﻿ Double Angles and Half Angles Worksheet | Problems & Solutions

# Double Angles and Half Angles Worksheet

Double Angles and Half Angles Worksheet
• Page 1
1.
Find the value of cot $x$ - .
 a. tan $x$ b. sin $x$ c. cos $x$ d. cot $x$

#### Solution:

cot x - cos 2xsin x cos x

= cot x - (cos²  x - sin² x)sin x cos x
[Use cos 2x = cos² x - sin² x.]

cot x - cos² xsin x cos x + sin² xsin x cos x
[Seperate the terms.]

= cot x - cot x + tan x
[Simplify.]

= tan x

Therefore, the value of cot x - cos 2xsin x cos x = tan x.

2.
If cos $x$ = 0.425 then cos ($\frac{x}{2}$) =
 a. 0.716 b. 0.642 c. 0.844 d. 0.783

#### Solution:

2 cos2(x2) - 1 = cos x
[Use double-angle formula.]

2 cos2(x2) = 1 + cos x
[Simplify.]

2 cos2(x2) = 1 + 0.425 = 1.425
[Substitute the value of cos x.]

cos2(x2) = 0.7125
[Simplify.]

cos (x2) = 0.7125 = 0.844097 = 0.844
[Find 0.7125 and round it three decimal places.]

Therefore, cos x2 = 0.844.

3.
If cosec $\theta$ = , then find the value of cot ($\frac{1}{4}$$\pi$ + $\frac{1}{2}$θ ).
 a. $\sqrt{\frac{q}{p}}$ b. $\frac{p}{q}$ c. $p$$q$ d. $\sqrt{\frac{p}{q}}$

#### Solution:

cosec θ = p + qp - q

1sin θ = p+qp-q
[Reciprocal Identity.]

1+tan2θ22 tanθ2=p+qp-q
[Write sin θ in terms of tan θ .]

1+tan2θ/2+2tanθ/21+tan2θ/2-2tanθ/2 = p+q+p-qp+q-p+q=pq
[ab =cd if and only if a + ba - b=c + dc - d.]

[1+tanθ21-tanθ2]² = pq
[(a + b)² = a² + b² + 2ab and (a - b)² = a² + b² - 2ab.]

1+tanθ21-tanθ2 =pq
[Take square root on both the sides.]

cot (π4+θ2) = 1tan (π4+θ2)
[Reiprocal Identity.]

= 11 + tanθ21 - tanθ2
[Expand tan (π4 +θ2).]

= 1pq = qp

Therefore, the value of cot (π4 +θ2) = qp.

4.
Find the value of tan 180° using triple-angle formula.
 a. $\frac{1}{\sqrt{3}}$ b. - 1 c. 1

#### Solution:

Tan 180° = Tan 3(60)°

= 3tan 60 -tan3 601 - 3tan2 60
[tan 3A = 3tan A -tan3 A1 - 3tan2 A]

= 33 -(3)31 - 3(3)2
[tan 60° = 3]

= 0
[Simplify.]

5.
Find the value of tan 135° using triple-angle formula.
 a. 1 b. $\frac{1}{\sqrt{3}}$ c. - 1

#### Solution:

Tan 135° = Tan 3(45)°

= 3tan 45 -tan3 451 - 3tan2 45
[tan 3A = 3tan A -tan3 A1 - 3tan2 A]

= 3 - 11 - 3
[tan 45° = 1]

= - 1
[Simplify.]

6.
Find the value of sin 180° using triple-angle formula.
 a. $\frac{1}{\sqrt{2}}$ b. - 1 c. 1

#### Solution:

Sin 180° = Sin 3(60)°

= 3sin 60 - 4sin360
[sin 3A = 3sin A - 4sin3 A]

= 3(32) - 4(32)3
[sin 60° = 32]

= 332 - 332
[Simplify.]

= 0

7.
Find the value of sin 270° using triple-angle formula.
 a. $\frac{1}{\sqrt{2}}$ b. 1 c. - 1

#### Solution:

Sin 270° = Sin 3(90)°

= 3sin 90 - 4sin390
[sin 3A = 3sin A - 4sin3 A]

= 3(1) - 4(1)
[sin 90° = 1]

= 3 - 4

= - 1
[Simplify.]

8.
Find the value of tan 360° using triple-angle formula.
 a. - 1 b. $\sqrt{3}$ c. 1

#### Solution:

Tan 360° = Tan 3(120)°

3tan 120 -tan3 1201 - 3tan2 120
[tan 3A = 3tan A -tan3 A1 - 3tan2 A]

= 3(-3) -(-3)31 - 3(-3)2
[tan 120° = - 3]

= 0
[Simplify.]

9.
Find the value of sin 135° using triple-angle formula.
 a. 1 b. - 1 c. $\frac{1}{\sqrt{2}}$

#### Solution:

Sin 135° = Sin 3(45)°

= 3sin 45 - 4sin345
[sin 3A = 3sin A - 4sin3 A]

= 3(12) - 4(12)3
[sin 45° = 12]

= 32 - 22

= 12
[Simplify.]

10.
Find the value of cos 180° using triple-angle formula.
 a. $\frac{1}{\sqrt{2}}$ b. - 1 c. 1

#### Solution:

Cos 180° = Cos 3(60)°

= 4cos360 - 3cos 60
[cos 3A = 4cos3 A - 3cos A]

= 4(12)3 - 3(12)
[cos 60° = 12]

= 12 - 32

= - 1
[Simplify.]