prove that sinA+CosA=√2cos(45°-A)
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Answered by
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 )
√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 )
sabir4:
thanks bro
Answered by
23
Solution:
RHS = √2cos(45°-A)
= √2[cos45°cosA+sin45°sinA]
/* Since ,
*/
=
/* cos45° = sin45° = */
=
=
=LHS
••••
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