Pramod's question at Yahoo Answers regarding angle sum/difference identities

[MarkFL]

Here is the question:

If an angle α be divided into two parts such that the tangent of one ......?


If an angle α be divided into two parts such that the tangent of one part is m times the tangent of the other then prove that their difference β is obtained from the equation :

sin β = [ ( m - 1 ) / ( m + 1 ) ] sin α.

I have posted a link there to this thread so the OP may view my work.

 

[MarkFL]

Hello Pramod,

Let's divide the angle $\alpha$ into the two angles $\alpha_1$ and $\alpha_2$. Hence:

\(\displaystyle \alpha=\alpha+1+\alpha_2\)

And let's define:

\(\displaystyle \beta=\alpha_2-\alpha_1\)

\(\displaystyle \tan\left(\alpha_2 \right)=m\tan\left(\alpha_1 \right)\)

which implies:

\(\displaystyle \sin\left(\alpha_2 \right)\cos\left(\alpha_1 \right)=m\sin\left(\alpha_1 \right)\cos\left(\alpha_2 \right)\)

Now, let's then state:

\(\displaystyle \sin(\beta)=k\sin(\alpha)\)

Using the definitions, we may write:

\(\displaystyle \sin\left(\alpha_2-\alpha_1 \right)=k\sin\left(\alpha+1+\alpha_2 \right)\)

Using the angle sum and difference identities for sine, we obtain:

\(\displaystyle \sin\left(\alpha_2 \right)\cos\left(\alpha_1 \right)-\cos\left(\alpha_2 \right)\sin\left(\alpha_1 \right)=k\sin\left(\alpha_2 \right)\cos\left(\alpha_1 \right)+k\cos\left(\alpha_2 \right)\sin\left(\alpha_1 \right)\)

\(\displaystyle (1-k)\sin\left(\alpha_2 \right)\cos\left(\alpha_1 \right)=(1+k)\sin\left(\alpha_1 \right)\cos\left(\alpha_2 \right)\)

Using \(\displaystyle \sin\left(\alpha_2 \right)\cos\left(\alpha_1 \right)=m\sin\left(\alpha_1 \right)\cos\left(\alpha_2 \right)\) we obtain:

\(\displaystyle m(1-k)=1+k\)

\(\displaystyle m-mk=1+k\)

\(\displaystyle m-1=k(m+1)\)

\(\displaystyle k=\frac{m-1}{m+1}\)

Hence, we may state:

\(\displaystyle \sin(\beta)=\frac{m-1}{m+1}\sin(\alpha)\)