Question

Solve the equation, \[\sin \theta +\cos \theta =\sqrt{2}\].

Answer 1

Hint: At first divide the whole equation by \[\sqrt{2}\], then use the knowledge that, \[\cos \dfrac{\pi }{4}=\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]. Then use the identity, \[\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B\]. Hence, find the general solutions.

Complete step-by-step answer:

In the question we are given an equation which is \[\sin \theta +\cos \theta =\sqrt{2}\] and we have to find general solutions or values of \[\theta \] for which equation satisfies.

We are given that,

\[\sin \theta +\cos \theta =\sqrt{2}\]

Now we will divide the whole equation by \[\sqrt{2}\] so we get,

\[\dfrac{1}{\sqrt{2}}\sin \theta +\dfrac{1}{\sqrt{2}}\cos \theta =1\]

We know that, \[\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\] and \[\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]. So, we can write equation as,

\[\cos \dfrac{\pi }{4}\cos \theta +\sin \dfrac{\pi }{4}\sin \theta =1\]

We know a identity which is \[\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B\] and we will use it as, \[\cos A\cos B+\sin A\sin B=\cos \left( A-B \right)\] and use A as \[\theta \] and B as \[\dfrac{\pi }{4}\]. So, we get,

\[\cos \left( x-\dfrac{\pi }{4} \right)=1\]

Now, we know that, \[\cos 0=1\]. So, we can write the equation as,

\[\cos \left( x-\dfrac{\pi }{4} \right)=\cos 0\]

So, we can write, \[x-\dfrac{\pi }{4}=2n\pi \pm 0\] or \[x=2n\pi +\dfrac{\pi }{4}\], by using formula if \[\cos x=\cos \alpha \], then \[x=2n\pi \pm \alpha \], where n is any integer.

Hence, the general solution is \[2n\pi +\dfrac{\pi }{4}\].

Note: Students can also do it by taking \[\cos \theta \] to the other side and then square both sides. The change \[{{\sin }^{2}}\theta \] to \[1-{{\cos }^{2}}\theta \]. Thus find values of \[\cos \theta \] for which equation satisfies.