Question

If $\sin \theta +\cos \theta =1$ then the general value of $\theta $ is
A. \[2n\pi \]
B. \[n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{4}\]
C. \[2n\pi +\frac{\pi }{2}\]
D. None of these.

Answer 1

Hint: To derive the general value of $\theta $, we will take the given equation and divide it by $\sqrt{2}$ on both sides. Then we will simplify it and rewrite it using trigonometric table values in a way to form of formula of $\sin (a+b)$.

Formula Used:$\sin (a+b)=\sin a\cos b+\cos a\sin b$

Complete step by step solution:We are given $\sin \theta +\cos \theta =1$ and we have to determine the value of $\theta $.
We will first take given equation $\sin \theta +\cos \theta =1$ and divide it by $\sqrt{2}$ on both sides.
$\begin{align}
  & \frac{\sin \theta +\cos \theta }{\sqrt{2}}=\frac{1}{\sqrt{2}} \\
 & \frac{\sin \theta }{\sqrt{2}}+\frac{\cos \theta }{\sqrt{2}}=\frac{1}{\sqrt{2}} \\
 & \frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta =\frac{1}{\sqrt{2}}
\end{align}$
As we know that $\sin \frac{\pi }{4}=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$, we will substitute this in the equation in a way to form the formula of $\sin (a+b)$.
$\begin{align}
  & \frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta =\frac{1}{\sqrt{2}} \\
 & \sin \theta \cos \frac{\pi }{4}+\cos \theta \sin \frac{\pi }{4}=\frac{1}{\sqrt{2}} \\
 & \sin \left( \frac{\pi }{4}+\theta \right)=\sin \frac{\pi }{4}
\end{align}$
As we know that if $\sin \theta =\sin \alpha $then $\theta =n\pi +{{(-1)}^{n}}\alpha $. So we will apply this in the equation,
$\begin{align}
  & \theta +\frac{\pi }{4}=n\pi +{{(-1)}^{n}}\frac{\pi }{4} \\
 & \theta =n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{4}
\end{align}$



Option ‘B’ is correct

Note: The general solution of $\sin \theta =\sin \alpha $is $\theta =n\pi +{{(-1)}^{n}}\alpha $ where $n\in Z$. We will show how the value of $\theta $ is derived from $\sin \theta =\sin \alpha $.
$\begin{align}
  & \sin \theta =\sin \alpha \\
 & \sin \theta -\sin \alpha =0
\end{align}$
Now we will use the formula $\sin C-\sin D=2\sin \frac{C-D}{2}\cos \frac{C+D}{2}$.
$\begin{align}
  & \sin \theta -\sin \alpha =0 \\
 & 2\sin \frac{\theta -\alpha }{2}\cos \frac{\theta +\alpha }{2}=0 \\
 & \sin \frac{\theta -\alpha }{2}\cos \frac{\theta +\alpha }{2}=0
\end{align}$
Now,
$\sin \frac{\theta -\alpha }{2}=0$ or $\cos \frac{\theta +\alpha }{2}=0$
$\frac{\theta -\alpha }{2}=m\pi $ or $\frac{\theta +\alpha }{2}=(2m+1)\frac{\pi }{2}$
We will now derive the value of $\theta $ for both.
First we will take,
$\frac{\theta -\alpha }{2}=m\pi $
$\theta =2m\pi +\alpha \,\,,m\in Z$. It means that value of $\theta $ will be any even multiple of $\pi $ in addition of another angle$\alpha $.
Now $\frac{\theta +\alpha }{2}=(2m+1)\frac{\pi }{2}$,
$\theta =(2m+1)\pi +\alpha \,\,\,,m\in Z$. It means that value of $\theta $ will be any odd multiple of $\pi $ with another angle$\alpha $subtracted from it.
We will now combine both the value of the angle $\theta $,
$\theta =n\pi +{{(-1)}^{n}}\alpha $, where $n\in Z$.