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

Oct 7, 2016

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

=$2 \sin \frac{2 \theta}{\sin} \left(2 \theta\right)$

$= 2$

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

Oct 7, 2016

See below.

#### Explanation:

According to de Moivre's identity

$\sin \theta = \frac{{e}^{i \theta} - {e}^{- i \theta}}{2 i}$ and
$\cos \theta = \frac{{e}^{i \theta} + {e}^{- i \theta}}{2}$

so following

$\frac{{a}^{3} - {b}^{3}}{a - b} = {a}^{2} + a b + {b}^{2}$

$\frac{{e}^{i 3 \theta} - {e}^{- i 3 \theta}}{{e}^{i \theta} - {e}^{- i \theta}} = {e}^{i 2 \theta} + 1 + {e}^{- i 2 \theta}$

and

$\frac{{e}^{i 3 \theta} + {e}^{- i 3 \theta}}{{e}^{i \theta} + {e}^{- i \theta}} = {e}^{i 2 \theta} - 1 + {e}^{- i 2 \theta}$

so

$\frac{{e}^{i 3 \theta} - {e}^{- i 3 \theta}}{{e}^{i \theta} - {e}^{- i \theta}} - \frac{{e}^{i 3 \theta} + {e}^{- i 3 \theta}}{{e}^{i \theta} + {e}^{- i \theta}} = 2$