Math Doubts

Simplify $\frac{\sin(n+1)\alpha -\sin(n-1)\alpha}{\cos(n+1)\alpha +2\cos n\alpha +\cos(n-1)\alpha}$

$\alpha$ and $n \alpha$ are two angles and they formed compound angles $(n+1)\alpha$ and $(n-1)\alpha$ by sum and difference. Sine and cosine functions formed a fractional function to represent a quantity in general trigonometric form.

$\dfrac{\sin(n+1)\alpha -\sin(n-1)\alpha}{\cos(n+1)\alpha +2\cos n\alpha +\cos(n-1)\alpha}$

Step: 1

The two sine functions with compound angles are in subtraction form. The subtraction of them can be simplified by the sum to product transformation trigonometric identity.

$= \dfrac{2\cos \Bigg[\dfrac{(n+1)\alpha + (n-1)\alpha}{2}\Bigg] \sin \Bigg[\dfrac{(n+1)\alpha -(n-1)\alpha}{2}\Bigg]}{\cos(n+1)\alpha +\cos(n-1)\alpha +2\cos n\alpha}$

Step: 2

Similarly, express the sum of cosine functions in terms of product form by the sum to product transformation trigonometric identity.

$= \dfrac{2\cos \Bigg[\dfrac{(n+1)\alpha + (n-1)\alpha}{2}\Bigg] \sin \Bigg[\dfrac{(n+1)\alpha -(n-1)\alpha}{2}\Bigg]}{2\cos \Bigg[\dfrac{(n+1)\alpha + (n-1)\alpha}{2}\Bigg] \cos \Bigg[\dfrac{(n+1)\alpha -(n-1)\alpha}{2}\Bigg] +2\cos n\alpha}$

$= \dfrac{2\cos \Bigg[\dfrac{n\alpha +\alpha + n\alpha -\alpha}{2}\Bigg] \sin \Bigg[\dfrac{n\alpha +\alpha -n\alpha +\alpha}{2}\Bigg]}{2\cos \Bigg[\dfrac{n\alpha +\alpha + n\alpha -\alpha}{2}\Bigg] \cos \Bigg[\dfrac{n\alpha +\alpha -n\alpha +\alpha}{2}\Bigg] +2\cos n\alpha}$

$\require{cancel} = \dfrac{2\cos \Bigg[\dfrac{n\alpha +\cancel{\alpha} + n\alpha -\cancel{\alpha}}{2}\Bigg] \sin \Bigg[\dfrac{\cancel{n\alpha} +\alpha -\cancel{n\alpha} +\alpha}{2}\Bigg]}{2\cos \Bigg[\dfrac{n\alpha +\cancel{\alpha} + n\alpha -\cancel{\alpha}}{2}\Bigg] \cos \Bigg[\dfrac{\cancel{n\alpha }+\alpha -\cancel{n\alpha} +\alpha}{2}\Bigg] +2\cos n\alpha}$

$= \dfrac{2\cos \Bigg[\dfrac{n\alpha +n\alpha}{2}\Bigg] \sin \Bigg[\dfrac{\alpha +\alpha}{2}\Bigg]}{2\cos \Bigg[\dfrac{n\alpha + n\alpha}{2}\Bigg] \cos \Bigg[\dfrac{\alpha +\alpha}{2}\Bigg] +2\cos n\alpha}$

$= \dfrac{2\cos \Bigg[\dfrac{2n\alpha}{2}\Bigg] \sin \Bigg[\dfrac{2\alpha}{2}\Bigg]}{2\cos \Bigg[\dfrac{2n\alpha}{2}\Bigg] \cos \Bigg[\dfrac{2\alpha}{2}\Bigg] +2\cos n\alpha}$

$\require{cancel} = \dfrac{2\cos \Bigg[\dfrac{\cancel{2}n\alpha}{\cancel{2}}\Bigg] \sin \Bigg[\dfrac{\cancel{2}\alpha}{\cancel{2}}\Bigg]}{2\cos \Bigg[\dfrac{\cancel{2}n\alpha}{\cancel{2}}\Bigg] \cos \Bigg[\dfrac{\cancel{2}\alpha}{\cancel{2}}\Bigg] +2\cos n\alpha}$

$= \dfrac{2\cos n\alpha \sin \alpha}{2\cos n\alpha \cos \alpha +2\cos n\alpha}$

Step: 3

$\cos n \alpha$ is a common term in expression of the denominator. Take it common from them to simplify it further.

$= \dfrac{2\cos n\alpha \sin \alpha}{2\cos n\alpha (\cos \alpha +1)}$

$\require{cancel} = \dfrac{\cancel{2\cos n\alpha} \sin \alpha}{\cancel{2\cos n\alpha} (\cos \alpha +1)}$

$= \dfrac{\sin \alpha}{\cos \alpha +1}$

$= \dfrac{\sin \alpha}{1+\cos \alpha}$

Step: 4

Expand the sine of angle alpha and also express $1+cos \alpha$.

$= \dfrac{2\sin \Big(\dfrac{\alpha}{2}\Big)\cos \Big(\dfrac{\alpha}{2}\Big) }{2cos^2 \Big(\dfrac{\alpha}{2}\Big)}$

$\require{cancel} = \dfrac{\cancel{2}\sin \Big(\dfrac{\alpha}{2}\Big) \cancel{\cos \Big(\dfrac{\alpha}{2}\Big)} }{\cancel{2cos^2 \Big(\dfrac{\alpha}{2}\Big)}}$

$= \dfrac{\sin \Big(\dfrac{\alpha}{2}\Big)}{cos \Big(\dfrac{\alpha}{2}\Big)}$

$= \tan \Big(\dfrac{\alpha}{2}\Big)$

Therefore, it is the required solution for this trigonometric problem.

Latest Math Topics
Latest Math Problems
Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved