Q. 154.3( 15 Votes )

# 2(sin^{6}θ + cos^{6}θ) − 3(sin^{4}θ + cos^{4}θ) is equal to

A. 0

B. 1

C. −1

D. None of these

Answer :

To find: 2(sin^{6} θ + cos^{6} θ) – 3(sin^{4} θ + cos^{4} θ)

First, we consider

sin^{6} θ + cos^{6} θ = (sin^{2} θ)^{3} + (cos^{2} θ)^{3}

Now, as (a + b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}

⇒ a^{3} + b^{3} = (a + b)^{3} – 3a^{2}b – 3ab^{2}

⇒ sin^{6} θ + cos^{6} θ

= (sin^{2} θ)^{3} + (cos^{2} θ)^{3}

= (sin^{2} θ + cos^{2} θ)^{3} – 3 (sin^{2} θ)^{2} cos^{2} θ – 3 sin^{2} θ (cos^{2} θ)^{2}

= 1 – 3 sin^{4} θ cos^{2} θ – 3 sin^{2} θ cos^{4} θ [∵ sin^{2} θ + cos^{2} θ = 1]

= 1 – 3 sin^{2} θ cos^{2} θ (sin^{2} θ + cos^{2} θ)

= 1 – 3 sin^{2} θ cos^{2} θ [∵ sin^{2} θ + cos^{2} θ = 1] ………(i)

Next, we consider

sin^{4} θ + cos^{4} θ = (sin^{2} θ)^{2} + (cos^{2} θ)^{2}

Now, as (a + b)^{2} = a^{2} + b^{2} + 2ab

⇒ a^{2} + b^{2} = (a + b)^{2} – 2ab

⇒sin^{4} θ + cos^{4} θ

= (sin^{2} θ)^{2} + (cos^{2} θ)^{2}

= (sin^{2} θ + cos^{2} θ)^{2} – 2 sin^{2} θ cos^{2} θ

= 1 – 2 sin^{2} θ cos^{2} θ [∵ sin^{2} θ + cos^{2} θ = 1] ………(ii)

Now, using (i) and (ii), we have

2(sin^{6} θ + cos^{6} θ) – 3(sin^{4} θ + cos^{4} θ)

= 2(1 – 3 sin^{2} θ cos^{2} θ) – 3(1 – 2 sin^{2} θ cos^{2} θ)

= 2 – 6 sin^{2} θ cos^{2} θ – 3 + 6 sin^{2} θ cos^{2} θ

= 2 – 3 = – 1

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