Question

Find the value of $si{n}^4{735}^{\circ }$ + $co{s}^4{735}^{\circ }$ - $si{n}^2{735}^{\circ }$ $co{s}^2{735}^{\circ }$.

Answer 1

Is this expression looks familiar.

Something ($a^2$ + $b^2$ - ab)

We know, $a^3$ + $b^3$ = (a+b) ($a^2$ + $b^2$ - ab)

a = $si{n}^2{735}^{\circ }$

b = $co{s}^2{735}^{\circ }$

So, ($si{n}^4{735}^{\circ }$ + $co{s}^4{735}^{\circ }$ - $si{n}^2{735}^{\circ }$ $co{s}^2{735}^{\circ }$)

1 x ($si{n}^4{735}^{\circ }$ + $co{s}^4{735}^{\circ }$ - $si{n}^2{735}^{\circ }$$co{s}^2{735}^{\circ }$)

($si{n}^2{735}^{\circ }$ + $co{s}^2{735}^{\circ }$) ($si{n}^4{735}^{\circ }$ + $co{s}^4{735}^{\circ }$ - $si{n}^2{735}^{\circ }$$co{s}^4{735}^{\circ }$)

= $(si{n}^2{735}^{\circ }{)}^3$ + $(co{s}^2{735}^{\circ }{)}^3$

= $si{n}^6{735}^{\circ }$ + $co{s}^6{735}^{\circ }$

Option C is correct.