Question

Prove that:cos^2(45+A)+cos (45-A)=1​

Answer 1

Step-by-step explanation:

$\boxed{cos^2x=\frac{1-cos2x}{2}}\\cos^2(45+A)+cos^2(45-A)=\frac{1-cos2(45+A)}{2}+\frac{1-cos2(45-A}{2}\\=\frac{1 - cos(90 +2A) }{2} + \frac{1 - cos(90 - 2A) }{2} \\ = \frac{2- ( - sin 2A) - sin2A}{2} \\ = \frac{2 + sin2A -sin2A }{2} \\ = \frac{2}{2} \\ = 1$

Answer 2

Step-by-step explanation:

Prove that

$\cos^2(45+A)+\cos^2(45-A) =1$

We know that

$\cos (\alpha \pm \beta) = \cos \alpha\cos \beta \mp \sin \alpha \sin\ beta)$

We can then write

$\cos (45+A)=\cos 45\cos A - \sin 45\sin A$

$\:\:\:\:\:\:\:\:= \frac{\sqrt{2}}{2}(\cos A - \sin A)$

Taking the square of the above expression, we get

$\cos^2(45+A) = \frac{1}{2}(\cos^2A - 2\sin A \cos A + \sin^2A)$

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

Similarly, we can write

$\cos^2(45-A) =\frac{1}{2}(1 + 2\sin A\cos A)\:\:\;\:\:\:\:(2)$

Combining (1) and (2), we get

$\cos^2(45+A)+\cos^2(45-A)$

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

$= 1$