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Prove the Following Identity : Sin 8 θ − Cos 8 θ = ( Sin 2 θ − Cos 2 θ ) ( 1 − 2 Sin 2 θ Cos 2 θ ) - Mathematics

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प्रश्न

Prove the following identity : 

`sin^8θ - cos^8θ = (sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`

योग

उत्तर

LHS = `sin^8θ - cos^8θ`

 = `(sin^4θ)^2 - (cos^4θ)^2`

= `(sin^4θ - cos^4θ)(sin^4θ + cos^4θ)`

= `(sin^2θ - cos^2θ)(sin^2θ + cos^2θ)(sin^4θ + cos^4θ)`

= `(sin^2θ - cos^2θ)(sin^4θ + cos^4θ)`

= `(sin^2θ - cos^2θ)((sin^2θ)^2 + (cos^2θ)^2 + 2sin^2θcos^2θ - 2sin^2θcos^2θ)`

= `(sin^2θ - cos^2θ)((sin^2θ + cos^2θ)^2 - 2sin^2θcos^2θ)`

= `(sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`

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अध्याय 21: Trigonometric Identities - Exercise 21.1

APPEARS IN

फ्रैंक Mathematics - Part 2 [English] Class 10 ICSE
अध्याय 21 Trigonometric Identities
Exercise 21.1 | Q 6.07
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