Question

​Prove that cos x + sin(3pi/2 + x)-sin(3pi/2 - x) + cos(pi + x)=0

Answer 1

Answer:

Step-by-step explanation:

pi value ?

Answer 2

Answer:

We have

L.H.S.=cosx+sin(3π2+x)−sin(3π2−x)+cos(π+x)

As sin(3π2+x)=sin(3π2−x)=−cosx

and cos(π+x)=−cosx,

hence we get

L.H.S.=cosx−cosx−(−cos−cosx=cosx−cosx+cosx−cosx=0 =R.H.S.

[Proved]