Trignometric Equations,SERIOUS HELP NEEDED?

Question

Given;
A = 2sin30°, B = 3cos30°
Solve;
A + B = asinb where b is an angle and a a constant.

Answer 1

All you need to do is rewrite 2sin30 + 3cos30 into a sin b right?

Use the trigonometric identity sin(x + y) = sinxcosy + cosxsiny.

A + B = asinb
2sin30 + 3cos30 = asinb

Hence, 2 = cosy, 3 = siny

You need to do this since sin and cos have restrictions from -1 to 1.
2^2 + 3^2 = hypotenuse^2
4 + 9 = hypotenuse^2
sqrt 13 = hypotenuse.

2sin30 + 3cos30 = asinb

Divide by sqrt 13

2 / sqrt 13 (sin30) + 3 /sqrt 13 cos30 = a/ sqrt 13 sin b

the angle is approximately 33.69 degrees.

So:

Adding that, you have:

1 / sqrt 13 sin ( 30 + 33.69)
sqrt 13 / 13 sin (63.69)

Answer 2

A+B = 2 sin 30 +3 cos 30 = a*sin(30+t) = a sin 30 cos t+a cos 30 sin t
a cos t =2
a sin t=3 square and sum
a=sqrt(13)
tant = 3/2 so t=56.31
A+B= sqrt(13) * sin (86.31)