The value of cos 15^circ - sin 15^circ is equ | vedclass

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(A) We know that $\cos 	heta - \sin 	heta = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos 	heta - \frac{1}{\sqrt{2}} \sin 	heta ight) = \sqrt{2} \cos(

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$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(A) We know that $\cos 	heta - \sin 	heta = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos 	heta - \frac{1}{\sqrt{2}} \sin 	heta ight) = \sqrt{2} \cos(	heta + 45^\circ)$.Substituting $	heta = 15^\circ$:$\cos 15^\circ - \sin 15^\circ = \sqrt{2} \cos(15^\circ + 45^\circ)$$= \sqrt{2} \cos 60^\circ$$= \sqrt{2} 	imes \frac{1}{2} = \frac{1}{\sqrt{2}}$.

The value of $\cos 15^\circ - \sin 15^\circ$ is equal to

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