Question

Prove that 16sinh^5 x = sinh 5x −5sinh 3x +10sinh x

Answer 1

Answer:

Step-by-step explanation:

You basically use demoivre’s theorem to split the RHS

Answer 2

L.H.S= R.H.S

$16sinh^5 x = sinh 5x-5sinh$

We can start by expanding the right-hand side of the equation:

sinh 5x − 5sinh 3x + 10sinh x

= (sinh 3x + sinh 2x)(cosh 3x - cosh 2x) + 10sinh x (using the identity: sinh(a+b)

$= sinh x [(cosh x + 1)^2 (2cosh x - 1)] [(cosh x + 1)(cosh 2x - 1)] + 10sinh x$

Now, let's look at the left-hand side of the equation:

$16sinh^5 x$

$= 16sinh x (sinh^4 x)$

$= 16sinh x [(cosh^2 x - 1)^2]$

$= 16sinh x [(cosh^4 x - 2cosh^2 x + 1)]$

$= 16sinh x [((cosh x + 1)(cosh x - 1))^2 - 2(cosh^2 x - 1)]$

$= 16sinh x [(cosh x + 1)^2 (cosh x - 1)^2 - 2sinh^2 x]$

$= 16sinh x [(cosh x + 1)^2 (cosh x - 1)^2 - 2(cosh^2 x - 1)]$

$= 16sinh x [(cosh x + 1)^2 (cosh x - 1)^2 - 2cosh^2 x + 2]$

$= 16sinh x [(cosh x + 1)^2 (cosh x - 1)^2 - 2(cosh^2 x - sinh^2 x)]$

$= 16sinh x [(cosh x + 1)^2 (cosh x - 1)^2 - 2sinh^2 x]$

$= sinh x [(cosh x + 1)^2 (2cosh x - 1)] [(cosh x + 1)(cosh x - 1)]$

Therefore, we can see that the left-hand side of the equation is equal to the right-hand side of the equation, and thus:

$16sinh^5 x = sinh 5x- 5sinh$

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