Question

if sin(A-B)=sinAcosB-cosAsinB and cos(A-B)=cosAcosB+sinAsinB, find value of sin 15 degree and cos 15 degree.

Answer 1

hello users.....

we have to find the value of 
sin 15° and cos 15° = ? 

solution:-
we know that
sin(A-B)=sinAcosB-cosAsinB
and
 cos(A-B)=cosAcosB+sinAsinB

now,
(1) sin 15° = sin ( 45° - 30°)
= sin 45°cos 30° - cos 45° sin 30°
=1/√2 × √3/2 - 1/√2 × 1/2
= √3 / 2√2 - 1 / 2√2
= ( √3 - 1) / 2√2 answer 

(2) cos 15° = cos(45° - 30°)
= cos 45° cos 30° + sin 45° sin 30°
= 1 / 2√2 × √3 / 2  + 1 / √2 × 1 / 2
= √3 / 2√2 + 1 / 2√2 
= ( √3 + 1) / 2√2 answer

✰✰ hope it helps ✰✰ 

Answer 2

sin(A-B)=sinAcosB-cosAsinB

sin15° can be written as

sin(45°-30°)

now,

=sin45°cos30°-cos45°sin30°

=1/√2×√3/2-1/√2×1/2

=√3/2√2-1/2√2

By taking LCM

=√3-1/2√2

Now,

cos15° can be written as

cos(45°-30°)

=cos45°cos30°+sin45°sin30°

=1/√2×√3/2+1/√2×1/2

=√3/2+1/2√2

by taking LCM we get

=√3+1/2√2