prove [sin(x) - sin(3x)] / [cos(x) + cos(3x)] = -tan(x)

[baselramjet]

[sin(x) - sin(3x)] / [cos(x) + cos(3x)] = -tan(x)

 

[soroban]

Heloo, baselramjet!

You may be expected to know some sum-to-product identities:

. . \(\displaystyle \sin A\,-\,\sin B\:=\:2\cdot\cos\left(\frac{A\,+\,B}{2}\right)\cdot\sin\left(\frac{A\,-\,B}{2}\right)\)

. . \(\displaystyle \cos A\,+\,\cos B\:=\:2\cdot\cos\left(\frac{A\,+\,B}{2}\right)\cdot\cos\left(\frac{A\,-\,B}{2}\right)\)

\(\displaystyle \L\frac{\sin x\,-\,\sin 3x}{\cos x\,+\,\cos3x}\:=\:-\tan x\)

The numerator is: \(\displaystyle \:\sin x\,-\,\sin 3x \;=\;2\cdot\cos\left(\frac{x\,+\,3x}{2}\right)\cdot\sin\left(\frac{x\,-\,3x}{2}\right) \;=\;2\cdot\cos\left(\frac{4x}{2}\right)\cdot\sin\left(\frac{-2x}{2}\right)\)

. . \(\displaystyle =\;2\cdot\cos 2x\cdot\sin(-x) \;=\;-2\cdot\cos2x\cdot\sin x\)


The denominator is: \(\displaystyle \:\cos x\,+\,\cos3x\;=\;2\cdot\cos\left(\frac{x\,+\,3x}{2}\right)\cdot\cos\left(x\,-\,3x}{2}\right) \;=\;2\cdot\cos\left(\frac{4x}{2}\right)\cdot\cos\left(\frac{-2x}{2}\right)\)

. . \(\displaystyle = \;2\cdot\cos 2x\cdot\cos(-x)\;=\;2\cdot\cos2x\cdot cos x\)


The problem becomes: \(\displaystyle \L\:\frac{\sin x\,-\,\sin3x}{\cos x\,+\,\cos3x}\;=\;\frac{-2\cdot\cos2x\cdot\sin x}{2\cdot\cos2x\cdot\cos x}\;=\;\frac{-\sin x}{\cos x}\)\(\displaystyle \;= \;-\tan x\)

 

[baselramjet]

Ok,

So in the example

[sin(x) - sin(3x)] / [cos(x) - cos(3x)] = -cot2x

the sum-product identity for

[cos(x) - cos(3x)] would be -2sin(x+y/2)*sin(x-y/2)

So in that example the numerator would be the same (as the previous example) and the denominator would be:

cosx-cos3x = -2sin(x+3x/2)*sin(x-3x/2)=-2sin(4x/2)*sin(-2x/2) = -2*sin2x*sin(-x) = -2*sin2x*sinx

Am I correct?

 

[soroban]

Hello, baselramjet!

So in the example: \(\displaystyle \L\:\frac{\sin(x)\,-\,\sin(3x)}{\cos(x)\,-\,\cos(3x)}\)\(\displaystyle \: = \;-\cot(2x)\)

The sum-product identity for: \(\displaystyle \,\cos(x)\,-\,\cos(y)\)
. . would be: \(\displaystyle \,-2\cdot\sin\left(\frac{x+y}{2}\right)\cdot\sin\left(\frac{x-y}{2}\right)\;\;\) Right!

So in that example the numerator would be the same as the previous example
and the denominator would be:

\(\displaystyle \cos(x)\,-\,\cos(3x) \:=\:-2\cdot\sin\left(\frac{x+3x}{2}\right)\cdot\sin\left(\frac{x-3x}{2}\right) \:=\:-2\cdot\sin\left(\frac{4x}{2}\right)\cdot\sin\left(\frac{-2x}{2}\right)\)

. . \(\displaystyle = \:-2\cdot\sin(2x)\cdot\sin(-x)\: = \:\)+\(\displaystyle 2\cdot\sin(2x)\cdot\sin(x)\)

Am I correct? . Yes!

 

[baselramjet]

Thank you Soroban! You guys in this forum are helping me understand this better than my teacher can explain it!!!

Thanks!

Ashley...