Question

prove that sin 60° cos 30° + Cos 60° sin 30° i= 1​

Answer 1

Step-by-step explanation:

Sin(A+B) = SinA CosB + cos A sinB

A = 60° , B = 30°

sin(90°)= sin60°cos 30° + cos 60° sin30°

1 = sin60° cos30° + cos60° sin30°

Answer 2

General results:-)

$sin60 \degree = \frac{ \sqrt{3} }{2} \\ \\ cos 30 \degree = \frac{ \sqrt{3} }{2} \\ \\sin 30\degree=\frac{1}{2}\\ \\cos 60\degree=\frac{1}{2}$

LHS:-

$sin60 \degree \: cos30 \degree + cos60 \degree \: sin30 \degree \\ \\ = \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2} + \frac{1}{2} \times \frac{1}{2} \\ \\ = \frac{3}{4} + \frac{1}{4} \\ \\ = \frac{3 + 1}{4} \\ \\ = \frac{4}{4} \\ \\ = 1$

= RHS.

Hence, the given result is proved.

OR

It can be solved using identity of sin(A + B)as

sin(A + B) = sinA cosB + cosA sinB.

in above question A = 60°

and

B = 30°

So,

LHS

sin 60° cos 30° + Cos 60° sin 30°

= sin(60 + 30)

= sin(90°)

= 1 = RHS.