Question

Find the value of (i) sin ${75}^0$ (ii) tan ${15}^0$.

Answer 1

i) sin ${75}^0$ = sin ${45}^0$ cos ${30}^0$ + cos ${45}^0$ sin ${30}^0$

[$\because$ sin(A+B) = sin A cos B + sin B cos A]

= $\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}$

= $\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}$

(ii) tan ${15}^0$ = tan (${45}^0-{30}^0$)

= $\frac{\text{tan}{45}^0-\text{tan}{30}^0}{1+\text{tan}{45}^0\text{tan}{30}^0}$

$[\because \text{tan}(A-B)=\frac{\text{tan A- tan B}}{1+\text{tan A tan B}}]$

= $\frac{1-\frac{1}{\sqrt{3}}}{1+1\times \frac{1}{\sqrt{3}}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}$

= $\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$

= $\frac{3+1-2\sqrt{3}}{3-1}=\frac{4-2\sqrt{3}}{2}$

= $2-\sqrt{3}$