Double Angles and Half Angles Worksheet

Double Angles and Half Angles Worksheet
  • Page 1
 1.  
Find the value of cot x - cos 2xsin x cos 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.


Correct answer : (1)
 2.  
If cos x = 0.425 then cos (x2) =
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.


Correct answer : (3)
 3.  
If cosec θ = p + qp - q, then find the value of cot (1 4π + 1 2θ ).
a.
qp
b.
pq
c.
pq
d.
pq


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.


Correct answer : (1)
 4.  
Find the value of tan 180° using triple-angle formula.
a.
13
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.]


Correct answer : (2)
 5.  
Find the value of tan 135° using triple-angle formula.
a.
1
b.
13
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.]


Correct answer : (4)
 6.  
Find the value of sin 180° using triple-angle formula.
a.
12
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


Correct answer : (3)
 7.  
Find the value of sin 270° using triple-angle formula.
a.
12
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.]


Correct answer : (4)
 8.  
Find the value of tan 360° using triple-angle formula.
a.
- 1
b.
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.]


Correct answer : (3)
 9.  
Find the value of sin 135° using triple-angle formula.
a.
1
b.
- 1
c.
12


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.]


Correct answer : (4)
 10.  
Find the value of cos 180° using triple-angle formula.
a.
12
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.]


Correct answer : (4)

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