Question

The value of $4(\cos ²60°+\cos ²45°)-\sin ²30°$ is

Answer 1

Put the value of $\cos 60°,\cos 45°,\sin 30°$ in $4(\cos ²60°+\cos ²45°)-\sin ²30°$ and simplify to get the result

We know that $\cos 60°=\frac{1}{2},\cos 45°=\frac{1}{\sqrt{2}},\sin 30°=\frac{1}{2}$

$\begin{array}{rcl}4(\cos ²60°+\cos ²45°)-\sin ²30°& =& 4\left[{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2\right]-{\left(\frac{1}{2}\right)}^2\\ & =& 4\left[\frac{1}{4}+\frac{1}{2}\right]-\frac{1}{4}\\ & =& 4\left[\frac{1+2}{4}\right]-\frac{1}{4}\\ & =& 4\left[\frac{3}{4}\right]-\frac{1}{4}\\ & =& \frac{12}{4}-\frac{1}{4}\\ & =& \frac{12-1}{4}\\ & =& \frac{11}{4}\\ \therefore 4(\cos ²60°+\cos ²45°)-\sin ²30°& =& \frac{11}{4}\end{array}$