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

Answer 1

A = B = 45^o
Now, A + B = 45^o + 45^o​ = 90^o
Also, A − B = 45^o − 45^o = 0^o

(i) sin (A + B) = sin 90^o = 1
sin A cos B + cos A sin B = sin 45^o cos 45^o + cos 45^o sin 45^o
= $\left(\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\right)=\left(\frac{1}{2}+\frac{1}{2}\right)=1$
∴ sin (A + B) = sin A cos B + cos A sin B

(ii) sin (A − B) = sin 0^o = 0
sin A cos B − cos A sin B = sin 45^o cos 45^o − cos 45^o sin 45^o
= $\left(\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\right)=\left(\frac{1}{2}-\frac{1}{2}\right)=0$
∴ sin (A − B) = sin A cos B − cos A sin B

(iii) cos (A + B) = cos 90^o = 0
cos A cos B − sin A sin B = cos 45^o cos 45^o − sin 45^o sin 45^o
= $\left(\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\right)=\left(\frac{1}{2}-\frac{1}{2}\right)=0$
∴​ cos (A + B) = cos A cos B − sin A sin B

(iv) cos (A − B) = cos 0^o = 1
cos A cos B + sin A sin B = cos 45^o cos 45^o + sin 45^o sin 45^o
= $\left(\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\right)=\left(\frac{1}{2}+\frac{1}{2}\right)=1$
∴​ cos (A − B) = cos A cos B + sin A sin B