Question

Answer plch !!!! !!!​

Answer 1

Answer:

It can be easily done by

Sin (45-30)

Root3/2-1/root2

=(root3-1)/(2-root 2)

And CosA= root of 1+ain square A

Similarly we can get tge values of sin15 degree and cos 15 degrees .

Hope it helps u

Mark it as BRAINLIEAST please i request

Answer 2

Explanation:

$\sin(A-B) = \sin A \cos B - \cos A \sin B \\\sin(15°) \\ = \sin(45°-30°) \\ = \sin 45° \cos 30° - \cos 45° \sin 30° \\ = \frac{1}{ \sqrt{2} } \times \frac{ \sqrt{3} }{2} - \frac{1}{ \sqrt{2} } \times \frac{1}{2} \\ \boxed{\sin 15°= \frac{ \sqrt{3} - 1}{2 \sqrt{2} } } \\ \cos(A-B) = \cos A \cos B+\sin A \sin B \\ \cos 15° \\ = \cos(45°-30°) \\ = \cos 45° \cos 30°+\sin 45° \sin 30° \\ = \frac{1}{ \sqrt{2} } \times \frac{ \sqrt{3} }{2} + \frac{1}{ \sqrt{2} } \times \frac{1}{2} \\ \boxed{\cos15° = \frac{ \sqrt{3} + 1}{2 \sqrt{2} } } \\ \\ \red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}$