Question

How do you prove `sin(3theta)/sintheta-cos(3theta)/costheta=2`?

Answer 1

Starting with the LHS subtract using the rule for subtraction of fractions, and then simplify using Trig identities.

Explanation:

`(sin(3theta)cos(theta)-sin(theta)cos(3theta))/(sin(theta)cos(theta)`

`=(sin(3theta-theta))/(1/2xx2sinthetacostheta`

=`2sin(2theta)/sin(2theta)`

`=2`

with problems like these recognition of the basic Trig. identities is vital.

Answer 2

See below.

Explanation:

According to de Moivre's identity

`sintheta = (e^(itheta)-e^(-itheta))/(2i)` and
`costheta = (e^(itheta)+e^(-itheta))/(2)`

so following

`(a^3-b^3)/(a-b)=a^2+a b + b^2`

`(e^(i3theta)-e^(-i3theta))/ (e^(itheta)-e^(-itheta)) = e^(i2theta)+1+e^(-i2theta)`

and

`(e^(i3theta)+e^(-i3theta))/ (e^(itheta)+e^(-itheta)) = e^(i2theta)-1+e^(-i2theta)`

so

`(e^(i3theta)-e^(-i3theta))/ (e^(itheta)-e^(-itheta)) - (e^(i3theta)+e^(-i3theta))/ (e^(itheta)+e^(-itheta)) =2`