Question

If cos theta-sintheta=root2sinthetathen prove that costheta+sintheta=root2costheta

Answer 1

Answer:

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Answer 2

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

Step-by-step explanation:

We have,

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

Prove that, $\cos \theta+\sin \theta=\sqrt{2}\cos \theta$

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

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

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

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

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

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

Using the trigonometric identity,

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

⇒ $2(1)=2\sin^2 \theta+x^2$

⇒ $2=2\sin^2 \theta+x^2$

⇒$x^2 =2-2\sin^2 \theta=2(1-\sin^2 \theta)$

Using the trigonometric identity,

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

⇒  $x^2 =2\cos^2 \theta=(\sqrt{2} \cos \theta)^2$

⇒ $x=\sqrt{2} \cos \theta$, proved.

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