Question

Find the value of $\theta$ in which $\sin \theta =\cos \theta$ where $\theta$ is an acute angle.

Hint: First of all square both sides of the given equation and convert the whole equation in terms of $\sin \theta$ and find the value of $\sin \theta$. Then if $\sin \theta =\sin \alpha$, then $\theta =n\pi +{{\left( -1 \right)}^{n}{\alpha }}$ where $\alpha \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ and $n\in I$. Then choose the acute $\theta$ among all.

Here, we have to find the values of $\theta$ if $\sin \theta =\cos \theta$ given that $\theta$ is acute. First of all, let us consider the equation given in the question, $\sin \theta =\cos \theta$
By squaring both sides of the above equation, we get
${{\sin }^{2}}\theta ={{\cos }^{2}}\theta$
Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ or ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$. By substituting the value of ${{\cos }^{2}}\theta$ in terms of ${{\sin }^{2}}\theta$ in the above equation, we get,
${{\sin }^{2}}\theta =1-{{\sin }^{2}}\theta$
By adding ${{\sin }^{2}}\theta$ on both sides of the above equation, we get,
${{\sin }^{2}}\theta +{{\sin }^{2}}\theta =1$
Or, $2{{\sin }^{2}}\theta =1$
By dividing 2 on both sides of the above equation, we get,
$\Rightarrow \dfrac{2{{\sin }^{2}}\theta }{2}=\dfrac{1}{2}$
By canceling the like terms in LHS of the above equation, we get,
$\Rightarrow {{\sin }^{2}}\theta =\dfrac{1}{2}$
By taking square root on both sides of the above equation, we get,
$\sqrt{{{\sin }^{2}}\theta }=\sqrt{\dfrac{1}{2}}$
We know that $\sqrt{{{a}^{2}}}=\pm a$ and $\sqrt{1}=1$. By applying these in the above equation, we get,
$\sin \theta =\pm \dfrac{1}{\sqrt{2}}$
Since we are given that $\theta$ is acute, so $\sin \theta >0$.
Therefore, we take $\sin \theta =\dfrac{1}{\sqrt{2}}$.
We know that $\sin {{45}^{o}}=\dfrac{1}{\sqrt{2}}$. So by substituting the value of $\dfrac{1}{\sqrt{2}}$ in terms of sine in the above equation, we get,
$\Rightarrow \sin \theta =\sin {{45}^{o}}=\sin \left( 45\times \dfrac{\pi }{180} \right)=\sin \dfrac{\pi }{4}$
We know that if $\sin \theta =\sin \alpha$, then
$\theta =n\pi +{{\left( -1 \right)}^{n}{\alpha }},\alpha \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\text{ and }n\in I$
By applying this in the above equation, we get,
$\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$
By substituting n = 0, we get
$\theta =0.\left( \pi \right)+{{\left( -1 \right)}^{0}}\dfrac{\pi }{4}$
We get, $\theta =\dfrac{\pi }{4}$
By substituting n = 1, we get
$\theta =\pi +{{\left( -1 \right)}^{1}}\dfrac{\pi }{4}$
$=\pi -\dfrac{\pi }{4}$
We get, $\theta =\dfrac{3\pi }{4}$
By substituting n = 2, we get,
$\theta =3\pi +{{\left( -1 \right)}^{2}}\dfrac{\pi }{4}$
$=3\pi +\dfrac{\pi }{4}$
We get, $\theta =\dfrac{13\pi }{4}$
So, we get the value of $\theta$ as $\dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{13\pi }{4}.....\text{so on}$
But in the question, we are given that $\theta$ must be acute that means $0<\theta <\dfrac{\pi }{2}$. So, only one value of $\theta$ i.e. $\theta =\dfrac{\pi }{4}$ is acceptable. Also, for $\theta =\dfrac{\pi }{4}$, we get $\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$.
So, we get $\theta =\dfrac{\pi }{4}={{45}^{o}}$.

Note: Some students often make this mistake of getting the result $\theta =\alpha$ initially by just looking at $\sin \theta =\sin \alpha$. But they must note that $\theta =\alpha$ is not the only result but it is one of the results. The general value of $\theta =n\pi +{{\left( -1 \right)}^{n}{\alpha }}$ where $\alpha \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ and $n\in I$ and there would be infinitely many values of $\theta$ for giving $\alpha$ not just $\theta =\alpha$. Also, students must check the value of $\theta$ by substituting it in the given equation that is $\sin \theta =\cos \theta$ and then only answer.