Question

cos (45° + A) + cos (45º - A) = √2cosA​

Answer 1

Answer:

we know that

Cos(A+B)=cosAcosB-sinAsinB

Cos( A-B)=cosAcosB+sinAsinB)

so

cos(45°+A) = cos45°cosA - sin45°sinA

=$\frac{1}{\sqrt{2}}$cosA - $\frac{1}{\sqrt{2}}$sinA

=$\frac{cosA}{\sqrt{2}}$ - $\frac{sinA}{\sqrt{2}}$

=$\frac{cosA-sinA}{\sqrt{2}}$

cos(45°-A) = cos45°cosA - sin45°sinA

= $\frac{1}{\sqrt{2}}$cosA + $\frac{1}{\sqrt{2}}$sinA

= $\frac{cosA}{\sqrt{2}}$ + $\frac{sinA}{\sqrt{2}}$

= $\frac{cosA+sinA}{\sqrt{2}}$

cos (45° + A) + cos (45º - A) = $\frac{cosA-sinA}{\sqrt{2}}$ + $\frac{cosA+sinA}{\sqrt{2}}$

= $\frac{cosA-sinA+cosA+sinA}{\sqrt{2}}$

= $\frac{2cosA}{\sqrt{2}}$

= rationalisng denominator $\sqrt{2}$

= $\frac{2\times\sqrt{2}cosA}{{2}}$

= $\sqrt{2}cosA$