Question

Find the value of the following:
$4{\sin }^2{60}^o+3{\tan }^2{30}^o-8\sin {45}^o\cdot \cos {45}^o$

Answer 1

We know that the values of the trignometric functions $\sin {60}^0=\frac{\sqrt{3}}{2}$, $\tan {30}^0=\frac{1}{\sqrt{3}}$, $\sin {45}^0=\frac{1}{\sqrt{2}}$ and $\cos {45}^0=\frac{1}{\sqrt{2}}$,

Therefore,

$4{\sin }^2{60}^0+3{\tan }^2{30}^0-8\sin {45}^0\cos {45}^0=4\times {\left(\frac{\sqrt{3}}{2}\right)}^2+3\times {\left(\frac{1}{\sqrt{3}}\right)}^2-(8\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}})$

$=(4\times \frac{3}{4})+(3\times \frac{1}{3})-\frac{8}{2}=3+1-4=0$

Hence, $4{\sin }^2{60}^0+3{\tan }^2{30}^0-8\sin {45}^0\cos {45}^0=0$.