Question

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

Answer 1

Hint: For these kinds of questions, the most obvious thing that we can do here is squaring on both sides. So the square root on the right hand side would not be there anymore. And also it will be easier for us to use the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. After using that, we will get our entire expression in only one trigonometric function on the left hand side and a constant without any square root on the right hand side. Then we can find the value of the angle A. And then find the value of ${{\cos }^{2}}A$.

Complete step by step solution:
So now, let us first square the equation that we have.
Upon doing so, we get the following :
\[\begin{align}
  & \Rightarrow \sin A+\cos A=\sqrt{2} \\
 & \Rightarrow {{\left( \sin A+\cos A \right)}^{2}}={{\left( \sqrt{2} \right)}^{2}} \\
\end{align}\]
We all know the formula of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Let us use this formula to expand further.
Upon using the formula, we get the following :
\[\begin{align}
  & \Rightarrow \sin A+\cos A=\sqrt{2} \\
 & \Rightarrow {{\left( \sin A+\cos A \right)}^{2}}={{\left( \sqrt{2} \right)}^{2}} \\
 & \Rightarrow {{\sin }^{2}}A+{{\cos }^{2}}A+2\sin A\cos A=2 \\
\end{align}\]
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Let us use this formula to solve further.
Upon using, we get the following :
\[\begin{align}
  & \Rightarrow \sin A+\cos A=\sqrt{2} \\
 & \Rightarrow {{\left( \sin A+\cos A \right)}^{2}}={{\left( \sqrt{2} \right)}^{2}} \\
 & \Rightarrow {{\sin }^{2}}A+{{\cos }^{2}}A+2\sin A\cos A=2 \\
 & \Rightarrow 1+2\sin A\cos A=2 \\
\end{align}\]
We know the expansion of $\sin 2\theta $ .
$\Rightarrow \sin 2\theta =2\sin \theta \cos \theta $ .
Let us use this formula to write everything in terms of $\sin $ trigonometric function.
Upon doing, we get the following :
\[\begin{align}
  & \Rightarrow \sin A+\cos A=\sqrt{2} \\
 & \Rightarrow {{\left( \sin A+\cos A \right)}^{2}}={{\left( \sqrt{2} \right)}^{2}} \\
 & \Rightarrow {{\sin }^{2}}A+{{\cos }^{2}}A+2\sin A\cos A=2 \\
 & \Rightarrow 1+2\sin A\cos A=2 \\
 & \Rightarrow 1+\sin 2A=2 \\
\end{align}\]
Let us send the $1$ which is present on the left hand side to the right hand side for further solving.
\[\begin{align}
  & \Rightarrow \sin A+\cos A=\sqrt{2} \\
 & \Rightarrow {{\left( \sin A+\cos A \right)}^{2}}={{\left( \sqrt{2} \right)}^{2}} \\
 & \Rightarrow {{\sin }^{2}}A+{{\cos }^{2}}A+2\sin A\cos A=2 \\
 & \Rightarrow 1+2\sin A\cos A=2 \\
 & \Rightarrow 1+\sin 2A=2 \\
 & \Rightarrow \sin 2A=1 \\
\end{align}\]
We know that for the value of $A=\dfrac{\pi }{4}$ , the value of $\sin 2A$ would be equal to $1$. Since the quadrant in which A is lying is not mentioned in the question, we can consider it to be the first quadrant.
Now let us substitute the value of A to get the value of ${{\cos }^{2}}A$.
Upon doing, we get the following :
\[\Rightarrow {{\cos }^{2}}A={{\cos }^{2}}\dfrac{\pi }{4}={{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}=\dfrac{1}{2}\] .

Note: It is very important to remember all the identities and formulae so as to solve the question quickly. We can also do that this question by dividing the entire equation with $\sqrt{{{a}^{2}}+{{b}^{2}}}$ where $a,b$ are the coefficients of $\sin A,\cos A$ respectively. And then we shrink the entire equation using either $\sin \left( A+B \right)$ formula or $\cos (A+B)$ formula . Either way, we would arrive at the same result. We should be careful while solving the question as there is a lot of scope for calculation errors.