Finding the Exact Value of sin[2arcsin(3/5)] with Inverse Trig Functions

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Homework Statement


Find the exact value of the expression:


Homework Equations


sin[2arcsin(3/5)]


The Attempt at a Solution


I know you're supposed to use sin2x=2sinxcosx somehow but not sure how to start.
 
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page13 said:

Homework Statement


Find the exact value of the expression:


Homework Equations


sin[2arcsin(3/5)]


The Attempt at a Solution


I know you're supposed to use sin2x=2sinxcosx somehow but not sure how to start.

Start by letting u = arcsin(3/5). Then your expression is sin(2u) = ?

It will be helpful to draw a right triangle where u is one of the acute angles. Label the sides and hypotenuse so that sin(u) = 3/5.
 
Ah, OK! 24/25 correct?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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