DOUBLE ANGLE FORMULAS FOR SINE COSINE AND TANGENT

Evaluate :

Problem 1 :

Solution:

Problem 2 :

Express as a single sine or cosine function.

a) 40sin x cos x

Solution:

= 40sin x cos x

= 20 × 2sin x cosx

[∴ sin2x = 2sin x cosx]

= 20 sin2x

b) cos²3x - sin²3x

Solution:

= cos²3x - sin²3x

  [∴ cos²x- sin²x= cos 2x] 

= cos(2 × 3x)

= cos 6x

c)

Solution:

Problem 3 :

If sin 2𝜃 = 3/4, then sin³𝜃 + cos³𝜃 = 

a) √5/8        b) √7/8          c) √11/8          d) 5√7/16

Solution:

Problem 4 :

cos 𝜃 = 1/3 and 0° < 𝜃 < 90°. Find sin 2𝜃.

Solution:

sin²θ + cos²θ = 1

Problem 5 :

cos 𝜃 =4/5 and 270° < 𝜃 < 360°. Find sin 2𝜃.

Solution:

cos 𝜃 =4/5

sin 2𝜃 = 2 sin𝜃 cos𝜃

sin²𝜃 = 1 - cos²𝜃

Since 𝜃 is in IV quadrant, sin 𝜃 is negative.

Problem 6 :

Solution:

cot 𝜃 = 4/3

t is an angle of a right triangle that has 3 sides.

Opposite = 3

Adjacent = 4

hypotenuse = 5

Since 𝜃 lies in the third quadrant, for the trigonometric ratios tangent and cotangent only we will be having positive, for the other trigonometric ratios will be having negative sign.

Problem 7 :

Solution:

Verifying Identities Double Angle Formulas

Problem 8 :

Verify that cos⁴x - sin⁴x = cos 2x

Solution:

cos⁴x - sin⁴x = cos 2x

Here we have LHS = cos⁴x - sin⁴x

cos⁴x - sin⁴x = (cos²x)² - (sin²x)²

= (cos²x + sin²x) (cos²x - sin²x)

= 1  (cos²x - sin²x)

= cos 2x

Hence, cos⁴x - sin⁴x = cos 2x is proved.

Problem 9 : 

Verify that sin 2x = -2 sin x  sin (x - 90)

Solution:

sin 2x = 2 sin x cos x -----(1)

We know that cos x = sin (90 - x)

cos x = sin (-(x - 90))

cos x = -sin (x - 90)

Applying the value of cos x above in (1), we get

sin 2x = 2 sin x(-sin (x - 90))

= -2 sin x sin (x - 90)

Hence it is proved.

Problem 10 :

cos 2𝜃 is not equal to 

a) 2 cos²𝜃 - 1     b) 1 - 2 sin²𝜃

Solution:

cos 2𝜃 = cos²𝜃 - sin²𝜃

sin²𝜃 + cos²𝜃 = 1

sin²𝜃 = 1 - cos²𝜃

cos²𝜃 = 1 - sin²𝜃

cos 2𝜃 = 1 - sin²𝜃 - sin²𝜃

= 1 - 2sin²𝜃

cos 2𝜃 = cos²𝜃 - (1 - cos²𝜃)

= cos²𝜃 - 1 + cos²𝜃

= 2 cos²𝜃 - 1

cos 2𝜃 = cos²𝜃 - sin²𝜃

So, option (C) is correct.

Problem 11 :

If sin A + cos A = 1, then sin 2A is equal to

a) 1     b) 2          c) 0        d) 1/2

Solution:

sin A + cos A = 1

Squaring on both sides.

(sin A + cos A)² = 1

sin²A + cos²A + 2sinA cosA = 1

1 + sin 2A = 1

sin 2A = 0

So, option (C) is correct.