DOUBLE ANGLE FORMULAS FOR SINE COSINE AND TANGENT

Double angle formula :

sin 2θ = 2 sin θ  cos θ

cos 2θ = cos2θ - sin2θ

cos 2θ = 2cos2θ - 1

cos 2θ = 1 - 2 sin2θ

tan 2θ = 2tan θ / (1 - tan2 θ)

sin 2θ = 2tan θ / (1 + tan2 θ)

cos 2θ = (1 - tan2 θ) / (1 + tan2 θ)

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

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.

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