Question

Please help it's just prove that...

Answer 1

Answer:

see below

Step-by-step explanation:

we would like to prove the following equality:

$\dfrac{1}{2} ( \sin(7\alpha) - \sin(3 \alpha) ) = \cos(5 \alpha )\sin(2 \alpha )$

well, In order to prove the equality, we can consider utilizing the sum and product identities of trigonometry.

we can prove this equality either from LHS or RHS.recall that,

$\displaystyle\sin( x ) \cos( x ) = \frac{1}{2} ( \sin( x + y ) + \sin( x- y) )$

Notice that This identity can be utilised in the RSH.So,

Assign variables:

$x \implies 2 \alpha \\ y \implies5 \alpha$

thus rewriting the RSH yields:

$\implies \displaystyle\dfrac{1}{2} ( \sin(7\alpha) - \sin(3 \alpha) ) \stackrel {?}{ = } \frac{1}{2} ( \sin( 2 \alpha + 5 \alpha ) + \sin( 2 \alpha - 5 \alpha ) )$

simplify:

$\implies \displaystyle\dfrac{1}{2} ( \sin(7\alpha) - \sin(3 \alpha) ) \stackrel {?}{ = } \frac{1}{2} ( \sin( 7 \alpha ) + \sin( - 3 \alpha ) )$

Call to mind that

• sin(-x) = -sin(x)

hence,

$\implies \displaystyle\dfrac{1}{2} ( \sin(7\alpha) - \sin(3 \alpha) ) \stackrel { \checkmark}{ = } \frac{1}{2} ( \sin( 7 \alpha ) - \sin( 3 \alpha ) )$

therefore,

• LSH=RSH

and we are done!