PRACTICE PROBLEMS ON DOUBLE ANGLE FORMULA TRIGONOMETRY

Use a double angle formula to rewrite the expression.

Problem 1 :

6 sin x cos x

Solution:

sin(2x) = 2 sin x cos x

6 sin x cos x = 3(2 sin x cos x)

= 3 sin 2x

Problem 2 :

5 - 10sin²x = 

Solution:

cos(2θ) = 1 − 2sin²θ

5 - 10sin²x = 5(1 - 2sin²x)

= 5 cos(2x)

Problem 3 :

cos²(5x) - sin²(5x) = 

Solution:

cos 2x = cos²x - sin²x

cos²(5x) - sin²(5x) = cos 2 (5x)

= cos 10x

Problem 4 :

14sin(3x) cos(3x) = 

Solution:

sin 2x = 2sinx cos x

14sin(3x) cos(3x) = 7(2sin(3x) cos(3x))

= 7(sin 6x)

Problem 5 :

If sin A = 1/3, what is the value of cos 2A?

a) -2/3     b) 2/3     c) -7/9        d) 7/9

Solution:

sin A = 1/3

cos 2A = 1 - 2sin²A

So, option (d) is correct.

Problem 6 :

The expression 2sin²A + cos 2A is equivalent to

a) 1    b) 2      c) sin²A     d) -sin²A

Solution:

So, option (a) is correct. 

Problem 7 :

If x is an acute angle, and cos x = 4/5, then cos 2x is equal to

a) 6/25       b) -1/25       c) 2/25       d) 7/25

Solution:

So, option (d) is correct.

Problem 8 :

a. csc 𝜃       b. sec 𝜃      c. cot 𝜃         d. sin 𝜃

Solution:

So, option (a) is correct.

Problem 9 :

The expression 1 - 2sin²30° has the same value as

a. sin 60°        b. cos 60°        c. cos 15°       d. sin 15°

Solution:

sin²x = 1 − cos²x

= 1 - 2sin²30°

= 1 - 2(1/2)²

= 1 - 2(1/4)

= 1/2

From the given options, cos 60° = 1/2

1 - 2sin²30° = cos 60°

Problem 10 :

The expression cos²θ - cos 2θ is equivalent to

a. sin²θ      b. -sin²θ          c. cos²θ + 1         d. -cos²θ - 1

Solution:

= cos²θ - cos 2θ

= cos²θ - 2cos²θ - 1

= -cos²θ - 1

So, option (d) is correct.

Problem 11 :

If sin A = 2/3 where 0° < A < 90°, what is the value of sin 2A?

Solution:

sin A = 2/3

opposite = 2

hypotenuse = 3

Adjacent = √5

So, option (c) is correct.

Problem 12 :

Given:

Solution:

a. sin(2B) 

sin(2B) = 2 sinB cosB

b. cos(2A)

cos(2A) = cos²A - sin²A

c. tan(2B)