Question #b2df1

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Question #b2df1

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Answer 1 · 2017-10-18T16:29:30

Proved.

Explanation:

We have to prove, ${(sin A + cos A)}^{2} {(sin A - cos A)}^{2} = 1 - {sin}^{2} (2 A)$ $(sin A + cos A)^2 (sin A - cos A)^2 = 1 - sin^2 (2A)$

R.H.S.

$\Rightarrow {(sin A + cos A)}^{2} {(sin A - cos A)}^{2}$ $rArr (sin A + cos A)^2 (sin A - cos A)^2$

$\Rightarrow ({sin}^{2} A + 2 sin A cos A + {cos}^{2} A) ({sin}^{2} A - 2 sin A cos A + {cos}^{2} A)$ $rArr ( sin^2 A + 2 sin A cos A + cos^2 A)(sin^2 A - 2 sin A cos A + cos^2 A)$

$\Rightarrow (1 + 2 sin A cos A) (1 - 2 sin A cos A)$ $rArr ( 1 + 2 sin A cos A)(1 - 2 sin A cos A)$ [As, ${sin}^{2} A + {cos}^{2} A = 1$ $sin^2 A+cos^2 A=1$ ]

$\Rightarrow (1 + sin 2 A) (1 - sin 2 a)$ $rArr (1 + sin 2A)(1 - sin 2a)$ [As, 2 sin A cos A = sin 2A]

$\Rightarrow 1 - {sin}^{2} 2 A$ $rArr 1 - sin^2 2A$ [ As formula, $(a + b) (a - b) = a^{2} - b^{2}$ $(a + b )( a - b) = a^2 - b^2$ ]

$\Rightarrow R . H . S .$ $rArr R.H.S.$

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Question #b2df1

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