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# Prove the following identities: (i) 2 (sin^6 θ + cos^6 θ) –3(sin^4 θ + cos^4 θ) + 1 = 0, (ii) (sin^8 θ – cos^8 θ) = (sin^2 θ – cos^2 θ) (1 – 2sin^2 θ cos^2 θ) - CBSE Class 10 - Mathematics

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#### Question

Prove the following identities:

(i) 2 (sin^6 θ + cos^6 θ) –3(sin^4 θ + cos^4 θ) + 1 = 0

(ii) (sin^8 θ – cos^8 θ) = (sin^2 θ – cos^2 θ) (1 – 2sin^2 θ cos^2 θ)

#### Solution

(i) We have,

LHS = 2 (sin^6 θ + cos^6 θ) –3(sin^4 θ + cos^4 θ) + 1

= 2 [(sin^2 θ)^3 + (cos^2 θ)^3 ] – [3 (sin^2 θ)^2 + (cos^2 θ)^2 ] + 1

= 2[(sin^2 θ + cos^2 θ) {(sin^2 θ)^2 + (cos^2 θ)^2 – sin^2 θ cos^2 θ)]}– 3[(sin^2 θ)^2 + (cos^2 θ)^2 + 2 sin^2 θ cos^2 θ –2 sin2 θ cos2 θ] + 1

= 2[(sin^2 θ)^2 + (cos^2 θ)^2 + 2 sin^2 θ cos^2 θ –3 sin^2 θ cos^2 θ]–3 [(sin^2 θ + cos^2 θ)^2 – 2 sin^2 θ cos^2 θ] + 1

= 2[(sin^2 θ + cos^2 θ)^2 – 3 sin^2 θ cos^2 θ] –3 [1 – 2 sin^2 θ cos^2θ] + 1

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

= 2 – 6 sin^2 θ cos^2 θ –3 + 6 sin^2 θ cos^2 θ + 1

= 0 = RHS

(ii) We have,

LHS = (sin^8 θ – cos^8 θ) = (sin^4 θ)^2 – (cos^4 θ)^2

= (sin^4 θ – cos^4 θ) (sin^4 θ + cos^4 θ)

= (sin^2 θ – cos^2 θ) (sin^2 θ + cos^2 θ) (sin^4 θ + cos^4 θ)

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

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

= (sin^2 θ – cos^2 θ) (1 – 2sin^2 θ cos^2 θ) = RHS

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Solution Prove the following identities: (i) 2 (sin^6 θ + cos^6 θ) –3(sin^4 θ + cos^4 θ) + 1 = 0, (ii) (sin^8 θ – cos^8 θ) = (sin^2 θ – cos^2 θ) (1 – 2sin^2 θ cos^2 θ) Concept: Trigonometric Identities.
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