Prove that Cot 4x (sin 5 x + sin 3 x) = Cot x (sin 5 x - sin 3 x) ?

2 Answers
May 29, 2018

#sin a + sin b = 2 sin( (a+b)/2 ) cos( (a-b)/2)#

#sin a - sin b = 2 sin( (a-b)/2 ) cos( (a+b)/2) #

Right side:

#cot x ( sin 5x - sin 3x) = cot x cdot 2 sin ( (5x-3x)/2) cos ( (5x+3x)/2) #

#= cos x / sin x cdot 2 sin x cos 4x = 2 cos x cos 4x#

Left side:

#cot(4x) ( sin 5x + sin 3x) =cot(4x) cdot 2 sin ( (5x+3x)/2) cos ( (5x-3x)/2)#

# = {cos 4x}/{sin 4x} cdot 2 sin 4x cos x = 2 cos x cos 4 x #

They're equal #quad sqrt#

May 29, 2018

Factor formula (Sum-to-Product and Product-to-Sum identities)

Explanation:

For this question, we can use the Sum-to-Product and Product-to-Sum identities.

I'm lazy, so here's a picture of the identities.
http://www.sosmath.com/trig/prodform/prodform.html

The product-to-sum formula above can be derived via compound angle identities.

Using the substitution #alpha = a + b# and #beta = a - b#, we can get the following product-to-sum formulas.

https://www.cliffsnotes.com/study-guides/trigonometry/trigonometric-identities/product-sum-and-sum-product-identities

So, now that we've got that sorted out, let's apply our formulas.

#cot (4x) (sin (5x) + sin (3x)) = cos(4x)/sin(4x) (2 sin((5x + 3x)/2)cos((5x - 3x)/2)) = cos(4x)/sin(4x)(2sin(4x)cos(x)) = 2cos(4x)cos(x) = cos(x)/sin(x)(2cos(4x)sin(x)) = cot(x)(sin(4x + x) - sin(4x - x)) = cot(x)(sin(5x) - sin(3x))#

Alternatively, you could also apply the sum-to-product formula on the right-hand side:

#cot(x)(sin(5x) - sin(3x)) = cos(x)/sin(x)(2 cos((5x + 3x)/2)sin((5x - 3x)/2)) = cos(x)/sin(x)(2cos(4x)sin(x)) = 2cos(4x)sin(x) = LHS.#

#QED#