1.

Find the value of 1 – 2 sin2θ + sin4θ.1). sin4θ2). cos4θ3). cosec4θ4). sec4θ

Answer»

We KNOW that,

sin2x + COS2x = 1

⇒ sin2θ = 1 – cos2θ

Squaring on both sides

(? (a + B)2 = a2 + 2ab + b2)

⇒ sin4θ = 1 + cos4θ – 2cos2θ

Now,

1 – 2 sin2θ + sin4θ

(? cos 2x = 2cos2x – 1 = 1 – 2sin2x)

= 2cos2θ – 1 + 1 + cos4θ – 2cos2θ

= cos4θ

∴ 1 – 2 sin2θ + sin4θ = cos4θ


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