Question

If $\sin A=\cos A$, then find the value of A in degrees.

Answer 1

There are many possible values of angle A.
This is a trigonometric equation of the form $A\sin x\pm B\cos x=C$. We define $\sin y=\dfrac{A}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$ and $\cos y=\dfrac{A}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$, so that $y={{\tan }^{-1}}\dfrac{A}{B}$.
Using the identity $\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B$, the equation simplifies to:
$\cos (x\pm y)=\dfrac{C}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$.
The solution is $x=2n\pi \pm {{\cos }^{-1}}\dfrac{C}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\pm y,n\in \mathbb{Z}$.
Recall that $\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}$, $\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}$ and $\tan 45{}^\circ =1$.

Complete step by step answer:
Given that $\sin A=\cos A$.
⇒ $\sin A-\cos A=0$ ... (1)
Comparing this with the equations of the type $A\sin x-B\cos x=C$, we have:
$A=1,B=1,C=0$
Let's say that $\sin y=\dfrac{A}{\sqrt{{{A}^{2}}+{{B}^{2}}}}=\dfrac{1}{\sqrt{{{1}^{2}}+{{(-1)}^{2}}}}=\dfrac{1}{\sqrt{2}}$ and $\cos y=\dfrac{B}{\sqrt{{{A}^{2}}+{{B}^{2}}}}=\dfrac{1}{\sqrt{{{1}^{2}}+{{(-1)}^{2}}}}=\dfrac{1}{\sqrt{2}}$, so that $y={{\tan }^{-1}}\dfrac{B}{A}={{\tan }^{-1}}\dfrac{1}{1}={{\tan }^{-1}}1=45{}^\circ$.
Dividing equation (1) by $\sqrt{2}$, we get:
⇒ $\dfrac{1}{\sqrt{2}}\sin A-\dfrac{1}{\sqrt{2}}\cos A=0$
Substituting $\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}$ and $\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}$, we get:
⇒ $\sin 45{}^\circ \sin A-\cos 45{}^\circ \cos A=0$
Multiplying by -1, we get:
⇒ $\cos 45{}^\circ \cos A-\sin 45{}^\circ \sin A=0$
Using the identity $\cos (A+B)=\cos A\cos B-\sin A\sin B$, we get:
⇒ $\cos (45{}^\circ +A)=0=\cos 90{}^\circ$
Since the period of $\cos \theta$ is $2\pi \text{ radians}=360{}^\circ$, we can write:
⇒ $\cos (45{}^\circ +A)=\cos (360{}^\circ n\pm 90{}^\circ ),n\in \mathbb{Z}$
⇒ $45{}^\circ +A=360{}^\circ n\pm 90{}^\circ,n\in \mathbb{Z}$
⇒ $A=360{}^\circ n+45{}^\circ$ OR $A=360{}^\circ n-135{}^\circ$, $n\in \mathbb{Z}$
Substituting $n=...,-2,-1,0,1,2,...$ etc., we get:
⇒ $A=...,-855{}^\circ,-675{}^\circ,-495{}^\circ -315{}^\circ,-135{}^\circ,45{}^\circ,225{}^\circ,405{}^\circ,585{}^\circ,765{}^\circ,...$

Note: The question can also be solved by using the fact that $\tan 45{}^\circ =1$ and that the period of $\tan \theta$ is $\pi$, i.e. $\tan \theta =\tan (n\pi +\theta ),n\in \mathbb{Z}$.
The value of $\sin \theta$ and $\cos \theta$ lies between -1 and 1.
$\sin (-\theta )=-\sin \theta$ and $\cos (-\theta )=\cos \theta$.

Trigonometric Ratios for Allied Angles:
$\sin \left( -\theta \right)=-\sin \theta$ $\cos \left( -\theta \right)=\cos \theta$
$\sin \left( 2n\pi +\theta \right)=\sin \theta$ $\cos \left( 2n\pi +\theta \right)=\cos \theta$
$\sin \left( n\pi +\theta \right)={{\left( -1 \right)}^{n}}\sin \theta$ $\cos \left( n\pi +\theta \right)={{\left( -1 \right)}^{n}}\cos \theta$
$\sin \left[ (2n+1)\dfrac{\pi }{2}+\theta \right]={{(-1)}^{n}}\cos \theta$ $\cos \left[ (2n+1)\dfrac{\pi }{2}+\theta \right]={{(-1)}^{n}}\left( -\sin \theta \right)$