Question #21f8d

2 Answers
Nghi N
Aug 7, 2017

Apply the trig identity:
#sin a + sin b = 2sin ((a + b)/2).cos ((a - b)/2)#
In this case:
#sin x + sin 3x = 2sin (2x).cos x#
From trig identity: sin 2a = 2sin a.cos a, we get:
#sin x + sin 3x = 4 sin x.cos x. cos x = 4sin x.cos^2 x#

Tazwar Sikder
Aug 8, 2017

See explanation below.

Explanation:

We have: #sin(x) + sin(3 x)#

#= sin(x) + sin(2 x + x)#

Let's apply the compound angle identity for #sin(x)#:

#= sin(x) + sin(2 x) cos(x) + cos(2 x) sin(x)#

Then, let's apply the double angle identities for #sin(x)# and #cos(x)#:

#= sin(x) + 2 sin(x) cos(x) cdot cos(x) + (1 - 2 sin^(2)(x)) cdot sin(x)#

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

One of the Pythagorean identities is #cos^(2)(x) + sin^(2)(x) = 1#.

We can rearrange it to get:

#Rightarrow cos^(2)(x) = 1 - sin^(2)(x)#

Let's apply this rearranged identity to our proof:

#= 2 sin(x) + 2 sin(x) cdot (1 - sin^(2)(x)) - 2 sin^(3)(x)#

#= 2 sin(x) + 2 sin(x) - 2 sin^(3)(x) - 2 sin^(3)(x)#

#= 4 sin(x) - 4 sin^(3)(x)#

#= 4 sin(x) cdot (1 - sin^(2)(x))#

Let's apply the rearranged identity again:

#= 4 sin(x) cdot cos^(2)(x)#

#= 4 sin(x) cos^(2)(x) " " " " "Q.E.D."#

Related questions
Impact of this question
305 views around the world
You can reuse this answer
Creative Commons License