Question

Prove that:cos^2(45+A)+cos (45-A)=1​

Answer 1

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`

`\:\:\:\:\:\:\:\:= (√(2))/(2)(\cos A - \sin A)`

Taking the square of the above expression, we get

`\cos^2(45+A) = (1)/(2)(\cos^2A - 2\sin A \cos A + \sin^2A)`

`= (1)/(2)(1 - 2\sin A\cos A)\:\:\;\:\:\:\:(1)`

Similarly, we can write

`\cos^2(45-A) =(1)/(2)(1 + 2\sin A\cos A)\:\:\;\:\:\:\:(2)`

Combining (1) and (2), we get

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

`= (1)/(2)(1 - 2\sin A\cos A) + (1)/(2)(1 + 2\sin A\cos A)`

`= 1`

Answer 2

Explanation:

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