Question

Please solve this question!!​

Answer 1

Let's start from LHS...

LHS = sin^4A = sin^2A * sin^2A

We can apply a few formulas here;

cos 2A = 1 - 2sin^2A

sin^2A = (1 - cos2A) / 2

cos2A = 2cos^2A - 1

cos^2A = (1 + cos2A) / 2

According to the second formula, sin^2A * sin^2A = (1 - cos2A) / 2 * (1 - cos2A) / 2.

(1 - cos2A) / 2 * (1 - cos2A) / 2 = 1/4(1 - cos2A)^2

We can simplify "(1 - cos2A)^2." Remember that (a - b)^2 = a^2 - 2ab + b^2. Similarly (1 - cos2A)^2 = (1 - 2cos 2A + cos^2 2A);

1/4(1 - cos2A)^2 = 1/4(1 - 2cos 2A + cos^2 2A)

According to the fourth formula, cos^2 2A = (1 + cos4A) / 2;

1/4(1 - 2cos 2A + cos^2 2A) = 1/4(1 - 2cos 2A + (1 + cos4A) / 2)

= (taking LCM) 1/4( (2 - 4cos 2A + 1 + cos4A)/2 )

= (simplified) 1/8(3 - 4cos 2A + cos 4A)

Answer 2

Answer: see proof below

Explanation:

Use the following Half-Angle Identities:

sin² A = (1 - cos 2A)/2

cos² A = (1 + cos 2A)/2

Proof LHS → RHS:

LHS: sin⁴ A

Expand: sin² A · sin² A

`\text{Half-Angle:}\qquad \qquad \bigg((1-\cos (2A))/(2)\bigg)\bigg((1-\cos (2A))/(2)\bigg)`

`\text{Distribute:}\qquad \qquad (1-2\cos (2A)+\cos^2 (2A))/(4)`

`\text{Half-Angle:}\qquad \qquad (1-2\cos (2A)+(1+\cos (2\cdot 2A))/(2))/(4)`

`\text{Simplify:}\qquad \qquad ((2)/(2)[1-2\cos (2A)]+(1+\cos (4A))/(2))/(4)`

`=(2-4\cos (2A) + 1 + \cos (4A))/(8)`

`=(1)/(8)\bigg(3-4\cos (2A)+\cos (4A)\bigg)`

LHS = RHS
`\checkmark`