Question

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

Answer 1

Hint: 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) $