Solve the equation \[\cos \theta +\sin \theta =cos2\theta +sin2\theta \].

Hint: We will apply the formula of \[\operatorname{cosA}+\operatorname{cosB}\] given by \[-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)\] and \[sinA-sinB\] given by \[2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)\] to simplify and solve the equation given in the question.

Complete step-by-step answer:

We will first consider the expression \[\cos \theta +\sin \theta =cos2\theta +sin2\theta .....(1)\]

Now we will take \[cos2\theta \] on the left side of the equality sign and \[\sin \theta \] to the right side of the equality sign.

Therefore, we have,

\[\cos \theta -cos2\theta =sin2\theta -\sin \theta \]

Now we will apply two trigonometry formulas, given by,

\[\operatorname{cosA}+\operatorname{cosB}=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)\] and

\[sinA-sinB=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)\].

Therefore, we have a new expression written as,

\[\begin{align}

  & -2\sin \left( \dfrac{\theta +2\theta }{2} \right)\sin \left( \dfrac{\theta -2\theta }{2} \right)=2\cos \left( \dfrac{2\theta +\theta }{2} \right)\sin \left( \dfrac{2\theta -\theta }{2} \right) \\

 & -\sin \left( \dfrac{3\theta }{2} \right)\sin \left( \dfrac{-\theta }{2} \right)=\cos \left( \dfrac{3\theta }{2} \right)\sin \left( \dfrac{\theta }{2} \right) \\

\end{align}\]

By applying the formula \[\sin (-\theta )=-\sin \theta \], thus we get,

\[\begin{align}

  & -\sin \left( \dfrac{3\theta }{2} \right)\left[ -\sin \left( \dfrac{\theta }{2} \right) \right]=\cos \left( \dfrac{3\theta }{2} \right)\sin \left( \dfrac{\theta }{2} \right) \\

 & \sin \left( \dfrac{3\theta }{2} \right)\sin \left( \dfrac{\theta }{2} \right)=\cos \left( \dfrac{3\theta }{2} \right)\sin \left( \dfrac{\theta }{2} \right) \\

 & \sin \left( \dfrac{3\theta }{2} \right)\sin \left( \dfrac{\theta }{2} \right)-\cos \left( \dfrac{3\theta }{2} \right)\sin \left( \dfrac{\theta }{2} \right)=0 \\

\end{align}\]

Now we will take the common term \[\sin \left( \dfrac{\theta }{2} \right)=0\] or \[\sin \left( \dfrac{3\theta }{2} \right)-\cos \left( \dfrac{3\theta }{2} \right)=0\].

First we will consider that \[\sin \left( \dfrac{\theta }{2} \right)=0\].

\[\Rightarrow \sin \left( \dfrac{\theta }{2} \right)=\sin {{0}^{\circ }}\]. This is because we know that \[0=\sin {{0}^{\circ }}\].

\[\begin{align}

  & \dfrac{\theta }{2}={{0}^{\circ }} \\

 & \theta ={{0}^{\circ }} \\

\end{align}\]

Also we know that \[sin\left( n\pi \right)=0\], where \[n=\pm 0,\pm 1...\]

\[sin\left( \dfrac{\theta }{2} \right)=\sin \left( n\pi \right)\], where \[n=\pm 0,\pm 1...\]

\[\left( \dfrac{\theta }{2} \right)=\left( n\pi \right)\], where \[n=\pm 0,\pm 1...\]

\[\theta =2n\pi \], where \[n=\pm 0,\pm 1...\]

Now we will consider \[\sin \left( \dfrac{3\theta }{2} \right)-\cos \left( \dfrac{3\theta }{2}

\right)=0\].

\[\sin \left( \dfrac{3\theta }{2} \right)=\cos \left( \dfrac{3\theta }{2} \right)....(2)\]

As we know that there is only one condition in which \[\sin \theta =\cos \theta \]. The angle \[\dfrac{\pi }{4}\] results in \[\sin \left( \dfrac{\pi }{4} \right)=\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\]. Since we have,

\[\begin{align}

  & \sin \left( \dfrac{\pi }{4} \right)=\cos \left( \dfrac{\pi }{4} \right) \\

 & \dfrac{\sin \left( \dfrac{\pi }{4} \right)}{\cos \left( \dfrac{\pi }{4} \right)}=1 \\

\end{align}\]

As we know that the value of \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. Therefore, we have,

\[\tan \left( \dfrac{\pi }{4} \right)=1....(3)\]

Now we will consider the equation (2). If we write, \[\sin \left( \dfrac{3\theta }{2} \right)=\cos \left( \dfrac{3\theta }{2} \right)\] as \[\dfrac{\sin \left( \dfrac{3\theta }{2} \right)}{\cos \left( \dfrac{3\theta }{2} \right)}=1\], then this will create contradiction, since \[\cos \left( \dfrac{3\theta }{2} \right)\ne 0\].

\[\begin{align}

  & \cos \left( \dfrac{3\theta }{2} \right)\ne 0 \\

 & \cos \left( \dfrac{3\theta }{2} \right)\ne \cos \left( \dfrac{\pi }{2} \right) \\

 & \left( \dfrac{3\theta }{2} \right)\ne \left( \dfrac{\pi }{2} \right) \\

\end{align}\]

If we choose \[\dfrac{3\theta }{2}=\dfrac{\pi }{4}\] which is not equal to \[\dfrac{\pi }{2}\], therefore we have that,


\[\dfrac{\sin \left( \dfrac{3\theta }{2} \right)}{\cos \left( \dfrac{3\theta }{2} \right)}=1\] when

\[\left( \dfrac{3\theta }{2} \right)\ne \left( \dfrac{\pi }{2} \right)\]

\[\tan \left( \dfrac{3\theta }{2} \right)=1.....(4)\]

By equation equations (3) and (4), we have that,

\[\tan \left( \dfrac{3\theta }{2} \right)=\tan \left( \dfrac{\pi }{4} \right)\]

\[\Rightarrow \left( \dfrac{3\theta }{2} \right)=\left( \dfrac{\pi }{4} \right)\] or \[\left( \dfrac{3\theta }{2} \right)=\dfrac{\pi }{4}+k\pi \].

\[\left( \dfrac{3\theta }{2} \right)=\dfrac{\pi }{4}+k\pi \], where \[k=0,\pm 1,\pm 2...\]

\[\theta =\dfrac{\pi }{6}+\dfrac{2}{3}k\pi \], where \[k=0,\pm 1,\pm 2...\]

Hence the solution of equation (1) is given by \[\theta =\dfrac{\pi }{6}+\dfrac{2}{3}k\pi \].

Here k = n.


Note: Whenever we get an expression of the form ab = ca, Don’t try to cancel the terms. First we will take ca to the left hand side of the equality, and then we will solve it as ab – ca = 0 or a (b-c) = 0. Here, automatically a = 0 but in trigonometric terms, this fact is considered as a concept. If we are writing any term in the denominator it should be checked for which value the denominator is not equal to 0. With the value at which the denominator is not 0 the proceedings of solutions can be done.