Question

Cos15 - sin15=1/under root 2

Answer 1

hope this answer helps you

Answer 2

Answer and Explanation:

To prove : $\cos 15^\circ-\sin 15^\circ=\frac{1}{\sqrt{2}}$

Proof :

Taking LHS,

$LHS=\cos 15^\circ-\sin 15^\circ$

We can write,

$LHS=\cos (45-30)^\circ-\sin (45-30)^\circ$

Applying formula,

$\cos (A-B)=\cos A\cos B+\sin A\sin B$

$\sin (A-B)=\sin A\cos B-\cos A\sin B$

$LHS=\cos 45\cos 30+\sin 45\sin 30-(\sin 45\cos 30-\cos 45\sin 30)$

We know,

$\cos 45=\frac{1}{\sqrt{2}}~~~~~~~~~\sin 45=\frac{1}{\sqrt{2}}\\\\\cos 30=\frac{\sqrt{3}}{2}~~~~~~~~~\sin 30=\frac{1}{2}$

Substituting the values,

$LHS=(\frac{1}{\sqrt{2}})(\frac{\sqrt{3}}{2})+(\frac{1}{\sqrt{2}})(\frac{1}{2})-((\frac{1}{\sqrt{2}})(\frac{\sqrt{3}}{2})-(\frac{1}{\sqrt{2}})(\frac{1}{2}))$

$LHS=(\frac{\sqrt3+1}{2\sqrt2})-(\frac{\sqrt3-1}{2\sqrt2})$

$LHS=\frac{3-1}{2\sqrt2}$

$LHS=\frac{2}{2\sqrt2}$

$LHS=\frac{1}{\sqrt2}$

$LHS=RHS$

Hence proved.