Question

3 sin-4 cos=0 then 3cos+4sin =?

Answer 1

Rewrite

4sinθ+3cosθ4sinθ=5=5−3cosθ.(1)

Square (1), yields

16sin2θ=25−30cosθ+9cos2θ.(2)

Use identity sin2θ=1−cos2θ and substitute to (2).

16(1−cos2θ)25cos2θ−30cosθ+9(5cosθ−3)2cosθ=25−30cosθ+9cos2θ=0=0=35.

Consequently, sinθ=45. Thus,