Question

If $\sin \left(\mathrm{x}\right)-\cos \left(\mathrm{x}\right)=0$ the value of ${\sin }^4\left(\mathrm{x}\right)+{\cos }^4\left(\mathrm{x}\right)$.

• A. $1$
• B. $\frac{3}{4}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C

$\frac{1}{2}$

Explanation for the correct option:

Given, $\sin \left(\mathrm{x}\right)-\cos \left(\mathrm{x}\right)=0$

$\Rightarrow \sin \left(\mathrm{x}\right)=\cos \left(\mathrm{x}\right)$

$\Rightarrow x$ is either in First Quadrant or Third Quadrant as both $\sin \left(\mathrm{x}\right)$ and $\cos \left(\mathrm{x}\right)$ have equal sign (either both positive or both negative).

$\Rightarrow x=\mathrm{n\pi }-\frac{3\mathrm{\pi }}{4},\mathrm{n}\in \mathrm{\mathbb{Z}}$

Here Quadrant does not matter as both $\sin \left(\mathrm{x}\right)$and $\cos \left(\mathrm{x}\right)$ have even powers in ${\sin }^4\left(\mathrm{x}\right)+{\cos }^4\left(\mathrm{x}\right)$ and ${\sin }^4\left(\mathrm{x}\right)$,${\cos }^4\left(\mathrm{x}\right)$ are positive in all quadrants.

So for simplicity let $\mathrm{n}=1$,

$\Rightarrow \mathrm{x}=\frac{\mathrm{\pi }}{4}$

$$\begin{gathered} \Rightarrow {\sin }^4\left(\frac{\mathrm{\pi }}{4}\right)+{\cos }^4\left(\frac{\mathrm{\pi }}{4}\right)={\left(\frac{1}{\sqrt{2}}\right)}^4+{\left(\frac{1}{\sqrt{2}}\right)}^4 \\ \Rightarrow {\sin }^4\left(\frac{\mathrm{\pi }}{4}\right)+{\cos }^4\left(\frac{\mathrm{\pi }}{4}\right)=\frac{1}{2} \end{gathered}$$

Hence, option C is correct.