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please help it's just prove that...


[tex] \frac{1}{2} (sin7 \alpha - sin3 \alpha ) = cos5 \alpha sin2 \alpha [/tex]
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Sagot :

Nayefx

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

  • sin(-x) = -sin(x)

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,

  • LSH=RSH

and we are done!

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