Question

Prove that Cos (∏/4 + x) + Cos(∏/4 – x) = 2 Cos x.

Answer 1

prove that,

cos(π/4 + x) + cos(π/4 - x) = √2cosx

LHS = cos(π/4 + x) + cos(π/4 - x)

we know,

• cos(A + B) = cosA.cosB - sinA.sinB
• cos(A + B) = cosA.cosB + sinA.sinB

LHS = cos(π/4).cosx - sin(π/4).sinx + cos(π/4).cosx + sin(π/4).sinx

= 2cos(π/4).cosx

= 2 × 1/√2 cosx

= √2cosx = RHS

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Answer 2

$\huge \fcolorbox{red}{pink}{Solution :)}$

We know that ,

$\large \mathtt{ \fbox{Cos(x) + Cos(y) = 2Cos (\frac{x + y}{2}) \times Cos (\frac{x - y}{2}) }}$

Thus ,

$\sf \hookrightarrow Cos (\frac{\pi}{4} + x) + Cos (\frac{4}{\pi} - x) \\ \\ \sf \hookrightarrow 2Cos \bigg(\frac{ (\frac{\pi}{4} + x) + (\frac{\pi}{4} - x)}{2}\bigg) \times Cos \bigg(\frac{ (\frac{\pi}{4} + x) - (\frac{\pi}{4} - x)}{2} \bigg) \\ \\ \sf \hookrightarrow 2Cos( \frac{\pi}{4} ) \times Cos(x) \\ \\ \sf \hookrightarrow 2 \times \frac{1}{ \sqrt{2} } \times Cos(x) \\ \\ \sf \hookrightarrow \sqrt{2} \times Cos(x)$

Hence proved

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