Question

Prove that ,Cos(60°+A)+cos(60°-A)-cosA=0

Answer 1

$\underline{\textsf{To prove:}}$

$\mathsf{cos(60^{\circ}+A)+cos(60^{\circ}-A)-cosA=0}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{cos(60^{\circ}+A)+cos(60^{\circ}-A)-cosA}$

$\textsf{Using the identities}$

$\boxed{\mathsf{cos(A+B)=cosA\,cosB-sinA\,sinB}}$

$\boxed{\mathsf{cos(A-B)=cosA\,cosB+sinA\,sinB}}$

$\mathsf{=cos60^{\circ}\,cosA-sin60^{\circ}\,sinA+cos60^{\circ}\,cosA+sin60^{\circ}\,sinA-cosA}$

$\mathsf{=2\,cos60^{\circ}\,cosA-cosA}$

$\mathsf{=2(\dfrac{1}{2})\,cosA-cosA}$

$\mathsf{=cosA-cosA}$

$\mathsf{=0}$

$\implies\boxed{\mathsf{cos(60^{\circ}+A)+cos(60^{\circ}-A)-cosA=0}}$

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