Question

If $\left(\sin \theta ,\cos \theta \right)$ and $\left(3,2\right)$ lie on the same side of the line $x+y=1$, then $\theta$ lies between

• A. $\left(0,\frac{\pi }{2}\right)$
• B. $\left(0,\pi \right)$
• C. $\left(\frac{\pi }{4},\frac{\pi }{2}\right)$
• D. $\left(0,\frac{\pi }{4}\right)$

Answer 1

The correct option is D

$\left(0,\frac{\pi }{4}\right)$

Explanation for the correct option:

Step 1: Requirements for the solution

Since both, the given points lie on the same side of the line $x+y=1$. Let us call this line $Q$.

After applying the coordinates $\left(3,2\right)$, we have

$\begin{array}{rcl}Q\left(3,2\right)& =& 3+2-1\\ & =& 4>0\end{array}$

Then after applying the coordinates $\left(\sin \theta ,\cos \theta \right)$ in the line the result will be greater than zero because both points lie on the same side, then

$Q\left(\sin \theta ,\cos \theta \right)=\sin \left(\theta \right)+\cos \left(\theta \right)-1>0$

Step 2: Finding where $\mathit{\theta }$ lies

As we know that $\sin \left(\theta \right)+\cos \left(\theta \right)-1>0$ then

$\sin \left(\theta \right)+\cos \left(\theta \right)>1$

Divide both the sides by $\sqrt{2}$,

$\begin{array}{rcl}\therefore \sin \left(\theta \right)\left(\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}\right)\cos \left(\theta \right)& >& \frac{1}{\sqrt{2}}\\ & \Rightarrow & \sin \left(\theta \right)\cos \left(\frac{\pi }{4}\right)+\sin \left(\frac{\pi }{4}\right)\cos \left(\theta \right)>\frac{1}{\sqrt{2}}\left[\because \sin \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}\text{and}\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}\right]\end{array}$

Now, apply the identity $\mathbf{sin}\left(A+B\right)\mathbf{=}\mathbf{sin}\left(A\right)\mathbf{cos}\left(B\right)\mathbf{+}\mathbf{sin}\left(B\right)\mathbf{cos}\left(A\right)$

$\sin \left(\theta +\frac{\pi }{4}\right)>\frac{1}{\sqrt{2}}$

We know that this can only happen when $\sin \left(A\right)$ is greater than $\frac{\pi }{4}$ but lesser than $\frac{\pi }{2}$, where $A=\theta +\frac{\pi }{4}$.

$\begin{array}{rcl}\therefore \frac{\pi }{4}& <& \theta +\frac{\pi }{4}<\frac{\pi }{2}\\ & \Rightarrow & \frac{\pi }{4}-\frac{\pi }{4}<\theta <\frac{\pi }{2}-\frac{\pi }{4}\\ & \Rightarrow & 0<\theta <\frac{\pi }{4}\end{array}$

Hence, the correct option is (D),