Question

If sin (A-B)=sin A cos B-Oos Asin B and cos (A-B) = cos Acos B + sin Asin B, find the
values of sin 15 and cos 15°.

Help me​

Answer 1

Step-by-step explanation:

$sin \: 15 \degree = sin \: (45 \degree - 30 \degree) \\ = sin \: 45 \degree \: cos 30 \degree - cos \: 45 \degree \: sin 30 \degree \\ = \frac{1} { \sqrt{2} } \times \frac{ \sqrt{3} }{2} - \frac{1} { \sqrt{2} } \times \frac{1}{2} \\ = \frac{ \sqrt{3} }{2 \sqrt{2} } - \frac{1 }{2 \sqrt{2} } \\ = \frac{ \sqrt{ 3} - 1}{2 \sqrt{2} } \\ thus \\ \huge \red{\boxed{ sin \: 15 \degree =\frac{ \sqrt { 3} - 1}{2 \sqrt{2} } }} \\ \\ cos \: 15 \degree = cos \: (45 \degree - 30 \degree) \\ = cos \: 45 \degree \: cos 30 \degree + sin \: 45 \degree \: sin 30 \degree \\ = \frac{1} { \sqrt{2} } \times \frac{ \sqrt{3} }{2} + \frac{1} { \sqrt{2} } \times \frac{1}{2} \\ = \frac{ \sqrt{3} }{2 \sqrt{2} } + \frac{1 }{2 \sqrt{2} } \\ = \frac{ \sqrt{ 3} + 1}{2 \sqrt{2} } \\ thus \\ \huge \purple{\boxed{ cos \: 15 \degree =\frac{ \sqrt { 3} + 1}{2 \sqrt{2} } }}$