Question

Prove it

cos(A+45°) cos(A-45°)= 1/2 (cos²A-sin²A)

Answer 1

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$\bold{\underline{Question-}}$

Prove it

cos(A+45°) cos(A-45°)= 1/2 (cos²A-sin²A)

$\bold{\underline{Answer-}}$

LHS = cos(A+45°)cos(A-45°)

= (cosA cos45°- sinA sin45°) (cosA cos45° + sinA sin 45°)

$= (cosA \times \frac{1}{ \sqrt{2} } - sinA \: \times \frac{1}{ \sqrt{2} } )(cosA \: \times \frac{1}{ \sqrt{2} } + sinA \times \frac{1}{ \sqrt{2} } )$

$= \frac{1}{ \sqrt{2} } (cosA - sinA). \frac{1}{ \sqrt{2} } (cosA + sinA)$

°•° (a+b)(a-b) = a² - b²

$= \frac{1}{2} ( {cos}^{2}A - sin {}^{2} A)$

Answer 2

the solution is here
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