Question

Prove that :
i) sin(45°+A) + cos (45°+A) = √2 cosA
ii) sin^2 (45°+A) + sin^2 45°-A)= 1

Answer 1

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Step-by-step explanation:

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proved

Answer 2

Step-by-step explanation:

(i) To prove sin(45°+A) + cos (45°+A) = √2 cosA

LHS = sin(45°+A) + cos (45°+A)

       = sin45°.cosA + cos45°.sinA + cos45°.cosA - sin45°.sinA

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

       = $\frac{1}{\sqrt{2} }$.cosA + $\frac{1}{\sqrt{2} }$.cosA

       = $\frac{1+1}{\sqrt{2} }$.cosA

       = $\frac{2}{\sqrt{2} }$.cosA

       = $\sqrt{2}$.cosA

       = RHS

=> LHS = RHS

hence proved

(ii) To prove sin²(45°+A) + sin²(45°-A)  = 1

take sin²(45°-A) = [ sin(45°-A) ]²

                          = [ sin45°.cosA - cos45°sinA ]²

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

                          = [cos45°.cosA - sin45°.sinA]²

                          = [cos(45°+A)]²

                          = cos²(45°+A)    ---------------------- (A)

LHS = sin²(45°+A) + sin²(45°-A)

       = sin²(45°+A) + cos²(45°+A)                    by (A)

       = 1      

       = RHS

=>  LHS = RHS

hence proved