit 1: Right triangles & trigonometry
About this unit.
Can you find the length of a missing side of a right triangle? You most likely can: if you are given two side lengths you can use the Pythagorean Theorem to find the third one. But, what if you are only given one side? Impossible? Cue sine, cosine, and tangent, which will help you solve for any side or any angle of a right triangle.
Ratios in right triangles
- Getting ready for right triangles and trigonometry (Opens a modal)
- Hypotenuse, opposite, and adjacent (Opens a modal)
- Side ratios in right triangles as a function of the angles (Opens a modal)
- Using similarity to estimate ratio between side lengths (Opens a modal)
- Using right triangle ratios to approximate angle measure (Opens a modal)
- Right triangles & trigonometry: FAQ (Opens a modal)
- Use ratios in right triangles Get 3 of 4 questions to level up!
Introduction to the trigonometric ratios
- Triangle similarity & the trigonometric ratios (Opens a modal)
- Trigonometric ratios in right triangles (Opens a modal)
- Trigonometric ratios in right triangles Get 3 of 4 questions to level up!
Solving for a side in a right triangle using the trigonometric ratios
- Solving for a side in right triangles with trigonometry (Opens a modal)
- Solve for a side in right triangles Get 3 of 4 questions to level up!
Solving for an angle in a right triangle using the trigonometric ratios
- Intro to inverse trig functions (Opens a modal)
- Solve for an angle in right triangles Get 3 of 4 questions to level up!
Sine and cosine of complementary angles
- Intro to the Pythagorean trig identity (Opens a modal)
- Sine & cosine of complementary angles (Opens a modal)
- Using complementary angles (Opens a modal)
- Trig word problem: complementary angles (Opens a modal)
- Trig challenge problem: trig values & side ratios (Opens a modal)
- Trig ratios of special triangles (Opens a modal)
- Relate ratios in right triangles Get 3 of 4 questions to level up!
Modeling with right triangles
- Right triangle word problem (Opens a modal)
- Angles of elevation and depression (Opens a modal)
- Right triangle trigonometry review (Opens a modal)
- Right triangle trigonometry word problems Get 3 of 4 questions to level up!
The reciprocal trigonometric ratios
- Reciprocal trig ratios (Opens a modal)
- Finding reciprocal trig ratios (Opens a modal)
- Using reciprocal trig ratios (Opens a modal)
- Trigonometric ratios review (Opens a modal)
- Reciprocal trig ratios Get 5 of 7 questions to level up!
9.1 Verifying Trigonometric Identities and Using Trigonometric Identities to Simplify Trigonometric Expressions
sin 2 θ − 1 tan θ sin θ − tan θ = ( sin θ + 1 ) ( sin θ − 1 ) tan θ ( sin θ − 1 ) = sin θ + 1 tan θ sin 2 θ − 1 tan θ sin θ − tan θ = ( sin θ + 1 ) ( sin θ − 1 ) tan θ ( sin θ − 1 ) = sin θ + 1 tan θ
This is a difference of squares formula: 25 − 9 sin 2 θ = ( 5 − 3 sin θ ) ( 5 + 3 sin θ ) . 25 − 9 sin 2 θ = ( 5 − 3 sin θ ) ( 5 + 3 sin θ ) .
9.2 Sum and Difference Identities
2 + 6 4 2 + 6 4
2 − 6 4 2 − 6 4
1 − 3 1 + 3 1 − 3 1 + 3
cos ( 5 π 14 ) cos ( 5 π 14 )
9.3 Double-Angle, Half-Angle, and Reduction Formulas
cos ( 2 α ) = 7 32 cos ( 2 α ) = 7 32
cos 4 θ − sin 4 θ = ( cos 2 θ + sin 2 θ ) ( cos 2 θ − sin 2 θ ) = cos ( 2 θ ) cos 4 θ − sin 4 θ = ( cos 2 θ + sin 2 θ ) ( cos 2 θ − sin 2 θ ) = cos ( 2 θ )
cos ( 2 θ ) cos θ = ( cos 2 θ − sin 2 θ ) cos θ = cos 3 θ − cos θ sin 2 θ cos ( 2 θ ) cos θ = ( cos 2 θ − sin 2 θ ) cos θ = cos 3 θ − cos θ sin 2 θ
10 cos 4 x = 10 ( cos 2 x ) 2 = 10 [ 1 + cos ( 2 x ) 2 ] 2 Substitute reduction formula for cos 2 x . = 10 4 [ 1 + 2 cos ( 2 x ) + cos 2 ( 2 x ) ] = 10 4 + 10 2 cos ( 2 x ) + 10 4 ( 1 + cos 2 ( 2 x ) 2 ) Substitute reduction formula for cos 2 x . = 10 4 + 10 2 cos ( 2 x ) + 10 8 + 10 8 cos ( 4 x ) = 30 8 + 5 cos ( 2 x ) + 10 8 cos ( 4 x ) = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x ) 10 cos 4 x = 10 ( cos 2 x ) 2 = 10 [ 1 + cos ( 2 x ) 2 ] 2 Substitute reduction formula for cos 2 x . = 10 4 [ 1 + 2 cos ( 2 x ) + cos 2 ( 2 x ) ] = 10 4 + 10 2 cos ( 2 x ) + 10 4 ( 1 + cos 2 ( 2 x ) 2 ) Substitute reduction formula for cos 2 x . = 10 4 + 10 2 cos ( 2 x ) + 10 8 + 10 8 cos ( 4 x ) = 30 8 + 5 cos ( 2 x ) + 10 8 cos ( 4 x ) = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x )
− 2 5 − 2 5
9.4 Sum-to-Product and Product-to-Sum Formulas
1 2 ( cos 6 θ + cos 2 θ ) 1 2 ( cos 6 θ + cos 2 θ )
1 2 ( sin 2 x + sin 2 y ) 1 2 ( sin 2 x + sin 2 y )
− 2 − 3 4 − 2 − 3 4
2 sin ( 2 θ ) cos ( θ ) 2 sin ( 2 θ ) cos ( θ )
tan θ cot θ − cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ tan θ cot θ − cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ
9.5 Solving Trigonometric Equations
x = 7 π 6 , 11 π 6 x = 7 π 6 , 11 π 6
π 3 ± π k π 3 ± π k
θ ≈ 1.7722 ± 2 π k θ ≈ 1.7722 ± 2 π k and θ ≈ 4.5110 ± 2 π k θ ≈ 4.5110 ± 2 π k
cos θ = − 1 , θ = π cos θ = − 1 , θ = π
π 2 , 2 π 3 , 4 π 3 , 3 π 2 π 2 , 2 π 3 , 4 π 3 , 3 π 2
9.1 Section Exercises
All three functions, F F , G G , and H H , are even.
This is because F ( − x ) = sin ( − x ) sin ( − x ) = ( − sin x ) ( − sin x ) = sin 2 x = F ( x ) F ( − x ) = sin ( − x ) sin ( − x ) = ( − sin x ) ( − sin x ) = sin 2 x = F ( x ) , G ( − x ) = cos ( − x ) cos ( − x ) = cos x cos x = cos 2 x = G ( x ) G ( − x ) = cos ( − x ) cos ( − x ) = cos x cos x = cos 2 x = G ( x ) and H ( − x ) = tan ( − x ) tan ( − x ) = ( − tan x ) ( − tan x ) = tan 2 x = H ( x ) . H ( − x ) = tan ( − x ) tan ( − x ) = ( − tan x ) ( − tan x ) = tan 2 x = H ( x ) .
When cos t = 0 , cos t = 0 , then sec t = 1 0 , sec t = 1 0 , which is undefined.
sin x sin x
sec x sec x
csc t csc t
sec 2 x sec 2 x
sin 2 x + 1 sin 2 x + 1
1 sin x 1 sin x
1 cot x 1 cot x
tan x tan x
− 4 sec x tan x − 4 sec x tan x
± 1 cot 2 x + 1 ± 1 cot 2 x + 1
± 1 − sin 2 x sin x ± 1 − sin 2 x sin x
Answers will vary. Sample proof:
cos x − cos 3 x = cos x ( 1 − cos 2 x ) = cos x sin 2 x cos x − cos 3 x = cos x ( 1 − cos 2 x ) = cos x sin 2 x
Answers will vary. Sample proof: 1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = sec 2 x + tan 2 x = tan 2 x + 1 + tan 2 x = 1 + 2 tan 2 x 1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = sec 2 x + tan 2 x = tan 2 x + 1 + tan 2 x = 1 + 2 tan 2 x
Answers will vary. Sample proof: cos 2 x − tan 2 x = 1 − sin 2 x − ( sec 2 x − 1 ) = 1 − sin 2 x − sec 2 x + 1 = 2 − sin 2 x − sec 2 x cos 2 x − tan 2 x = 1 − sin 2 x − ( sec 2 x − 1 ) = 1 − sin 2 x − sec 2 x + 1 = 2 − sin 2 x − sec 2 x
Proved with negative and Pythagorean identities
True 3 sin 2 θ + 4 cos 2 θ = 3 sin 2 θ + 3 cos 2 θ + cos 2 θ = 3 ( sin 2 θ + cos 2 θ ) + cos 2 θ = 3 + cos 2 θ 3 sin 2 θ + 4 cos 2 θ = 3 sin 2 θ + 3 cos 2 θ + cos 2 θ = 3 ( sin 2 θ + cos 2 θ ) + cos 2 θ = 3 + cos 2 θ
9.2 Section Exercises
The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures x , x , the second angle measures π 2 − x . π 2 − x . Then sin x = cos ( π 2 − x ) . sin x = cos ( π 2 − x ) . The same holds for the other cofunction identities. The key is that the angles are complementary.
sin ( − x ) = − sin x , sin ( − x ) = − sin x , so sin x sin x is odd. cos ( − x ) = cos ( 0 − x ) = cos x , cos ( − x ) = cos ( 0 − x ) = cos x , so cos x cos x is even.
6 − 2 4 6 − 2 4
− 2 − 3 − 2 − 3
− 2 2 sin x − 2 2 cos x − 2 2 sin x − 2 2 cos x
− 1 2 cos x − 3 2 sin x − 1 2 cos x − 3 2 sin x
csc θ csc θ
cot x cot x
tan ( x 10 ) tan ( x 10 )
sin ( a − b ) = ( 4 5 ) ( 1 3 ) − ( 3 5 ) ( 2 2 3 ) = 4 − 6 2 15 cos ( a + b ) = ( 3 5 ) ( 1 3 ) − ( 4 5 ) ( 2 2 3 ) = 3 − 8 2 15 sin ( a − b ) = ( 4 5 ) ( 1 3 ) − ( 3 5 ) ( 2 2 3 ) = 4 − 6 2 15 cos ( a + b ) = ( 3 5 ) ( 1 3 ) − ( 4 5 ) ( 2 2 3 ) = 3 − 8 2 15
cot ( π 6 − x ) cot ( π 6 − x )
cot ( π 4 + x ) cot ( π 4 + x )
sin x 2 + cos x 2 sin x 2 + cos x 2
They are the same.
They are the different, try g ( x ) = sin ( 9 x ) − cos ( 3 x ) sin ( 6 x ) . g ( x ) = sin ( 9 x ) − cos ( 3 x ) sin ( 6 x ) .
They are the different, try g ( θ ) = 2 tan θ 1 − tan 2 θ . g ( θ ) = 2 tan θ 1 − tan 2 θ .
They are different, try g ( x ) = tan x − tan ( 2 x ) 1 + tan x tan ( 2 x ) . g ( x ) = tan x − tan ( 2 x ) 1 + tan x tan ( 2 x ) .
− 3 − 1 2 2 , or − 0.2588 − 3 − 1 2 2 , or − 0.2588
1 + 3 2 2 , 1 + 3 2 2 , or 0.9659
tan ( x + π 4 ) = tan x + tan ( π 4 ) 1 − tan x tan ( π 4 ) = tan x + 1 1 − tan x ( 1 ) = tan x + 1 1 − tan x tan ( x + π 4 ) = tan x + tan ( π 4 ) 1 − tan x tan ( π 4 ) = tan x + 1 1 − tan x ( 1 ) = tan x + 1 1 − tan x
cos ( a + b ) cos a cos b = cos a cos b cos a cos b − sin a sin b cos a cos b = 1 − tan a tan b cos ( a + b ) cos a cos b = cos a cos b cos a cos b − sin a sin b cos a cos b = 1 − tan a tan b
cos ( x + h ) − cos x h = cos x cosh − sin x sinh − cos x h = cos x ( cosh − 1 ) − sin x sinh h = cos x cos h − 1 h − sin x sin h h cos ( x + h ) − cos x h = cos x cosh − sin x sinh − cos x h = cos x ( cosh − 1 ) − sin x sinh h = cos x cos h − 1 h − sin x sin h h
True. Note that sin ( α + β ) = sin ( π − γ ) sin ( α + β ) = sin ( π − γ ) and expand the right hand side.
9.3 Section Exercises
Use the Pythagorean identities and isolate the squared term.
1 − cos x sin x , sin x 1 + cos x , 1 − cos x sin x , sin x 1 + cos x , multiplying the top and bottom by 1 − cos x 1 − cos x and 1 + cos x , 1 + cos x , respectively.
a) 3 7 32 3 7 32 b) 31 32 31 32 c) 3 7 31 3 7 31
a) 3 2 3 2 b) − 1 2 − 1 2 c) − 3 − 3
cos θ = − 2 5 5 , sin θ = 5 5 , tan θ = − 1 2 , csc θ = 5 , sec θ = − 5 2 , cot θ = − 2 cos θ = − 2 5 5 , sin θ = 5 5 , tan θ = − 1 2 , csc θ = 5 , sec θ = − 5 2 , cot θ = − 2
2 sin ( π 2 ) 2 sin ( π 2 )
2 − 2 2 2 − 2 2
2 − 3 2 2 − 3 2
2 + 3 2 + 3
− 1 − 2 − 1 − 2
a) 3 13 13 3 13 13 b) − 2 13 13 − 2 13 13 c) − 3 2 − 3 2
a) 10 4 10 4 b) 6 4 6 4 c) 15 3 15 3
120 169 , – 119 169 , – 120 119 120 169 , – 119 169 , – 120 119
2 13 13 , 3 13 13 , 2 3 2 13 13 , 3 13 13 , 2 3
cos ( 74° ) cos ( 74° )
cos ( 18 x ) cos ( 18 x )
3 sin ( 10 x ) 3 sin ( 10 x )
− 2 sin ( − x ) cos ( − x ) = − 2 ( − sin ( x ) cos ( x ) ) = sin ( 2 x ) − 2 sin ( − x ) cos ( − x ) = − 2 ( − sin ( x ) cos ( x ) ) = sin ( 2 x )
sin ( 2 θ ) 1 + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) 1 + cos 2 θ − sin 2 θ tan 2 θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan 2 θ = sin ( θ ) cos θ tan 2 θ = cot ( θ ) tan 2 θ = tan 3 θ sin ( 2 θ ) 1 + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) 1 + cos 2 θ − sin 2 θ tan 2 θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan 2 θ = sin ( θ ) cos θ tan 2 θ = cot ( θ ) tan 2 θ = tan 3 θ
1 + cos ( 12 x ) 2 1 + cos ( 12 x ) 2
3 + cos ( 12 x ) − 4 cos ( 6 x ) 8 3 + cos ( 12 x ) − 4 cos ( 6 x ) 8
2 + cos ( 2 x ) − 2 cos ( 4 x ) − cos ( 6 x ) 32 2 + cos ( 2 x ) − 2 cos ( 4 x ) − cos ( 6 x ) 32
3 + cos ( 4 x ) − 4 cos ( 2 x ) 3 + cos ( 4 x ) + 4 cos ( 2 x ) 3 + cos ( 4 x ) − 4 cos ( 2 x ) 3 + cos ( 4 x ) + 4 cos ( 2 x )
1 − cos ( 4 x ) 8 1 − cos ( 4 x ) 8
3 + cos ( 4 x ) − 4 cos ( 2 x ) 4 ( cos ( 2 x ) + 1 ) 3 + cos ( 4 x ) − 4 cos ( 2 x ) 4 ( cos ( 2 x ) + 1 )
( 1 + cos ( 4 x ) ) sin x 2 ( 1 + cos ( 4 x ) ) sin x 2
4 sin x cos x ( cos 2 x − sin 2 x ) 4 sin x cos x ( cos 2 x − sin 2 x )
2 tan x 1 + tan 2 x = 2 sin x cos x 1 + sin 2 x cos 2 x = 2 sin x cos x cos 2 x + sin 2 x cos 2 x = 2 sin x cos x . cos 2 x 1 = 2 sin x cos x = sin ( 2 x ) 2 tan x 1 + tan 2 x = 2 sin x cos x 1 + sin 2 x cos 2 x = 2 sin x cos x cos 2 x + sin 2 x cos 2 x = 2 sin x cos x . cos 2 x 1 = 2 sin x cos x = sin ( 2 x )
2 sin x cos x 2 cos 2 x − 1 = sin ( 2 x ) cos ( 2 x ) = tan ( 2 x ) 2 sin x cos x 2 cos 2 x − 1 = sin ( 2 x ) cos ( 2 x ) = tan ( 2 x )
sin ( x + 2 x ) = sin x cos ( 2 x ) + sin ( 2 x ) cos x = sin x ( cos 2 x − sin 2 x ) + 2 sin x cos x cos x = sin x cos 2 x − sin 3 x + 2 sin x cos 2 x = 3 sin x cos 2 x − sin 3 x sin ( x + 2 x ) = sin x cos ( 2 x ) + sin ( 2 x ) cos x = sin x ( cos 2 x − sin 2 x ) + 2 sin x cos x cos x = sin x cos 2 x − sin 3 x + 2 sin x cos 2 x = 3 sin x cos 2 x − sin 3 x
1 + cos ( 2 t ) sin ( 2 t ) − cos t = 1 + 2 cos 2 t − 1 2 sin t cos t − cos t = 2 cos 2 t cos t ( 2 sin t − 1 ) = 2 cos t 2 sin t − 1 1 + cos ( 2 t ) sin ( 2 t ) − cos t = 1 + 2 cos 2 t − 1 2 sin t cos t − cos t = 2 cos 2 t cos t ( 2 sin t − 1 ) = 2 cos t 2 sin t − 1
( cos 2 ( 4 x ) − sin 2 ( 4 x ) − sin ( 8 x ) ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) + sin ( 8 x ) ) = = ( cos ( 8 x ) − sin ( 8 x ) ) ( cos ( 8 x ) + sin ( 8 x ) ) = cos 2 ( 8 x ) − sin 2 ( 8 x ) = cos ( 16 x ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) − sin ( 8 x ) ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) + sin ( 8 x ) ) = = ( cos ( 8 x ) − sin ( 8 x ) ) ( cos ( 8 x ) + sin ( 8 x ) ) = cos 2 ( 8 x ) − sin 2 ( 8 x ) = cos ( 16 x )
9.4 Section Exercises
Substitute α α into cosine and β β into sine and evaluate.
Answers will vary. There are some equations that involve a sum of two trig expressions where when converted to a product are easier to solve. For example: sin ( 3 x ) + sin x cos x = 1. sin ( 3 x ) + sin x cos x = 1. When converting the numerator to a product the equation becomes: 2 sin ( 2 x ) cos x cos x = 1 2 sin ( 2 x ) cos x cos x = 1
8 ( cos ( 5 x ) − cos ( 27 x ) ) 8 ( cos ( 5 x ) − cos ( 27 x ) )
sin ( 2 x ) + sin ( 8 x ) sin ( 2 x ) + sin ( 8 x )
1 2 ( cos ( 6 x ) − cos ( 4 x ) ) 1 2 ( cos ( 6 x ) − cos ( 4 x ) )
2 cos ( 5 t ) cos t 2 cos ( 5 t ) cos t
2 cos ( 7 x ) 2 cos ( 7 x )
2 cos ( 6 x ) cos ( 3 x ) 2 cos ( 6 x ) cos ( 3 x )
1 4 ( 1 + 3 ) 1 4 ( 1 + 3 )
1 4 ( 3 − 2 ) 1 4 ( 3 − 2 )
1 4 ( 3 − 1 ) 1 4 ( 3 − 1 )
cos ( 80° ) − cos ( 120° ) cos ( 80° ) − cos ( 120° )
1 2 ( sin ( 221° ) + sin ( 205° ) ) 1 2 ( sin ( 221° ) + sin ( 205° ) )
2 cos ( 31° ) 2 cos ( 31° )
2 cos ( 66.5° ) sin ( 34.5° ) 2 cos ( 66.5° ) sin ( 34.5° )
2 sin ( −1.5° ) cos ( 0.5° ) 2 sin ( −1.5° ) cos ( 0.5° )
2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 x ) = 2 ( sin ( 4 x ) cos ( 3 x ) + sin ( 3 x ) cos ( 4 x ) ) − 2 ( sin ( 4 x ) cos ( 3 x ) − sin ( 3 x ) cos ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) = 4 sin ( 3 x ) cos ( 4 x ) 2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 x ) = 2 ( sin ( 4 x ) cos ( 3 x ) + sin ( 3 x ) cos ( 4 x ) ) − 2 ( sin ( 4 x ) cos ( 3 x ) − sin ( 3 x ) cos ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) = 4 sin ( 3 x ) cos ( 4 x )
sin x + sin ( 3 x ) = 2 sin ( 4 x 2 ) cos ( − 2 x 2 ) = 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 4 sin x cos 2 x sin x + sin ( 3 x ) = 2 sin ( 4 x 2 ) cos ( − 2 x 2 ) = 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 4 sin x cos 2 x
2 tan x cos ( 3 x ) = 2 sin x cos ( 3 x ) cos x = 2 ( .5 ( sin ( 4 x ) − sin ( 2 x ) ) ) cos x = 1 cos x ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) ) 2 tan x cos ( 3 x ) = 2 sin x cos ( 3 x ) cos x = 2 ( .5 ( sin ( 4 x ) − sin ( 2 x ) ) ) cos x = 1 cos x ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) )
2 cos ( 35° ) cos ( 23° ) , 1.5081 2 cos ( 35° ) cos ( 23° ) , 1.5081
− 2 sin ( 33° ) sin ( 11° ) , − 0.2078 − 2 sin ( 33° ) sin ( 11° ) , − 0.2078
1 2 ( cos ( 99° ) − cos ( 71° ) ) , −0.2410 1 2 ( cos ( 99° ) − cos ( 71° ) ) , −0.2410
It is an identity.
It is not an identity, but 2 cos 3 x 2 cos 3 x is.
tan ( 3 t ) tan ( 3 t )
2 cos ( 2 x ) 2 cos ( 2 x )
− sin ( 14 x ) − sin ( 14 x )
Start with cos x + cos y . cos x + cos y . Make a substitution and let x = α + β x = α + β and let y = α − β , y = α − β , so cos x + cos y cos x + cos y becomes cos ( α + β ) + cos ( α − β ) = cos α cos β − sin α sin β + cos α cos β + sin α sin β = 2 cos α cos β cos ( α + β ) + cos ( α − β ) = cos α cos β − sin α sin β + cos α cos β + sin α sin β = 2 cos α cos β
Since x = α + β x = α + β and y = α − β , y = α − β , we can solve for α α and β β in terms of x and y and substitute in for 2 cos α cos β 2 cos α cos β and get 2 cos ( x + y 2 ) cos ( x − y 2 ) . 2 cos ( x + y 2 ) cos ( x − y 2 ) .
cos ( 3 x ) + cos x cos ( 3 x ) − cos x = 2 cos ( 2 x ) cos x − 2 sin ( 2 x ) sin x = − cot ( 2 x ) cot x cos ( 3 x ) + cos x cos ( 3 x ) − cos x = 2 cos ( 2 x ) cos x − 2 sin ( 2 x ) sin x = − cot ( 2 x ) cot x
cos ( 2 y ) − cos ( 4 y ) sin ( 2 y ) + sin ( 4 y ) = − 2 sin ( 3 y ) sin ( − y ) 2 sin ( 3 y ) cos y = 2 sin ( 3 y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y cos ( 2 y ) − cos ( 4 y ) sin ( 2 y ) + sin ( 4 y ) = − 2 sin ( 3 y ) sin ( − y ) 2 sin ( 3 y ) cos y = 2 sin ( 3 y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y
cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x
tan ( π 4 − t ) = tan ( π 4 ) − tan t 1 + tan ( π 4 ) tan ( t ) = 1 − tan t 1 + tan t tan ( π 4 − t ) = tan ( π 4 ) − tan t 1 + tan ( π 4 ) tan ( t ) = 1 − tan t 1 + tan t
9.5 Section Exercises
There will not always be solutions to trigonometric function equations. For a basic example, cos ( x ) = −5. cos ( x ) = −5.
If the sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide both sides of the equation by the leading coefficient. Then, if the number it is set equal to has an absolute value greater than one, the equation has no solution.
π 3 , 2 π 3 π 3 , 2 π 3
3 π 4 , 5 π 4 3 π 4 , 5 π 4
π 4 , 5 π 4 π 4 , 5 π 4
π 4 , 3 π 4 , 5 π 4 , 7 π 4 π 4 , 3 π 4 , 5 π 4 , 7 π 4
π 4 , 7 π 4 π 4 , 7 π 4
7 π 6 , 11 π 6 7 π 6 , 11 π 6
π 18 , 5 π 18 , 13 π 18 , 17 π 18 , 25 π 18 , 29 π 18 π 18 , 5 π 18 , 13 π 18 , 17 π 18 , 25 π 18 , 29 π 18
3 π 12 , 5 π 12 , 11 π 12 , 13 π 12 , 19 π 12 , 21 π 12 3 π 12 , 5 π 12 , 11 π 12 , 13 π 12 , 19 π 12 , 21 π 12
1 6 , 5 6 , 13 6 , 17 6 , 25 6 , 29 6 , 37 6 1 6 , 5 6 , 13 6 , 17 6 , 25 6 , 29 6 , 37 6
0 , π 3 , π , 5 π 3 0 , π 3 , π , 5 π 3
π 3 , π , 5 π 3 π 3 , π , 5 π 3
π 3 , 3 π 2 , 5 π 3 π 3 , 3 π 2 , 5 π 3
0 , π 0 , π
π − sin − 1 ( − 1 4 ) , 7 π 6 , 11 π 6 , 2 π + sin − 1 ( − 1 4 ) π − sin − 1 ( − 1 4 ) , 7 π 6 , 11 π 6 , 2 π + sin − 1 ( − 1 4 )
1 3 ( sin − 1 ( 9 10 ) ) 1 3 ( sin − 1 ( 9 10 ) ) , π 3 − 1 3 ( sin − 1 ( 9 10 ) ) π 3 − 1 3 ( sin − 1 ( 9 10 ) ) , 2 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) 2 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , π − 1 3 ( sin − 1 ( 9 10 ) ) π − 1 3 ( sin − 1 ( 9 10 ) ) , 4 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) 4 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , 5 π 3 − 1 3 ( sin − 1 ( 9 10 ) ) 5 π 3 − 1 3 ( sin − 1 ( 9 10 ) )
π 6 , 5 π 6 , 7 π 6 , 11 π 6 π 6 , 5 π 6 , 7 π 6 , 11 π 6
3 π 2 , π 6 , 5 π 6 3 π 2 , π 6 , 5 π 6
0 , π 3 , π , 4 π 3 0 , π 3 , π , 4 π 3
There are no solutions.
cos − 1 ( 1 3 ( 1 − 7 ) ) cos − 1 ( 1 3 ( 1 − 7 ) ) , 2 π − cos − 1 ( 1 3 ( 1 − 7 ) ) 2 π − cos − 1 ( 1 3 ( 1 − 7 ) )
tan − 1 ( 1 2 ( 29 − 5 ) ) tan − 1 ( 1 2 ( 29 − 5 ) ) , π + tan − 1 ( 1 2 ( − 29 − 5 ) ) π + tan − 1 ( 1 2 ( − 29 − 5 ) ) , π + tan − 1 ( 1 2 ( 29 − 5 ) ) π + tan − 1 ( 1 2 ( 29 − 5 ) ) , 2 π + tan − 1 ( 1 2 ( − 29 − 5 ) ) 2 π + tan − 1 ( 1 2 ( − 29 − 5 ) )
0 , 2 π 3 , 4 π 3 0 , 2 π 3 , 4 π 3
sin − 1 ( 3 5 ) , π 2 , π − sin − 1 ( 3 5 ) , 3 π 2 sin − 1 ( 3 5 ) , π 2 , π − sin − 1 ( 3 5 ) , 3 π 2
cos − 1 ( − 1 4 ) , 2 π − cos − 1 ( − 1 4 ) cos − 1 ( − 1 4 ) , 2 π − cos − 1 ( − 1 4 )
π 3 π 3 , cos − 1 ( − 3 4 ) cos − 1 ( − 3 4 ) , 2 π − cos − 1 ( − 3 4 ) 2 π − cos − 1 ( − 3 4 ) , 5 π 3 5 π 3
cos − 1 ( 3 4 ) cos − 1 ( 3 4 ) , cos − 1 ( − 2 3 ) cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( − 2 3 ) 2 π − cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( 3 4 ) 2 π − cos − 1 ( 3 4 )
0 , π 2 , π , 3 π 2 0 , π 2 , π , 3 π 2
π 3 π 3 , cos −1 ( − 1 4 ) cos −1 ( − 1 4 ) , 2 π − cos −1 ( − 1 4 ) 2 π − cos −1 ( − 1 4 ) , 5 π 3 5 π 3
π + tan −1 ( −2 ) π + tan −1 ( −2 ) , π + tan −1 ( − 3 2 ) π + tan −1 ( − 3 2 ) , 2 π + tan −1 ( −2 ) 2 π + tan −1 ( −2 ) , 2 π + tan −1 ( − 3 2 ) 2 π + tan −1 ( − 3 2 )
2 π k + 0.2734 , 2 π k + 2.8682 2 π k + 0.2734 , 2 π k + 2.8682
π k − 0.3277 π k − 0.3277
0.6694 , 1.8287 , 3.8110 , 4.9703 0.6694 , 1.8287 , 3.8110 , 4.9703
1.0472 , 3.1416 , 5.2360 1.0472 , 3.1416 , 5.2360
0.5326 , 1.7648 , 3.6742 , 4.9064 0.5326 , 1.7648 , 3.6742 , 4.9064
sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) , 3 π 2 sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) , 3 π 2
π 2 , 3 π 2 π 2 , 3 π 2
7.2 ∘ 7.2 ∘
5.7 ∘ 5.7 ∘
82.4 ∘ 82.4 ∘
31.0 ∘ 31.0 ∘
88.7 ∘ 88.7 ∘
59.0 ∘ 59.0 ∘
36.9 ∘ 36.9 ∘
Review Exercises
sin − 1 ( 3 3 ) sin − 1 ( 3 3 ) , π − sin − 1 ( 3 3 ) π − sin − 1 ( 3 3 ) , π + sin − 1 ( 3 3 ) π + sin − 1 ( 3 3 ) , 2 π − sin − 1 ( 3 3 ) 2 π − sin − 1 ( 3 3 )
sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 )
cos ( 4 x ) − cos ( 3 x ) cos x = cos ( 2 x + 2 x ) − cos ( x + 2 x ) cos x = cos ( 2 x ) cos ( 2 x ) − sin ( 2 x ) sin ( 2 x ) − cos x cos ( 2 x ) cos x + sin x sin ( 2 x ) cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + sin x ( 2 ) sin x cos x cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + 2 sin 2 x cos 2 x = cos 4 x − 2 cos 2 x sin 2 x + sin 4 x − 4 cos 2 x sin 2 x − cos 4 x + cos 2 x sin 2 x + 2 sin 2 x cos 2 x = sin 4 x − 4 cos 2 x sin 2 x + cos 2 x sin 2 x = sin 2 x ( sin 2 x + cos 2 x ) − 4 cos 2 x sin 2 x = sin 2 x − 4 cos 2 x sin 2 x cos ( 4 x ) − cos ( 3 x ) cos x = cos ( 2 x + 2 x ) − cos ( x + 2 x ) cos x = cos ( 2 x ) cos ( 2 x ) − sin ( 2 x ) sin ( 2 x ) − cos x cos ( 2 x ) cos x + sin x sin ( 2 x ) cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + sin x ( 2 ) sin x cos x cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + 2 sin 2 x cos 2 x = cos 4 x − 2 cos 2 x sin 2 x + sin 4 x − 4 cos 2 x sin 2 x − cos 4 x + cos 2 x sin 2 x + 2 sin 2 x cos 2 x = sin 4 x − 4 cos 2 x sin 2 x + cos 2 x sin 2 x = sin 2 x ( sin 2 x + cos 2 x ) − 4 cos 2 x sin 2 x = sin 2 x − 4 cos 2 x sin 2 x
tan ( 5 8 x ) tan ( 5 8 x )
− 24 25 , − 7 25 , 24 7 − 24 25 , − 7 25 , 24 7
2 ( 2 + 2 ) 2 ( 2 + 2 )
2 10 , 7 2 10 , 1 7 , 3 5 , 4 5 , 3 4 2 10 , 7 2 10 , 1 7 , 3 5 , 4 5 , 3 4
cot x cos ( 2 x ) = cot x ( 1 − 2 sin 2 x ) = cot x − cos x sin x ( 2 ) sin 2 x = − 2 sin x cos x + cot x = − sin ( 2 x ) + cot x cot x cos ( 2 x ) = cot x ( 1 − 2 sin 2 x ) = cot x − cos x sin x ( 2 ) sin 2 x = − 2 sin x cos x + cot x = − sin ( 2 x ) + cot x
10 sin x − 5 sin ( 3 x ) + sin ( 5 x ) 8 ( cos ( 2 x ) + 1 ) 10 sin x − 5 sin ( 3 x ) + sin ( 5 x ) 8 ( cos ( 2 x ) + 1 )
− 2 2 − 2 2
1 2 ( sin ( 6 x ) + sin ( 12 x ) ) 1 2 ( sin ( 6 x ) + sin ( 12 x ) )
2 sin ( 13 2 x ) cos ( 9 2 x ) 2 sin ( 13 2 x ) cos ( 9 2 x )
3 π 4 , 7 π 4 3 π 4 , 7 π 4
0 , π 6 , 5 π 6 , π 0 , π 6 , 5 π 6 , π
3 π 2 3 π 2
No solution
0.2527 , 2.8889 , 4.7124 0.2527 , 2.8889 , 4.7124
1.3694 , 1.9106 , 4.3726 , 4.9137 1.3694 , 1.9106 , 4.3726 , 4.9137
Practice Test
sec ( θ ) sec ( θ )
− 1 2 cos θ + 3 2 sin θ − 1 2 cos θ + 3 2 sin θ
1 − cos ( 64 ∘ ) 2 1 − cos ( 64 ∘ ) 2
2 cos ( 3 x ) cos ( 5 x ) 2 cos ( 3 x ) cos ( 5 x )
4 sin ( 2 θ ) cos ( 6 θ ) 4 sin ( 2 θ ) cos ( 6 θ )
x = cos –1 ( 1 5 ) x = cos –1 ( 1 5 )
3 5 , – 4 5 , – 3 4 3 5 , – 4 5 , – 3 4
tan 3 x – tan x sec 2 x = tan x ( tan 2 x – sec 2 x ) = tan x ( tan 2 x – ( 1 + tan 2 x ) ) = tan x ( tan 2 x – 1 – tan 2 x ) = – tan x = tan ( – x ) = tan ( – x ) tan 3 x – tan x sec 2 x = tan x ( tan 2 x – sec 2 x ) = tan x ( tan 2 x – ( 1 + tan 2 x ) ) = tan x ( tan 2 x – 1 – tan 2 x ) = – tan x = tan ( – x ) = tan ( – x )
sin ( 2 x ) sin x – cos ( 2 x ) cos x = 2 sin x cos x sin x – 2 cos 2 x – 1 cos x = 2 cos x – 2 cos x + 1 cos x = 1 cos x = sec x = sec x sin ( 2 x ) sin x – cos ( 2 x ) cos x = 2 sin x cos x sin x – 2 cos 2 x – 1 cos x = 2 cos x – 2 cos x + 1 cos x = 1 cos x = sec x = sec x
Amplitude: 1 4 1 4 , period: 1 60 1 60 , frequency: 60 Hz
Amplitude: 8, fast period: 1 500 1 500 , fast frequency: 500 Hz, slow period: 1 10 1 10 , slow frequency: 10 Hz
D ( t ) = 20 ( 0.9086 ) t cos ( 4 π t ) D ( t ) = 20 ( 0.9086 ) t cos ( 4 π t ) , 31 second
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Access for free at https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites
- Authors: Jay Abramson
- Publisher/website: OpenStax
- Book title: Algebra and Trigonometry
- Publication date: Feb 13, 2015
- Location: Houston, Texas
- Book URL: https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites
- Section URL: https://openstax.org/books/algebra-and-trigonometry/pages/chapter-9
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Name: Date: Unit 8: Right Triangles & Trigonometry Homework 5: Trigonometry: Finding Sides and Angles ** This is a 2-page document! ** -tan 39 X: 33,3 Directions: Solve for x. Round to the nearest tenth. Cos 143 = 52 = Cos 16: fin X = 5 X: COS-I (£9 @ Gina Wilson (All Things Algebraø, LLC), 2014-2018. Name: Date: Unit 8: Right Triangles ...
Unit 8 - Right Triangles & Trigonometry. Directions: Use the Law of Cosines to solve for x. Round your answer to the nearest tenth. - - = 8105, 121 = cosx COS X cosx 2q{u -2.0 18 2.1131 46. A utility pole is supported by two wires, one on each side going in the opposite direction. The two wires form a 75' angle at the utility pole.
Geometry questions and answers. Name: Date: Unit 8: Right Triangles & Trigonometry Homework 9: Law of Sines & Law of Cosines; + Applications ** This is a 2-page document ** Per Directions: Use the Law of Sines and/or the Law of Cosines to solve each triangle. Round to the nearest tenth when necessary 1. OR 19 mZP P 85 13 R MZO - 2.
Add-on. U08.AO.01 - Terminology Warm-Up for the Trigonometric Ratios (Before Lesson 2) RESOURCE. ANSWER KEY. EDITABLE RESOURCE. EDITABLE KEY.
Solutions Key 8 Right Triangles and Trigonometry CHAPTER ARE YOU READY? PAGE 515 1. D 2. C 3. A 4. E 5. PR___ RT ... 8-1 SIMILARITY IN RIGHT TRIANGLES, PAGES 518-523 CHECK IT OUT! PAGES 518-520 1. ... b. x = (10)(30) = 300 x √= 300 = 10 √3 c. x 2 = (8)(9) = 72 x = √ 72 = 6 ...
Unit 8 Similarity and Trigonometry Target 8.1: Solve problems using the Pythagorean Theorem 8.1a - Applying the Pythagorean Theorem 8.1b - Converse of the Pythagorean Theorem Target 8.2: Solve problems using similar right triangles 8.2a- Use Similar Right Triangles 8.2b- Special Right Triangles (45-45-90 & 30-60-90 Triangles)
Identify if the triangle is a right triangle or not. 20, 48, 52 By the converse of Pythagorean theorem, check the sum of squares of smaller sides with the square of largest side i.e., 220+482=400+2304=2704 252=2704 → 202+482= 522 The triangle is a right triangle. 3. The longest side in a right triangle is: e. hypotenuse f. adjacent g. opposite h.
45-45-90 Special Right Triangle. *For all isosceles right triangles, the length of the hypotenuse = the length of the leg times the square root of two. *If given the hypotenuse length, divide by the square root of two in order to find the length of the leg. 30-60-90 Special Right Triangle.
10 of 10. Quiz yourself with questions and answers for Unit 8 Test: Right Triangles & Trigonometry, so you can be ready for test day. Explore quizzes and practice tests created by teachers and students or create one from your course material.
Unit 8 Part 1 - Pythagorean Triples, Pythagorean Theorem and its Converse, Special Right Triangles. Flashcards; Learn; Test; ... Right Triangles and Trigonometry Unit Test. 16 terms. austin222345. Preview. Compaction of DNA into Chromatin . 65 terms. n_tessier. Preview. Unit 7 Review. 43 terms. mmirafuentes.
Geometry questions and answers. Name: Cayce Date: Per: Unit 8: Right Triangles & Trigonometry Homework 4: Trigonometric Ratios & Finding Missing Sides SOH CAH TOA ** This is a 2-page document! ** 1. 48/50 Р sin R = Directions: Give each trig ratio as a fraction in simplest form. 14/50 48 sin Q = 48150 cos 14/48 tan Q = Q 14150 14 .
Click here 👆 to get an answer to your question ️ unit 8: right triangles & trigonometry homework 9 law of sines and law of cosines + applications on que ... Final answer: To solve this homework, you could use the Law of Sines or the Law of Cosines, given that enough information about the sides and angles of the triangle are provided in the ...
Trigonometry 4 units · 36 skills. Unit 1 Right triangles & trigonometry. Unit 2 Trigonometric functions. Unit 3 Non-right triangles & trigonometry. Unit 4 Trigonometric equations and identities. Course challenge. Test your knowledge of the skills in this course. Start Course challenge. Math.
Trigonometry; Trigonometry questions and answers; Name: Unit 8: Right Triangles & Trigonometry Homework 8: Law of Cosines Date: Per ** This is a 2-page documenti ** Directions: Use the Law of Cosines to find each missing side. Round to the nearest tenth 1. 10 122 19 2. 14 67 8 15 38 13 34 26 21 Oina Won Althings Age 2014-2018
Introduction to Further Applications of Trigonometry; 10.1 Non-right Triangles: Law of Sines; 10.2 Non-right Triangles: Law of Cosines; 10.3 Polar Coordinates; 10.4 Polar Coordinates: Graphs; 10.5 Polar Form of Complex Numbers; 10.6 Parametric Equations; 10.7 Parametric Equations: Graphs; 10.8 Vectors
Trigonometric Ratios. The ratio of the length of two sides of a right triangle. Sine. opposite/hypotenuse. Cosine. adjacent/hypotenuse. Tangent. opposite/adjacent. Study with Quizlet and memorize flashcards containing terms like Special Right Triangles, Trigonometric Ratios, Sine and more.
x=. 35.6. Solve for the missing angle x, round to the nearest tenth. 41.2. Solve for the missing angle x, round to the nearest tenth. 43.9. Solve for the missing angle x, round to the nearest tenth. Unit 8: Trigonometry REVIEW. 2.5.
Unit 8: Right Triangles & Trigonometry Homework 4: Trigonometric Ratios & Finding Missing Sides Per: This is a 2-page document. Directions: Give each trig ratio as a fraction in simplest form. sin Q = 48/50 sin R = 49/50 cos Q = 14/42 cos R = 42/14 tan Q = tan R = Directions: Solve for x. Round to the nearest tenth: TAN Cos6l = 0.24 2X - 7.10 ...
This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. Question: Unit 8: Right Triangles & Trigonometry Homework 5: Trigonometry: Finding Sides and Angles ge document! ** enth. 2. TAN 39 - = 27 399 X х 27 X=27 Tana9 X=33.34 159 33 х 6. 5 27° х.
Transcribed image text: Name: Unit 7: Right Triangles & Trigonometry Date: Per: Homework 9: Law of Sines & Law of Cosines; + Applications ** This is a 2-page document! Directions: Use the Law of Sines and/or the Law of Cosines to solve each triangle. Round to the nearest tenth when necessary. 1. OR = 19 MZP = P 85 R 13 m29- 2.
Geometry questions and answers; Name: Unit 8: Right Triangles & Trigonometry Date: Per: Homework 4: Trigonometric Ratios & Finding Missing Sides ** This is a 2-page document ** Directions: Give eachtrig ratio as a fraction in simplest form. 1. . • sin = • sin R 14 50 . • cos Q- cos R= . tan R • tan = Directions: Solve for x.
Question: Name: Unit 8: Right Triangles & Trigonometry Homework 7: Law of Sines Per Date: ** This is a 2-page document! Directions: Use the Law of Sines to find each missing side or angle. Round to the nearest tenth. 1. 2. 22 5 65 46 29 53 3. т 73 59 1280 18 12 15 5. 191 75 32 26 28 7. 9 514 52 70 16 Gna Wilson ( Ang Algebra C, 2014-2018