Question

If ${\sin }^4\theta +{\cos }^4\theta =\frac{1}{2}$ then the value of $\sin \theta \cos \theta$ is

• A. $\pm \frac{1}{8}$
• B. $\pm \frac{1}{4}$
• C. $\pm 1$
• D. $\pm \frac{1}{2}$

Answer 1

The correct option is D $\pm \frac{1}{2}$

We know that $a^4+b^4={(a^2+b^2)}^2-2a^2{b}^2$

So, ${\sin }^4\theta +{\cos }^{4}0={({\sin }^2\theta +{\cos }^2\theta )}^2-2({\sin }^2\theta {)}^2{\left({\cos }^2\theta \right)}^2$

But ${\sin }^2\theta +{\cos }^2\theta =1$

So, ${\sin }^4\theta +{\cos }^4\theta =1^2-2(\sin \theta \cos \theta {)}^2$

$\Rightarrow \frac{1}{2}=1-2(\sin \theta \cos \theta {)}^2$

$\Rightarrow 2(\sin \theta \cos \theta {)}^2=\frac{1}{2}$

$\Rightarrow (\sin \theta \cos \theta {)}^2=\frac{1}{4}$

$\Rightarrow \sin \theta \cos \theta =\pm \frac{1}{2}$