Respuesta :
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
= [ [tex]\frac{1}{\sqrt{2} }[/tex] cosA - [tex]\frac{1}{\sqrt{2} }[/tex] sinA ]² + [ [tex]\frac{1}{\sqrt{2} }[/tex] cosA + [tex]\frac{1}{\sqrt{2} }[/tex] sinA ]²
= [tex]\frac{1}{2}[/tex]cos²A - sinAcosA + [tex]\frac{1}{2}[/tex] sin²A + [tex]\frac{1}{2}[/tex] cos²A + sinAcosA + [tex]\frac{1}{2}[/tex] sin²A
= cos²A + sin²A
= 1
= right side , then proven
Answer:
Step-by-step explanation:
cos 2x=cos²x-sin²x=cos²x-(1-cos²x)=cos²x-1+cos²x=2cos²x-1
2cos²x=1+cos2x
[tex]cos^2x=\frac{1}{2}(1+cos2x)[/tex]
cos²(45+A)+cos²(45-A)
[tex]=\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[/tex]
cos (90-x)=sin x
cos (90+x)=-sin x
