How sin(90°+ θ) is equal to M'P'/OP' or Cos θ? I'm learning Trigonometry right now with myself and at current I'm understanding how to find the trigonometric ratio of the angle (90°+ θ) in those of θ. I'm little bit confused right now in the Figure, I didn't get how angle sin(90°+ θ) is equal to M'P'/OP'.And how sin(90°+θ)=sin θ of triangle P'OM']1 
[]2Please help. Sorry If I'm asking foolish question. Thankyou in advance.
Just another way we know $sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$ so here $a=\theta,b=90$ thus $$\sin(\theta+90)=\sin(\theta)\cos(90)+\cos(\theta)\sin(90)=\cos(\theta)$$
Just another way we know $sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$ so here $a=\theta,b=90$ thus $$\sin(\theta+90)=\sin(\theta)\cos(90)+\cos(\theta)\sin(90)=\cos(\theta)$$