Question

Prove that: cos^2 (45+A)+cos^2 (45-A)=1​

Answer 1

Answer:

see explanation

Step-by-step explanation:

Using the cosine addition formula

cos(A ± B ) = cosAcosB ∓ sinAsinB

Then considering the left side

cos²(45 + A) + cos²(45 - A)

= [ cos45cosA - sin45sinA ]² + [cos45cosA + sin45sinA]]²

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

= $\frac{1}{2}$cos²A - sinAcosA + $\frac{1}{2}$ sin²A + $\frac{1}{2}$ cos²A + sinAcosA + $\frac{1}{2}$ sin²A

= cos²A + sin²A

= 1

= right side , then proven

Answer 2

Answer:

Step-by-step explanation:

cos 2x=cos²x-sin²x=cos²x-(1-cos²x)=cos²x-1+cos²x=2cos²x-1

2cos²x=1+cos2x

$cos^2x=\frac{1}{2}(1+cos2x)$

cos²(45+A)+cos²(45-A)

$=\frac{1}{2}(1+cos(90+2A))+\frac{1}{2}(1+cos(90-2A))\\=\frac{1}{2} (1-sin2A)+\frac{1}{2} (1+sin 2A)\\=\frac{1}{2} (1-sin2A+1+sin 2A)\\=\frac{1}{2} \times2\\=1$

cos (90-x)=sin x

cos (90+x)=-sin x