$\sin^26x-\sin^24x=\sin2x\sin10x$
Please help me to solve the above trigonometric function as I am trying to solve this question since last an hour.
$\sin^26x-\sin^24x=\sin2x\sin10x$
Please help me to solve the above trigonometric function as I am trying to solve this question since last an hour.
$\sin^26x-\sin^24x=\sin(6x+4x)\sin(6x-4x)=\sin2x\sin10x$ applying $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$
Alternatively,
$\sin^26x-\sin^24x=\frac{1}{2}(2\sin^26x-2\sin^24x)$ $=\frac{1}{2}(1-\cos12x-(1-\cos8x))$ as $\cos2A=1-2sin^2A$
So, $\sin^26x-\sin^24x=\frac{\cos8x-\cos12x}{2}=\sin10x\sin2x$ applying $\cos C - \cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}=2\sin\frac{C+D}{2}\sin\frac{D-C}{2}$
$sin^26x-sin^24x=(sin6x+sin4x)(sin6x-sin4x)$
,if you know the Difference product formula,
$sinα+sinβ=2sin(\frac{α+β}{2})·cos(\frac{α-β}{2}),sinα-sinβ=2cos(\frac{α+β}{2})·sin(\frac{α-β}{2})$
$sin6x+sin4x=2sin(5x)cos(x),sin6x-sin4x=2cos(5x)sin(x)$
so,$sin^26x-sin^24x=2sin(5x)cos(5x)*2sin(x)cos(x)=sin(10x)sin(2x)$