Question

If (cos0 + sin0) =√2sin0, prove that (sin0 - cos0) = √2cos0.​

Answer 1

proved! xyztrigonometryxyz

Answer 2

$\sin \theta-\cos \theta=\sqrt{2} \cos \theta$, proved.

Step-by-step explanation:

We have,

$\cos \theta + \sin \theta =\sqrt{2}\sin \theta$            ............ (1)

Let $\sin \theta-\cos \theta=x$                 ............ (2)

To prove that, $\sin \theta-\cos \theta=\sqrt{2} \cos \theta$.  

Squaring and adding (1) and (2), we get

$(\cos \theta + \sin \theta)^2+(\sin \theta-\cos \theta)^2=(\sqrt{2}\sin \theta)^2+x^2$

⇒ $(\cos^2 \theta + \sin^2 \theta+2\sin \theta\cos \theta)+(\cos^2 \theta + \sin^2 \theta-2\sin \theta\cos \theta)=2\sin^2 \theta+x^2$

⇒ $\cos^2 \theta + \sin^2 \theta+\cos^2 \theta + \sin^2 \theta=2\sin^2 \theta+x^2$

⇒ $2(\cos^2 \theta + \sin^2 \theta)=2\sin^2 \theta+x^2$

Using the algebraic identity,

$\cos^2 A + \sin^2 A =1$

⇒ $2(1)=2\sin^2 \theta+x^2$

⇒ $2=2\sin^2 \theta+x^2$

⇒ $x^2=2-2\sin^2 \theta$

⇒ $x^2=2(1-\sin^2 \theta)$

⇒ $x^2=2\cos^2 \theta$

⇒ x = $\sqrt{2} \cos \theta$

$\sin \theta-\cos \theta=\sqrt{2} \cos \theta$, proved.

Thus, $\sin \theta-\cos \theta=\sqrt{2} \cos \theta$, proved.