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  4. Find the value of cos 15°- sin 15° ?

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Q. Find the value of $\cos 15^{\circ}-\sin 15^{\circ}$ ?

Trigonometry

Solution:

$\cos 15^{\circ} =\cos (45^{\circ}-30^{\circ})$
$ =\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ}$
$ =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}=\frac{\sqrt{3}+1}{2 \sqrt{2}} $
$\sin 15^{\circ} =\sin (45^{\circ}-30^{\circ}) $
$ =\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ} $
$ =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
$\text { So, } \cos 15^{\circ}-\sin 15^{\circ}$
$ =\frac{\sqrt{3}+1}{2 \sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}=\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}}$