If sin theta + cos theta = sqrt{2} cos theta, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given: $\sin 	heta + \cos 	heta = \sqrt{2} \cos 	heta$ .... $(1)$Let $x = \cos 	heta - \sin 	heta$ .... $(2)$Squaring both equations:$(\sin \

Vedclass · Accepted Answer

$\sqrt{2} \sin 	heta$

Vedclass · Answer

(B) Given: $\sin 	heta + \cos 	heta = \sqrt{2} \cos 	heta$ .... $(1)$Let $x = \cos 	heta - \sin 	heta$ .... $(2)$Squaring both equations:$(\sin 	heta + \cos 	heta)^2 = (\sqrt{2} \cos 	heta)^2 \Rightarrow \sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = 2 \cos^2 	heta$$1 + 2 \sin 	heta \cos 	heta = 2 \cos^2 	heta$ .... $(3)$$x^2 = (\cos 	heta - \sin 	heta)^2 = \cos^2 	heta + \sin^2 	heta - 2 \sin 	heta \cos 	heta$$x^2 = 1 - 2 \sin 	heta \cos 	heta$ .... $(4)$Adding $(3)$ and $(4)$:$1 + 2 \sin 	heta \cos 	heta + x^2 = 2 \cos^2 	heta + 1 - 2 \sin 	heta \cos 	heta$$1 + x^2 = 2 \cos^2 	heta + 1$$x^2 = 2 \cos^2 	heta$Wait,let's re-evaluate: From $(1)$,$\sin 	heta = (\sqrt{2} - 1) \cos 	heta$.Then $\cos 	heta - \sin 	heta = \cos 	heta - (\sqrt{2} - 1) \cos 	heta = \cos 	heta (1 - \sqrt{2} + 1) = \cos 	heta (2 - \sqrt{2}) = \sqrt{2} \cos 	heta (\sqrt{2} - 1)$.Alternatively,using the identity $(\sin 	heta + \cos 	heta)^2 + (\cos 	heta - \sin 	heta)^2 = 2(\sin^2 	heta + \cos^2 	heta) = 2$.$(\sqrt{2} \cos 	heta)^2 + x^2 = 2$$2 \cos^2 	heta + x^2 = 2$$x^2 = 2(1 - \cos^2 	heta) = 2 \sin^2 	heta$$x = \sqrt{2} \sin 	heta$.

If $\sin \theta + \cos \theta = \sqrt{2} \cos \theta$,then the value of $(\cos \theta - \sin \theta)$ is

Similar Questions

$\tan 1^\circ \tan 2^\circ \tan 3^\circ \tan 4^\circ \dots \tan 89^\circ = $

If $\alpha + \beta - \gamma = \pi,$ then $\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = $

If $f(x) = \cos^2 x + \sec^2 x$,then

If $A$ lies in the second quadrant and $3\tan A + 4 = 0,$ the value of $2\cot A - 5\cos A + \sin A$ is equal to

The radius of the circle whose arc of length $15 \ cm$ makes an angle of $3/4$ radian at the centre is ..... $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module