Question

y = $\frac{si{n}^2{1500}^{\circ }+4si{n}^4{750}^{\circ }-4si{n}^2{750}^{\circ }co{s}^2{750}^{\circ }}{4-si{n}^2{1500}^{\circ }+4si{n}^2{750}^{\circ }}$ Find the value of 9y.

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Answer 1

Solution: We need to calculate the values of $sin{750}^{\circ },cos{750}^{\circ }andsin{1500}^{\circ }$
(i) $Sin{750}^{\circ }=sin(2\times 360+{30}^{\circ })=sin{30}^{\circ }$ = $\frac{1}{2}$
(ii) $Cos{750}^{\circ }=cos(2\times 360+{30}^{\circ })=cos{30}^{\circ }$ = $\frac{\sqrt{3}}{2}$
(iii) $Sin{1500}^{\circ }=sin(360\times 4+{60}^{\circ })=sin{60}^{\circ }$ = $\frac{\sqrt{3}}{2}$
Substituting these values in the given equation
y = $\frac{si{n}^2{1500}^{\circ }+4si{n}^4{750}^{\circ }-4si{n}^2{750}^{\circ }co{s}^2{750}^{\circ }}{4-si{n}^2{1500}^{\circ }+4si{n}^2{750}^{\circ }}$
= $\frac{{\left(\frac{\sqrt{3}}{2}\right)}^2+4.\frac{1}{16}-4.{\left(\frac{1}{2}\right)}^2{\left(\frac{\sqrt{3}}{2}\right)}^2}{4-{\left(\frac{\sqrt{3}}{2}\right)}^2-4.{\left(\frac{1}{2}\right)}^2}$
= $\frac{\frac{3}{4}+4.\frac{1}{16}-4\times \frac{1}{4}\times \frac{1}{3}}{4-\frac{3}{4}-4.\frac{1}{4}}$
= $\frac{\frac{3}{4}+\frac{1}{4}-\frac{3}{4}}{4-\frac{3}{4}-1}$
= $\frac{\frac{1}{4}}{\frac{9}{4}}$ = $\frac{1}{9}$
9y = 9 $\times$ $\frac{1}{9}$ = 1.