Write 8sin3t9cos3t in the form A sin Bt using sum or differe

Write 8sin(3t)+9cos(3t) in the form A sin (Bt+) using sum or difference formulas.

Solution

-8sin(3t) + 9cos(3t) = (9^2 + 8^2)[ (-8/(9^2 + 8^2))sin 3t + (9/(9^2 + 8^2))cos 3t ]
= 145 * [ (-8/145)sin 3t + (9/145)cos 3t ]
and we choose :
A = 145
= artan(-9/8) <--- artan is an odd function
then :
sin() = 9/145 and
cos() = -8/145
so the get :
-8sin(3t) + 9cos(3t) = A [ cos() *sin 3t + sin() * cos 3t ]
= A sin( 3t + )

conclusion :
A = 145
B =3
= artan(-9/8) # -48.4° = 311.6 degree