Math, asked by sabir4, 1 year ago

prove that sinA+CosA=√2cos(45°-A)

Answers

Answered by ajayaj
30
multiplying and divide LHS √2
√2 (1/√2 sin A + 1/√2 cos A)
and sin 45 = cos 45 = 1/√2
=> √2 ( sin 45 sin A + cos 45 cos A)
=> √2 cos(45-A )

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Answered by mysticd
23

Solution:

RHS = 2cos(45°-A)

= 2[cos45°cosA+sin45°sinA]

/* Since ,

\boxed {Cos(X-Y) \\= CosXCosY + SinXSinY}

*/

=\sqrt{2}[\frac{1}{\sqrt{2}}cosA+\frac{1}{\sqrt{2}}sinA]

/* cos45° = sin45° =\frac{1}{\sqrt{2}} */

=\frac{\sqrt{2}}{\sqrt{2}}(cosA+sinA)

= sinA+cosA

=LHS

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