Question

If $y=\frac{{\sin }^2{1500}^{\circ }+4{\sin }^4{750}^{\circ }-4{\sin }^2{750}^{\circ }{\cos }^2{750}^{\circ }}{4-{\sin }^2{1500}^{\circ }-4{\sin }^2{750}^{\circ }}$. Then the value of $9y$ is ___

Answer 1

We see the angles in the trigonometric functions as ${750}^{\circ }\&{1500}^{\circ }$. So let's assume $\theta ={750}^{\circ }$, which would make $2\theta ={1500}^{\circ }$ and see if we can simplify the expression first before jumping on to calculating it's value.
$\therefore y=\frac{{\sin }^{2}2\theta +4{\sin }^4\theta -4{\sin }^2\theta {\cos }^2\theta }{4-{\sin }^{2}2\theta -4{\sin }^2\theta }\Rightarrow y=\frac{(2\sin \theta \cos \theta {)}^2+4{\sin }^4\theta -4{\sin }^2\theta {\cos }^2\theta }{4-(2\sin \theta \cos \theta {)}^2-4{\sin }^2\theta }\Rightarrow y=\frac{{\sin }^4\theta }{1-{\sin }^2\theta {\cos }^2\theta -{\sin }^2\theta }\Rightarrow y=\frac{{\sin }^4\theta }{(1-{\sin }^2\theta )-{\sin }^2\theta {\cos }^2\theta }\Rightarrow y=\frac{{\sin }^4\theta }{{\cos }^2\theta (1-{\sin }^2\theta )}\Rightarrow y=\frac{{\sin }^4\theta }{{\cos }^4\theta }\Rightarrow y={\tan }^4\theta$
Now since $\theta ={750}^{\circ }$, all we need is the value of $\tan {750}^{\circ }$.

$\tan {750}^{\circ }=\tan ({720}^{\circ }+{30}^{\circ })=\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$
$\therefore y={\tan }^4{750}^{\circ }={\left(\frac{1}{\sqrt{3}}\right)}^4=\frac{1}{9}\therefore 9y=1$