Use this formula for the numerator sin x + sin 3x.

sin p + sin q = 2 sin ((p+q)/2)*cos ((p-q)/2)

sin x + sin 3x = 2 sin ((x+3x)/2)*cos ((x-3x)/2)

sin x + sin 3x = 2 sin (2x)*cos(-2x)

The cosine function is even, therefore cos(-2x) = cos 2x

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

Use this formula for the denominator cos x - cos 3x.

cos p - cos q = -2 sin ((p+q)/2)*sin ((p-q)/2)

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

(sin x + sin 3x)/(cos x - cos 3x) = 2 sin (2x)*cos(2x)/2 sin (2x)*sin (2x)

Reduce terms: (sin x + sin 3x)/(cos x - cos 3x) =cos(2x)/sin (2x)

(sin x + sin 3x)/(cos x - cos 3x) = cot (2x)

**ANSWER: The last line proves the identity (sin x + sin 3x)/(cos x - cos 3x) = cot (2x)**

sin(A+B)= sin(A)cos(B) + cos(A)sin(B)

sin(A-B) = sin(A)cos(B) - cos(A)sin(B)

sin(A+B) + sin(A-B) = 2sin(A)cos(B)

cos(A+B) = cos(A)cos(B) - sin(A)sin(B)

cos(A-B) = cos(A)cos(B) + sin(A)sin(B)

Subtracting

cos(A-B) - cos(A+B) = 2sin(A)sin(B)

So

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

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

So

(2sin(2x)cos(x))/(2sin(2x)sin(x)) = cos(x)/sin(x) = cot(x)

So this identity is not true.

cos(A-B) + cos(A+B) = 2cos(A)cos(B)

cos(x) + cos(3x) = cos(2x - x) + cos(2x + x) = 2cos(2x)cos(x)

(sin(3x)+sin(x))/(cos(x)+cos(3x)) = tan(2x)

The given expression -cot(2x) = sin x + sin(3x) / cos x - cos(3x) is not an identity.

For example for x = `pi/6` .

`-cot(2*pi/6)` = -0.5773` `

On the other hand `sin(pi/6) + (sin (pi/2))/(cos (pi/6)) - cos (pi/2)`

=> 1.654

**The two are not equal showing that the expression is not an identity.**

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