Question

If $\sin A+\cos A=\sqrt{2}$, then the value of ${\cos }^{2}A$ is

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

Answer 1

The correct option is B

$\frac{1}{2}$

Explanation for the correct option.

Step 1: Find the value of $A$

Given that, $\sin A+\cos A=\sqrt{2}$

Divide both sides by $\sqrt{2}$ we have:
$\begin{array}{rcl}\frac{{\displaystyle (\mathrm{sinA}+\mathrm{cosA})}}{\sqrt{2}}& =& \frac{\sqrt{2}}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\mathrm{sinA}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}\mathrm{cosA}& =& 1\end{array}$

We know that, $\sin \frac{\mathrm{\pi }}{4}=\cos \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}\mathrm{and}\sin \left(\mathrm{a}+\mathrm{b}\right)=\mathrm{sinacosb}+\mathrm{cosasinb}$

So we have
$\begin{array}{rcl}\cos \frac{\mathrm{\pi }}{4}\mathrm{sinA}+\sin \frac{\mathrm{\pi }}{4}\mathrm{cosA}& =& 1\\ & \Rightarrow & \sin \left(A+\frac{\mathrm{\pi }}{4}\right)=\sin \frac{\mathrm{\pi }}{2}\\ & \Rightarrow & A+\frac{\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{2}\\ & \Rightarrow & A=\frac{\mathrm{\pi }}{4}\end{array}$

Step 2: Find the value of ${\cos }^{2}A$

Now substitute $A=\frac{\mathrm{\pi }}{4}$ in ${\cos }^{2}A$.
$$\begin{gathered} \Rightarrow {\cos }^{2}A={\cos }^2\begin{array}{l}\frac{\mathrm{\pi }}{4}\end{array} \\ ={\left(\frac{1}{\sqrt{2}}\right)}^2 \\ =\frac{1}{2} \end{gathered}$$
Hence, option B is correct.