Question

If $\sin \theta -\cos \theta =\frac{1}{2}$then find the value of $\sin \theta +\cos \theta$.

Answer 1

Find the value of the given expression.

Given: $\sin \theta -\cos \theta =\frac{1}{2}$.

To find: $\sin \theta +\cos \theta$

On squaring both sides of the given equation we get,

$$\begin{gathered} {\left(\sin \theta -\cos \theta \right)}^2={\left(\frac{1}{2}\right)}^2 \\ \Rightarrow {\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta =\frac{1}{4} \\ \Rightarrow 1-2\sin \theta \cos \theta =\frac{1}{4}[\because {\sin }^2\theta +{\cos }^2\theta =1] \\ \Rightarrow 2\sin \theta \cos \theta =1-\frac{1}{4} \\ \Rightarrow 2\sin \theta \cos \theta =\frac{3}{4} \end{gathered}$$

Now, on squaring the expression $\sin \theta +\cos \theta$.we get,

$$\begin{gathered} {\left(\sin \theta +\cos \theta \right)}^2={\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta \\ \Rightarrow \sin \theta +\cos \theta =\sqrt{{\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta } \\ \Rightarrow \sin \theta +\cos \theta =\sqrt{1+2\sin \theta \cos \theta }[\because {\sin }^2\theta +{\cos }^2\theta =1] \\ \Rightarrow \sin \theta +\cos \theta =\sqrt{1+\frac{3}{4}} \\ \Rightarrow \sin \theta +\cos \theta =\sqrt{\frac{7}{4}} \\ \Rightarrow \sin \theta +\cos \theta =\frac{\sqrt{7}}{2} \end{gathered}$$

Hence, the value of $\sin \theta +\cos \theta$ is $\frac{\sqrt{7}}{2}$.