Question

In a right angled triangle $ABC$, $\angle B={90}^o$. If $\sin A=\frac{\sqrt{3}}{2}$ then verifying the following results:
(i) $\sin (A+C)=\sin A\cos C+\cos A.\sin C$
(ii) $\cos (A+C)=\cos A\cos C-\sin A\sin C$

Answer 1

From the identity, $si{n}^{2}A+co{s}^{2}A=1$, and given $sinA=\frac{\sqrt{3}}{2}$, we obtain $cosA=\frac{1}{2}$, given $\angle B={90}^o,weobtain\angle A=si{n}^{-1}\frac{\sqrt{3}}{2}={60}^o$

Hence $\angle C={30}^o$.

By substituting the obtained values in (i), $sin\left({90}^o\right)=\frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}+\frac{1}{2}.\frac{1}{2}$

$1=1$

For verifying (ii), substitute values in (ii), $cos\left({90}^o\right)=\frac{1}{2}.\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}.\frac{1}{2}$

$0=0$