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Question

# If A = B = 45°, verify that: (i) sin (A + B) = sin A cos B + cos A sin B (ii) sin (A − B) = sin A cos B − cos A sin B (iii) cos (A + B) = cos A cos B − sin A sin B (iv) cos (A − B) = cos A cos B + sin A sin B

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Solution

## A = B = 45o Now, A + B = 45o + 45o​ = 90o Also, A − B = 45o − 45o = 0o (i) sin (A + B) = sin 90o = 1 sin A cos B + cos A sin B = sin 45o cos 45o + cos 45o sin 45o = $\left(\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}\right)=\left(\frac{1}{2}+\frac{1}{2}\right)=1$ ∴ sin (A + B) = sin A cos B + cos A sin B (ii) sin (A − B) = sin 0o = 0 sin A cos B − cos A sin B = sin 45o cos 45o − cos 45o sin 45o = $\left(\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}\right)=\left(\frac{1}{2}-\frac{1}{2}\right)=0$ ∴ sin (A − B) = sin A cos B − cos A sin B (iii) cos (A + B) = cos 90o = 0 cos A cos B − sin A sin B = cos 45o cos 45o − sin 45o sin 45o = $\left(\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}\right)=\left(\frac{1}{2}-\frac{1}{2}\right)=0$ ∴​ cos (A + B) = cos A cos B − sin A sin B (iv) cos (A − B) = cos 0o = 1 cos A cos B + sin A sin B = cos 45o cos 45o + sin 45o sin 45o = $\left(\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}\right)=\left(\frac{1}{2}+\frac{1}{2}\right)=1$ ∴​ cos (A − B) = cos A cos B + sin A sin B

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