Question

If an angle Theta is divided into two parts alpha and beta such that tan a: tan b = x: y, show that sin (alpha-beta) =


`(x - y)/(x + y) sin \: theta`
You can use the symbols of alpha beta and theta too​

Answer 1

See below for proof.

Explanation:

Trigonometric identity

`\boxed{\tan A=(\sin A)/(\cos A)}`

Given that tan α : tan β = x : y, and using the above identity, rewrite the ratio in terms of sine and cosine, and isolate sin α cos β:

`\begin{aligned}\tan \alpha : \tan \beta &= x:y\\\\(\tan \alpha)/(\tan \beta) &= (x)/(y)\\\\y\tan \alpha&=x\tan \beta\\\\(y\sin\alpha)/(\cos \alpha)&=(x\sin \beta)/(\cos \beta)\\\\y\sin\alpha\cos \beta&=x\cos \alpha\sin \beta\\\\\sin \alpha \cos \beta&=(x\cos \alpha\sin \beta)/(y)\end{aligned}`

`\hrulefill`

Trigonometric identity

`\boxed{\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B}`

Given that θ = α + β, and using the above identity and the expression from the previous calculation, write sin θ in terms of cos α, sin β, x and y:

`\begin{aligned}\sin \theta&=\sin (\alpha + \beta)\\\\\implies \sin \theta&=\sin \alpha \cos \beta +\cos \alpha \sin \beta\\\\\implies \sin \theta&=(x\cos \alpha\sin \beta)/(y)+\cos \alpha \sin \beta\\\\\implies \sin \theta&=(x\cos \alpha\sin \beta)/(y)+(y\cos \alpha \sin \beta)/(y)\\\\\implies \sin \theta&=((x+y)\cos \alpha\sin \beta)/(y)\\\\\implies (\sin \theta)/((x+y))&=(\cos \alpha\sin \beta)/(y)\end{aligned}`

`\hrulefill`

Trigonometric identity

`\boxed{\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B}`

Using the above identity, rewrite sin(α - β):

`\implies \sin(\alpha - \beta)=\sin \alpha \cos \beta-\cos \alpha \sin \beta`

Substitute the found expression for sin α cos β (from part 1):

`\implies \sin(\alpha - \beta)=(x\cos \alpha\sin \beta)/(y)-\cos \alpha \sin \beta`

Simplify:

`\implies \sin(\alpha - \beta)=(x\cos \alpha\sin \beta)/(y)-(y\cos \alpha \sin \beta)/(y)`

`\implies \sin(\alpha - \beta)=(x\cos \alpha\sin \beta-y\cos \alpha \sin \beta)/(y)`

`\implies \sin(\alpha - \beta)=((x-y)\cos \alpha\sin \beta)/(y)`

`\implies \sin(\alpha - \beta)=(x-y) \cdot (\cos \alpha\sin \beta)/(y)`

Substitute the found expression for (cos α sin β) / y (from Part 2):

`\implies \sin(\alpha - \beta)=(x-y) \cdot (\sin \theta)/((x+y))`

`\implies \sin(\alpha - \beta)=((x-y))/((x+y))\sin \theta`