COMPOUND ANGLES WORKSHEET

1. Find the value of cos15°.

2. Find the value of sin75°.

3. Find the value of tan15°.

4. Find the value of tan165°.

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

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

1. 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

2. 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

3. 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

4. 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)

5. 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

6. Answer :

sinA = 3/5

cosA = √(1 - sin2x)

√(1 - (3/5)2)

√(1 - (9/25))

√(25 - 9)/25

√(16/25)

cosA = 4/5

cosB = 9/41

sinB = √(1 - cos2B)

√(1 - (9/41)2)

√(1 - (81/1681)

√(1681 - 81)/1681

√(1600/1681)

sinB = 40/41

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

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