Respuesta :

happy2438

Step-by-step explanation:

[tex] \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[/tex]

Trancescient

Step-by-step explanation:

Prove that

[tex]\cos^2(45+A)+\cos^2(45-A) =1[/tex]

We know that

[tex]\cos (\alpha \pm \beta) = \cos \alpha\cos \beta \mp \sin \alpha \sin\ beta)[/tex]

We can then write

[tex]\cos (45+A)=\cos 45\cos A - \sin 45\sin A[/tex]

[tex]\:\:\:\:\:\:\:\:= \frac{\sqrt{2}}{2}(\cos A - \sin A)[/tex]

Taking the square of the above expression, we get

[tex]\cos^2(45+A) = \frac{1}{2}(\cos^2A - 2\sin A \cos A + \sin^2A)[/tex]

[tex]= \frac{1}{2}(1 - 2\sin A\cos A)\:\:\;\:\:\:\:(1)[/tex]

Similarly, we can write

[tex]\cos^2(45-A) =\frac{1}{2}(1 + 2\sin A\cos A)\:\:\;\:\:\:\:(2)[/tex]

Combining (1) and (2), we get

[tex]\cos^2(45+A)+\cos^2(45-A)[/tex]

[tex]= \frac{1}{2}(1 - 2\sin A\cos A) + \frac{1}{2}(1 + 2\sin A\cos A)[/tex]

[tex]= 1[/tex]

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