Question

Prove the following identities
$2(si{n}^6\theta +co{s}^6\theta )-3(si{n}^4\theta +co{s}^4\theta )+1=0$

Answer 1

$2(sin⁶\theta +cos⁶\theta )-3(sin⁴\theta +cos⁴\theta )+1=0$

$LHS=2(sin⁶\theta +cos⁶\theta )-3(sin⁴\theta +cos⁴\theta )+1$

$=2\left[\right(sin²\theta )³+(cos²\theta \left)³\right]-3sin⁴\theta -3cos⁴\theta +1$

$=2(sin²\theta +cos²\theta )\left[\right(sin²\theta )²+(cos²\theta )²-sin²\theta cos²\theta )]-3sin⁴\theta -3cos⁴\theta +1$.....As$[a³+b³=(a+b\left)\right(a²+b²-a-b]$

$=2\left(1\right)[sin⁴\theta +cos⁴\theta -sin²\theta cos²\theta ]-3sin⁴\theta -3cos⁴\theta +1$ ...........As$[sin²\theta +cos²\theta =1]$

$=2sin⁴\theta +2cos⁴\theta -2sin²\theta cos²\theta -3sin⁴\theta -3cos⁴\theta +1$

$=2sin⁴\theta -3sin⁴\theta +2cos⁴\theta -3cos⁴\theta -2sin²\theta cos²\theta +1$

$=-sin⁴\theta -cos⁴\theta -2sin²\theta cos²\theta +1$

$=1-[sin⁴\theta +cos⁴\theta +2sin²\theta (cos²\theta \left)\right]$

$=1-\left[\right(sin²\theta )²+(cos²\theta )²+2sin²\theta cos²\theta ]$

$=1-(sin²\theta +cos²\theta )²$..............As$[a²+b²+2ab=(a+b\left)²\right]$

$=1-\left(1\right)²$

$LHS=1-1=0$

$LHS=RHS$

$2(sin⁶\theta +cos⁶\theta )-3(sin⁴\theta +cos⁴\theta )+1=0$