Question

If $sin\theta cos\theta =\frac{1}{2}$,then what is the value of $si{n}^6\theta +co{s}^6\theta$ ?

• A. 1
• B. 1/4
• C. 1/2
• D. 2

Answer 1

The correct option is B

1/4

$sin\theta cos\theta =\frac{1}{2}.....\left(1\right)(sin\theta cos\theta {)}^2=\frac{1}{4}$
Or, $si{n}^2\theta co{s}^2\theta =\frac{1}{4}$
Or, $si{n}^2\theta (1-si{n}^2\theta )=\frac{1}{4}$
Or, $si{n}^2\theta -si{n}^4\theta =\frac{1}{4}$
Let $si{n}^2\theta =x$
Or, $4x-4x^2+1=0$
Or, $4x^2-4x-1=0$
Or, $4x^2-2x-2x-1=0$
Or, $2x(2x-1)+1(2x-1)=0$
$x=\frac{1}{2}$ and $x=-\frac{1}{2}$
Since, $x=si{n}^2\theta$, it can't be negative. So, $x=\frac{1}{2}$
Hence, $sin\theta =\pm \frac{1}{\sqrt{2}}$
Putting the value of sin x in equation (1)
We get $cos\theta =\pm \frac{1}{\sqrt{2}}$
So, $si{n}^6\theta +co{s}^6\theta ={(\pm \frac{1}{\sqrt{2}})}^6+{(\pm \frac{1}{\sqrt{2}})}^6$
$=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$