Question

Prove that 8 sin⁴ θ = 3 - 4 cos 2θ + cos 4θ​

Answer 1

Question : -

Prove that 8 sin⁴ x = 3-4 cos 2x + cos 4x

ANSWER

Given : -

8 sin⁴ x = 3-4 cos 2x + cos 4x

Required to prove : -

• LHS = RHS

Proof : -

Before solving this question we need to derive some formulae so, let's derive them quickly !

Let's find the value of cos 2x

Using the formula;

cos (x+y) = cos x cos y - sin x sin y

This implies;

Here,

• x = x
• y = x

cos (x+x)

cos 2x =

cos x cos x - sin x sin x

cos²x - sin² x

we know that;

• sin² x + cos² x = 1
• sin² x + cos² x = 1 => cos² x = 1 - sin² x

(1-sin² x) - sin² x

1 - sin² x - sin² x

1 - 2sin² x

Hence,

• cos 2x = 1 - 2sin² x

Similarly,

Using the formula;

sin (x+y) = sin x cos y + sin y cos x

This implies;

Here,

• x = x
• y = x

sin (x+x)

sin 2x =

sin x cos x + sin x cos x

2 sin x cos x

Hence,

• sin 2x = 2 sin x cos x

Now,

Let's find the value of cos 4x

cos (2x+2x)

cos 4x =

cos 2x cos 2x - sin 2x sin 2x

(1 - 2sin² x)(1 - 2sin² x) - (2 sin x cos x)(2 sin x cos x)

(1 - 2sin² x)² - (2 sin x cos x)²

Using the identity;

• (a-b)² = a² + b² - 2ab

Here,

• a= 1
• b = 2sin² x

(1)²+(2sin² x)²-2(1)(2sin² x) - 4 sin² x cos² x

1 + 4sin⁴ x - 4 sin² x - 4 sin² x cos² x

Hence

• cos 4x = 1+4sin⁴ x - 4sin² x - 4 sin² x cos² x

Now,

Let's come back to solve this question !

Consider the RHS part

3 - 4 cos 2x + cos 4x

Substituting the values of cos 2x & cos 4x

3 - 4 (1-2sin² x) + (1+4sin⁴ x - 4sin² x - 4 sin² x cos² x)

3 - 4 + 8sin² x + 1 + 4sin⁴ x - 4sin² x - 4 sin² x (1 - sin² x)

3 - 4 + 8sin² x + 1 + 4sin⁴ x - 4 sin² x - 4 sin² x + 4 sin⁴ x

4 - 4 + 8 sin² x + 4 sin² x - 4 sin² x - 4 sin² x + 4 sin⁴ x

8 sin² x + 4 sin² x - 4 sin² x - 4 sin² x + 4 sin⁴ x

8 sin² x - 8 sin² x + 8 sin⁴ x

Cancelling 8 sin² x on both sides

8 sin⁴ x

LHS = RHS

Hence Proved !

Answer 2

Answer:

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