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 :

1-2sin25𝜋8

Solution:

=1-2sin25𝜋8cos 2𝜃=1-2sin2𝜃=cos2×5𝜋8=cos5𝜋4=cos𝜋+𝜋4=-cos𝜋4=-22=-222=-12

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)

8 sinx2 cosx2

Solution:

=8 sinx2cosx2=4×2sinx2cosx2=4×sin2×x2=4 sin x

Problem 3 :

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

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

Solution:

sin3𝜃=(sin 𝜃+cos 𝜃) sin2𝜃-sin𝜃 cos𝜃+cos2𝜃=(sin𝜃+cos𝜃) 1-sin 2𝜃2(sin𝜃+cos𝜃)2=sin2𝜃+cos2𝜃+2sin𝜃 cos𝜃=1+sin 2𝜃sin𝜃+cos𝜃=1+sin 2𝜃=1+sin 2𝜃1-sin 2𝜃2=1+341-342=72×58=5716

Problem 4 :

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

Solution:

sin2θ + cos2θ = 1

sin 𝜃=1-cos2𝜃=1-132=1-19=89sin 𝜃=223sin 2𝜃=2 sin𝜃 cos𝜃=2×13×223sin 2𝜃=429

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

=1-452=1-1625=925sin 𝜃=±35

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

sin 𝜃=-35sin 2𝜃=2 sin𝜃 cos𝜃=2-3545=-2425

Problem 6 :

cot 𝜃=43 and 𝜋<𝜃<3𝜋2. Find sin 2𝜃

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.

sin 𝜃=-35cos 𝜃=-45sin 2θ=2 sinθ cosθ=2-35-45=2425

Problem 7 :

cot 𝜃=43 and 𝜋<𝜃<3𝜋2. Find cot 2𝜃

Solution:

cot 2𝜃=cot2𝜃-12 cot𝜃=432-12×43=169-183=16-9983=79×38=724

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

c) 1+tan2𝜃1-tan2𝜃d) 1-tan2𝜃1+tan2𝜃

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

cos 2𝜃=cos2𝜃1-sin2𝜃cos2𝜃cos 2𝜃=1-tan2𝜃sec2𝜃=1-tan2𝜃1+tan2𝜃

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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