Question

sinx+sin2x+sin3x+sin4x/cosx+cos2x-cos3x-cos4x=cotx​

Answer 1

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Answer 2

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First let’s look at some trig identities.

sin(A+B)=sinAcosB+cosAsinB, sin(A-B)=sinAcosB-cosAsinB,

Therefore sin(A+B)+sin(A-B)=2sinAcosB.

cos(A+B)=cosAcosB-sinAsinB, cos(A-B)=cosAcosB+sinAsinB,

Therefore cos(A-B)-cos(A+B)=2sinAsinB.

sin(x)+sin(2x)+sin(3x)+sin(4x) is the same as:

(sin(x)+sin(3x))+(sin(2x)+sin(4x)).

If A-B=x and A+B=3x, then 2A=4x, A=2x and B=x, and:

sin(3x)+sin(x)=2sin(2x)cos(x).

If A-B=2x and A+B=4x, then 2A=6x, A=3x and B=x, and:

sin(4x)+sin(2x)=2sin(3x)cos(x).

So, putting these together:

sin(x)+sin(2x)+sin(3x)+sin(4x)=2cos(x)(sin(2x)+sin(3x)).

Similarly, we can write the denominator as:

(cos(x)-cos(3x))+(cos(2x)-cos(4x)).

Let A-B=x and A+B=3x (as before) then:

cos(x)-cos(3x)=2sin(2x)sin(x).

And let A-B=2x and A+B=4x, then:

cos(2x)-cos(4x)=2sin(3x)sin(x).

Putting these together:

cos(x)+cos(2x)-cos(3x)-cos(4x)=2sin(x)(sin(2x)+sin(3x)).

When we divide we are left with

$\bf \: \frac{cos(x)}{sin(x)} =cot(x)$

QED, because the term 2(sin(2x)+sin(3x)) cancels out.