Solve 4sin(7x)+3cos(7x)=Asin(7x+B)?

Question

Solve this problem for A and B. show steps.

4sin(7x)+3cos(7x)=Asin(7x+B)

Answer 1

LET 4 = r cos t
3 = r sin t
square and add t = 5

t = tan ^1- 3/4

4 sin 7x + 3 cos 7 x = 5 sin 7x cos t + 5 cos 7x sin t
= 5 sin (7x+t)

comapring with A (7x +B)

A = 5 and B= t or B= tan^-1 3/4

Answer 2

Asin(7x+B) = 4sin(7x)+3cos(7x) ...(1)

While
Asin(7x+B) = Asin7x cosB + Acos 7x sinB ...(2)

and comparing both coefficient of sin7x and cos7x of both (1) & (2)

Therefore A cosB = 4 ...(3)
and AsinB = 3 ...(4)

Squaring both (3) and (4) and add it.
A^2 = 25 or A = +/- 5

Then divide (4) by (3)
tanB = 3/4
B = 36.87 (in degree)

Answer 3

4sin(7x)+3cos(7x)=Asin(7x+B)
or, 4/5sin(7x) + 3/5cos(7x) = A/5sin(7x+B)
or, sin (7x + C) = A/5sin(7x+B) [ Let 4/5 = cos C so, sinC should be 3/5 right?]
Now, comparing both sides we get A/5 =1 or, A =5
and, B = C where C = arcsin (3/5) = B

Answer 4

B = arctan(3/4) = 36.870°
A = √(16 + 9) = 5