Question

sin^2 6x - sin^2 4x = sin 2x sin 10x
pls prove it

Answer 1

=(sin6x+sin4x)(sin6x-sin4x)
sin6x=sin(5x+x)=sin5xcosx+cos5xsinx
sin4x=sin(5x-x)=sin5xcosx-cos5xsinx
from that we got= 2sin5xcosx(2cos5xsinx)
=2sin5xcos5x(2sinxcosx)=sin10xsin2x

Answer 2

thank u infauzan

Answer 3

You can solve this problem in this way;
You can see that the left side is a difference of squares that always returns the product:
a^2 - b^2 = (a-b)(a+b)
(sin 6x)^2 - (sin 4x)^2 = (sin 6x - sin 4x)(sin 6x + sin 4x)
We'll transform the algebraic sums into products:
sin 6x - sin 4x = 2 cos [(6x+4x)/2]*sin [(6x-4x)/2]
sin 6x - sin 4x = 2 cos(5x)*sin (x)
sin 6x + sin 4x = 2 sin [(6x+4x)/2]*cos [(6x-4x)/2]
sin 6x + sin 4x = 2 sin (5x)*cos (x)
(sin 6x - sin 4x)(sin 6x + sin 4x) = 2 cos(5x)*sin (5x)*2cos (x)*sin (x)
We'll recall the double angle formula:
(sin 6x - sin 4x)(sin 6x + sin 4x) = [sin 2*(5x)]*[sin 2(x)]
(sin 6x)^2 - (sin 4x)^2 =(sin 6x - sin 4x)(sin 6x + sin 4x) =(sin 10 x)*(sin 2x) q.e.d.

Answer 4

thank u for ur concern, giorgia

Answer 5

Better to findmore ways to solve any problem