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Prove that: Sin4θ - cos4θ = 1 - 2cos2θ

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प्रश्न

Prove that:

Sin4θ - cos4θ = 1 - 2cos2θ

योग
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उत्तर

Sin4θ – cos4θ = 1 – 2cos2θ

LHS = Sin4θ – cos4θ

LHS = (Sin2θ)2 – (cos2θ)2

LHS = (Sin2θ + cos2θ)(Sin2θ - cos2θ)       ...[a2 – b2 = (a + b)(a – b)]

LHS = (Sin2θ – cos2θ).(1)         ...(Sin2θ + cos2θ = 1)

LHS = 1 – cos2θ – cos2θ          ...(1 – Sin2θ = cos2θ)

LHS = 1 – 2cos2θ

RHS = 1 – 2cos2θ 

LHS = RHS

Hence proved.

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अध्याय 6: Trigonometry - Practice Set 6.1 [पृष्ठ १३१]

APPEARS IN

बालभारती Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
अध्याय 6 Trigonometry
Practice Set 6.1 | Q 6.07 | पृष्ठ १३१

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