Question

prove that sin75° + sin15° =√6/2​

Answer 1

Answer:

We have a set of formulae to transform the sum of two sines or two cosines into the product of sines and cosines.

These are;

1. sinA + sinB = 2 sin[(A+B)/2]*cos[(A-B)/2].

2. sinA - sinB = 2 sin[(A-B)/2]*cos[(A+B)/2].

3. cosA + cosB = 2cos[(A+B)/2]*cos[(A-B)/2].

4. cosA - cosB = -2sin[(A+B)/2]*sin[(A-B)/2].

Now, coming to the question;

We can easily apply 1st formula above to find the answer of this question ....

We have;

sin 75 + sin 15

= 2 sin[(75+15)/2] * cos[(75-15)/2]

= 2 sin[90/2] * cos[60/2]

= 2 sin45 * cos30

= 2 * 1/(2^1/2) * (3^1/2)/2

= (3/2)^1/2.

i.e., root (3/2).

Step-by-step explanation:

Answer 2

Answer:

Use formula sinA + sinB = 2 sin[(A+B)/2]×cos[(A-B)/2]

Step-by-step explanation:

sin 75 + sin 15

= 2 sin[(75+15)/2] × cos[(75-15)/2]

= 2 sin[90/2] ×cos[60/2]

= 2 sin45 ×cos30

= 2 × 1/(2^1/2) ×(3^1/2)/2

= (3/2)^1/2.

=√3/2