Question

If $\sin \theta -\cos \theta =0$, then find the value of ${\sin }^4\theta +{\cos }^4\theta$

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. None of these

Answer 1

The correct option is B $\frac{1}{2}$

$\sin \theta -\cos \theta =0$ (Given)

To calculate values of ${\sin }^4\theta +{\cos }^4\theta$

Let $\sin \theta =a$ and $\cos \theta =b$

then $a^2+b^2=1$ $(\because {\sin }^2\theta +{\cos }^2\theta =1)$

Now, given

$a-b=0$ - $\left(2\right)$

Squaring both the sides, we get in eq $\left(1\right)$

$a^2+b^2-2ab=0$

$2ab=1$ - $\left(2\right)$

Again squaring both the sides in $\left(2\right)$ , we get

$[2a^2{b}^2=\frac{1}{2}]$ - $\left(3\right)$

To calculate $a^4+b^4$

we can write $a^4+b^4$ as

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

now using $a^2+b^2=1$ and result of equation $\left(3\right)$, we get

$a^4+b^4=1^2-\frac{1}{2}$

$a^4+b^4=\frac{1}{2}$

$i.e{\sin }^4\theta +{\cos }^4\theta =\frac{1}{2}$