Question

The value of \[\sin 45+\cos 45\]
\[\begin{align}
  & \text{A}\text{. 1} \\
 & \text{B}\text{. }\dfrac{1}{\sqrt{2}} \\
 & \text{C}\text{. }\dfrac{\sqrt{3}}{2} \\
 & \text{D}\text{. }\sqrt{2} \\
\end{align}\]

Answer 1

Hint: Let us assume the value of \[\sin 45\] is equal to A. Now we have to find the value of \[\sin 45\]. Let us assume this as equation (1). Let us assume the value of \[\cos 45\] is equal to B. Now we have to find the value of \[\cos 45\]. Let us assume this as equation (2). We must recollect the standard angle values and the results of both equations (1) and (2) will be the same. Let us assume the value of \[\sin 45+\cos 45\] is equal to C. Let us assume this as equation (3). Now let us substitute equation (1) and equation (2) in equation (3). In this way, we can find the value of \[\sin 45+\cos 45\].

Complete step-by-step solution:
Now from the question, we should find the value of \[\sin 45+\cos 45\].
Let us assume the value of \[\sin 45\] is equal to A.
\[\Rightarrow A=\sin 45\]
We know that \[\sin 45=\dfrac{1}{\sqrt{2}}\].
So, we can write
\[\Rightarrow A=\dfrac{1}{\sqrt{2}}.....(1)\]
From equation (1), it is clear that the value of A is equal to \[\dfrac{1}{\sqrt{2}}\].
Let us assume the value of \[\cos 45\] is equal to B.
\[\Rightarrow B=\cos 45\]
We know that \[\cos 45=\dfrac{1}{\sqrt{2}}\].
So, we can write
\[\Rightarrow B=\dfrac{1}{\sqrt{2}}.....(2)\]
From equation (2), it is clear that the value of B is equal to \[\dfrac{1}{\sqrt{2}}\].
Let us assume the value of \[\sin 45+\cos 45\] is equal to C.
\[\begin{align}
  & \Rightarrow C=\sin 45+\cos 45. \\
 & \Rightarrow C=A+B.....(3) \\
\end{align}\]
Now let us substitute equation (1) and equation (2) in equation (3), then we get
\[\begin{align}
  & \Rightarrow C=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} \\
 & \Rightarrow C=\dfrac{2}{\sqrt{2}} \\
\end{align}\]
Now let us multiply and divide with \[\sqrt{2}\].
\[\begin{align}
  & \Rightarrow C=\dfrac{2}{\sqrt{2}}\left( \dfrac{\sqrt{2}}{\sqrt{2}} \right) \\
 & \Rightarrow C=\dfrac{2\sqrt{2}}{2} \\
 & \Rightarrow C=\sqrt{2}....(4) \\
\end{align}\]
From equation (4), it is clear that \[\sin 45+\cos 45\] is equal to \[\sqrt{2}\].
Hence, option C is correct.

Note: This problem can be solved in an alternative method also.
We know that if \[0\le \theta \le 90\] then \[\sin \theta =\cos \left( 90-\theta \right)\].
Let us consider
\[\sin \theta =\cos \left( 90-\theta \right).................(1)\]
Now from the question, we should find the value of \[\sin 45+\cos 45\].
So, let us substitute \[\theta =45\] in equation (1), then we get
\[\begin{align}
  & \Rightarrow \sin 45=\cos \left( 90-45 \right) \\
 & \Rightarrow \sin 45=\cos 45................(2) \\
\end{align}\]
Let us assume the value of \[\sin 45+\cos 45\] is equal to C.
\[\Rightarrow C=\sin 45+\cos 45……….....(3)\]
Let us substitute equation (2) in equation (3), then we get
\[\Rightarrow C=2\sin 45……………...(4)\]
We know that \[\sin 45=\dfrac{1}{\sqrt{2}}\].
\[\begin{align}
  & \Rightarrow C=2\left( \dfrac{1}{\sqrt{2}} \right) \\
 & \Rightarrow C=\dfrac{2}{\sqrt{2}} \\
\end{align}\]
Now let us multiply and divide with \[\sqrt{2}\].
\[\begin{align}
  & \Rightarrow C=\dfrac{2}{\sqrt{2}}\left( \dfrac{\sqrt{2}}{\sqrt{2}} \right) \\
 & \Rightarrow C=\dfrac{2\sqrt{2}}{2} \\
 & \Rightarrow C=\sqrt{2}.....(5) \\
\end{align}\]
From equation (5), it is clear that \[\sin 45+\cos 45\] is equal to \[\sqrt{2}\].