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

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The discussion revolves around proving the relationship between the angles when an angle α is divided into two parts, where the tangent of one part is m times that of the other. The key equation derived is sin β = [(m - 1) / (m + 1)] sin α, which expresses the sine of the difference between the two angles in terms of the sine of the original angle. The proof utilizes the angle sum and difference identities for sine, leading to the conclusion that the relationship holds true under the given conditions. The mathematical derivation involves manipulating trigonometric identities and establishing a connection between the angles involved. This provides a clear understanding of the angle sum/difference identities in trigonometry.
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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.
 
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Hello Pramod,

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

$$\alpha=\alpha+1+\alpha_2$$

And let's define:

$$\beta=\alpha_2-\alpha_1$$

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

which implies:

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

Now, let's then state:

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

Using the definitions, we may write:

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

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

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

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

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

$$m(1-k)=1+k$$

$$m-mk=1+k$$

$$m-1=k(m+1)$$

$$k=\frac{m-1}{m+1}$$

Hence, we may state:

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