Question

Express (sin 135°- i cos 135°) in polar form​

Answer 1

Step-by-step explanation: Firstly change sin135° and cos135°

   sin(180°-45°) =sin45° = $\frac{1}{\sqrt{2} }$

  cos(180°-45°) =  -cos45°  =  -$\frac{1}{\sqrt{2} }$

  Z   = $\frac{1}{\sqrt{2} }$   - i $\frac{1}{\sqrt{2} }$       , here x is positive and  y  is negitive   θ = π - α

   tanθ  =   $\frac{y}{x}$   ⇒   θ = $1$  ⇒θ  =π /4       ⇒     θ  =   π   -  π/4  ⇒   θ = 3π / 4

  polar form ,  z   =  r (cosθ  + i sinθ )

                           =  1  ( cos 3π/4 + i sin3π/4)      (  r =$\sqrt{x^{2} +y2^{} }$ )

                                                                             ( modulus of complex number)