COMPOUND ANGLES WORKSHEET

1) Use an appropriate compound angle formula to express as a single trig function, and then determine an exact value for each

a)  sin 𝜋/4 cos (𝜋/12) + cos (𝜋/4) sin (𝜋/12)

b)  cos (2𝜋/9) cos (5𝜋/18) - sin (2𝜋/9) sin (5𝜋/18)

2)  Apply a compound angle formula, and then determine an exact value for each.

a)  tan (𝜋/4 + 𝜋)

b)  tan (𝜋/3 - 𝜋/6)

3. Find the value of cos15°.

4. Find the value of sin75°.

5. Find the value of tan15°.

6. Find the value of tan165°.

7. If sin A = 4/5 (in quadrant I) and cos B = -12/13 (in quadrant II), then find (i) sin (A - B), (i) cos(A- B). 

8. If sin A = 3/5 and cos B = 9/41 , 0 < A < π/2, 0 < B < π/2, find the value of (i) sin (A + B) (ii) cos (A − B).

9. Angles 𝑥 and 𝑦 are located in the first quadrant such that sin𝑥 = 3/5 and cos𝑦 = 5/13 . Determine exact values  for cos 𝑥 and sin 𝑦.

1. Answer :

a)  sin (𝜋/4) cos (𝜋/12) + cos (𝜋/4) sin (𝜋/12)

sin (A + B) = sin A cos B + cos A sin B

Here A = 𝜋/4 and B = 𝜋/12

Using the formula above,

= sin ((𝜋/4) + (𝜋/12))

= sin ((3𝜋 + 𝜋)/12)

= sin (4𝜋/12)

= sin (𝜋/3)

= √3/2

b)  cos (2𝜋/9) cos (5𝜋/18) - sin (2𝜋/9) sin (5𝜋/18)

cos (A + B) = cos A  cos B - sin A sin B

Here A = 2𝜋/9 and B = 5𝜋/18

Using the formula above,

= cos (2𝜋/9 + 5𝜋/18)

= cos ((4𝜋 + 5𝜋)/18)

= cos (9𝜋/18)

= cos (𝜋/2)

= 0

2. Answer :

a)  tan (𝜋/4 + 𝜋)

tan (A + B) = tan A + tan B / (1 - tan A tan B)

A = 𝜋/4 and B = 𝜋

= (tan 𝜋/4 + tan 𝜋) / (1 - tan 𝜋/4 tan 𝜋) ----(1)

Evaluating the value of tan 𝜋 :

tan 𝜋 = tan (𝜋/2 + 𝜋/2)

= cot 𝜋/2

= 1/tan 𝜋/2

= 1/∞

= 0

Applying these values in (1), we get

= (1 + 0)/(1 - 1(0))

= 1/1

= 1

b)  tan (𝜋/3 - 𝜋/6)

tan (A - B) = (tan A - tan B) / (1 + tan A tan B)

A = 𝜋/3 and B = 𝜋/6

= (tan 𝜋/3 - tan 𝜋/6) / (1 + tan 𝜋/3 tan 𝜋/6) ----(1)

Evaluating the value of tan 𝜋/3 and tan 𝜋/6 :

tan 𝜋/3 = √3

tan 𝜋/6 = 1/√3

Applying these values in (1), we get 

= (√3 - (1/√3)) / (1 + √3(1/√3))

= [(3 - 1)/√3] / 2

= (2/√3) x (1/2)

= 1/√3

3. Answer :

Write the given angle 15° in terms of sum or difference of two standard angles. 

15° = 45° - 30°

cos15° = cos(45° - 30°)

= cos45°cos30° + sin45°sin30°

Using the above trigonometric ratio table, we have

= (√2/2) ⋅ (√3/2) + (√2/2) ⋅ (1/2)

= (√6/4)  +  (√2/4)

= (√6 + √2)/4

4. Answer :

Write the given angle 75° in terms of  sum or difference of two standard angles. 

75° = 45° + 30°

sin75° = sin(45° + 30°)

= sin45°cos30° + cos45°sin30°

Using the above trigonometric ratio table, we have

= (√2/2) ⋅ (√3/2) + (√2/2) ⋅ (1/2)

= (√6/4) + (√2/4)

= (√6 + √2)/4

5. Answer :

Write the given angle 15° in terms of  sum or difference of two standard angles. 

15° = 45° - 30°

tan15° = tan(45° - 30°)

= (tan45° -  tan30°)/(1 + tan45°tan30°)

Using the above trigonometric ratio table, we have

= (1 - 1/√3)/(1 + 1 ⋅ 1/√3)

=  (1 - 1/√3)/(1 + 1/√3)

= (√3/√3 - 1/√3)/(√3/√3 + 1/√3)

= [(√3 - 1)/√3]/[(√3 + 1)/√3]

= [(√3 - 1)/√3] x [(√3/(√3 + 1)]

= (√3 - 1)/(√3 + 1)

By rationalizing the denominator, we get

= 2 - √3

6. Answer :

Write the given angle 165° in terms of  sum or difference of two standard angles. 

165° = 120° + 45°

tan165° = tan(120° + 45°)

= (tan120° + tan45°)/(1 - tan120°tan45°) ----(1)

tan120° = tan(180° - 60°) = -tan60° = -√3.

(1)----> = (-√3 + 1)/[1 - (-√3)(1)]

= (1 - √3)/(1 + √3)

7. Answer :

sin²A + cos²A = 1

cos²A = 1 - sin²A

cosA = ±√(1 - sin²A)

Substitute sinA = 4/5.

cosA = ±√[1 - (4/5)²]

= ±√[1 - 16/25]

= ±√[(25 - 16)/25]

= ±√(9/25)

= ± 3/5

In the first quadrant, cosA is always positive.

cosA = 3/5

And also, 

sin²B + cos²B = 1

sin²B = 1 - cos²B

sinB = ±√(1 - cos²B)

Substitute cosB = -12/13.

sinB = ±√[1 - (-12/13)²]

= ±√[1 - 144/169]

= ±√[(169 - 144)/169]

= ±√[25/169]

= ± 5/13

In the second quadrant sinB is always positive.

sinB = 5/13

sin(A - B) : 

= sinAcosB - cosAsinB

= (4/5)(-12/13) - (3/5)(5/13)

= -48/65 - 15/65

sin(A- B) = -63/65

cos(A - B) : 

= cosAcosB + sinAsinB

= (3/5)(-12/13) + (4/5)(5/13)

= -36/65 + 20/65

cos(A- B) = -16/65

8. Answer :

sin(A + B) : 

= sinAcosB + cosAsinB

 = (3/5)(9/41) + (4/15)(40/41)

= (27/205) + (160/205)

= (27 + 160)/205

= 187/205

cos(A - B) :

= cosAcosB + sinAsinB

= (4/5)(9/41) + (3/5)(40/41)

= (36/205) + (120/205)

= (36 + 120)/205

= 156/205

9. Answer :

sin 𝑥 = 3/5

 cos 𝑦 = 5/13

cos x and sin 𝑦

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