Prove Trigonometric Statement: Ideas & Solutions

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Have you tried mathematical induction?
 
Metaleer said:
Have you tried mathematical induction?

Hi thanks for the reply,

No, don't know why I didn't think of that,

Will give it a go now.
 
I don't think this is to be done with mathematical induction. I ended up with these massive identities and really long formulas,

Is there another way?
 
I managed to do it using

<br /> \cos a \sin b = (\sin (a+b) - \sin(a-b))/2<br />

PROBLEM SOLVED!
 
Have u proven the P1 case and related Pk with Pk+1?

Do u think the identity below will help ? :wink:

sin(x)cos[(2k+1)x]
=\frac{sin[((2k+2)x)-((2k)x)]}{2} + \frac{cos[((2k+2)x)+((2k)x)]}{2}
=\frac{1}{2}[sin((2k+2)x)-sin((2k)x))]
 
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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