3.Trigonometrical Ratios, Functions and Identities
easy

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

A

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

B

$\frac{1}{2}$

C

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

D

$0$

Solution

(a) $\cos {15^o} – \sin {15^o} = \sqrt 2 \,.\cos \,({45^o} + {15^o}) $

$= \sqrt 2 \,.\,\cos \,\,{60^o}$

$ = \sqrt 2 \,.\frac{1}{2} = \frac{1}{{\sqrt 2 }}$.

Standard 11
Mathematics

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