Question

Prove that (a) `sin(x-(3pi)/2)=cosx` and (b) `sin(theta-pi/3)+cos(theta-pi/6)=sintheta`?

Answer 1

Verified the above relations below.

Explanation:

(a) LHS
= `sin(x-(3pi)/2)`

= `sinxcos((3pi)/2)-cosxsin((3pi)/2)`

As `cos((3pi)/2)=0` and `sin((3pi)/2)=-1`

Substituting we have

= `sinx xx0-cosx xx(-1)`

= `0+cosx`

= `cosx` =RHS

(b) `LHS=sin(theta-pi/3)+cos(theta-pi/6)`

Expanding using addition formula

`sin (A-B)=sinA cosB - cosA sinB` and
`cos (A-B)=cosA cosB + sinA SinB`

LHS
= `sinthetacos(pi/3)-costhetasin(pi/3)+costhetacos(pi/6)+sinthetasin(pi/6)`

As `cos(pi/6)=sin(pi/3)=sqrt3/2` and `cos(pi/3)=sin (pi/6)=1/2`

Therefore LHS=`sinthetaxx1/2-costhetaxxsqrt3/2+costhetaxxsqrt3/2+sinthetaxx1/2`

= `sintheta(1/2+1/2)`

= `sintheta`

= RHS