Question

sin theta plus cos theta=
$\sqrt{2}$
​theta=?

Answer 1

Given,

$\longrightarrow \sin\theta+\cos\theta=\sqrt2$

Divide both sides by $\sqrt2.$

$\longrightarrow\dfrac{1}{\sqrt2}\sin\theta+\dfrac{1}{\sqrt2}\cos\theta=1$

Since $\dfrac{1}{\sqrt2}=\sin\dfrac{\pi}{4}=\cos\dfrac{\pi}{4},$

$\longrightarrow\sin\theta\cos\dfrac{\pi}{4}+\cos\theta\sin\dfrac{\pi}{4}=1$

We know that,

• $\sin(A+B)=\sin A\cos B+\cos A\sin B$

Thus we get,

$\longrightarrow\sin\left(\theta+\dfrac{\pi}{4}\right)=1$

Taking the general solution,

$\longrightarrow\theta+\dfrac{\pi}{4}=2n\pi+\dfrac{\pi}{2}$

$\longrightarrow\underline{\underline{\theta=2n\pi+\dfrac{\pi}{4}}}$

where $n\in\mathbb{Z}.$

Taking principal solutions,

$\longrightarrow\theta+\dfrac{\pi}{4}=\dfrac{\pi}{2}$

$\longrightarrow\underline{\underline{\theta=\dfrac{\pi}{4}}}$