Question

2 sin 75° cos15° = {(2+√3)
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Answer 1

Answer:

Sin75°=sin(45°+30°)

=Sin45°cos30°+cos45°sin30°

= >[Sin(A+B)=sinA cosB+cosA sinB]

=1/√2.√3/2+1/√2.1/2 [sin45°=cos45°=1/√2]

=√3+1/2√2 Ans.

Cos15°=cos(45°–30°)

=> [Cos(A-B)=cosAcosB+sinAsinB]

Cos(45°-30°)=cos45°cos30°+sin45°sin30°

=1/√2. √3/2+1/√2. 1/2

[sin30°=1/2 & cos30°=√3/2]

=√3/2√2+1/2√2

=√3+1/2√2

Now,

2 sin 75° cos15° = 2*√3+1/2√2

Answer 2

Answer:

Step-by-step explanation:

2 sin75 cos15=sin(75+15)+sin(75-15)

                       =sin90+sin60

                         =1+$\frac{\sqrt{3} }{2}$

                          =$\frac{2+\sqrt{3} }{2}$