Answer:
see below
Step-by-step explanation:
we would like to prove the following equality:
[tex] \dfrac{1}{2} ( \sin(7\alpha) - \sin(3 \alpha) ) = \cos(5 \alpha )\sin(2 \alpha )[/tex]
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,
[tex] \displaystyle\sin( x ) \cos( x ) = \frac{1}{2} ( \sin( x + y ) + \sin( x- y) )[/tex]
Notice that This identity can be utilised in the RSH.So,
Assign variables:
[tex]x \implies 2 \alpha \\ y \implies5 \alpha [/tex]
thus rewriting the RSH yields:
[tex] \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 ) )[/tex]
simplify:
[tex] \implies \displaystyle\dfrac{1}{2} ( \sin(7\alpha) - \sin(3 \alpha) ) \stackrel {?}{ = } \frac{1}{2} ( \sin( 7 \alpha ) + \sin( - 3 \alpha ) )[/tex]
Call to mind that
hence,
[tex] \implies \displaystyle\dfrac{1}{2} ( \sin(7\alpha) - \sin(3 \alpha) ) \stackrel { \checkmark}{ = } \frac{1}{2} ( \sin( 7 \alpha ) - \sin( 3 \alpha ) )[/tex]
therefore,
and we are done!
