Trigonometry identity sin(pi)cos(wpi)+cos(pi)sin(wpi)

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Homework Statement



Hi guys, How can sin(∏)cos(ω∏)-cos(∏)sin(ω∏) = sin(ω∏)? please guide me trigonometry identity to apply with this?

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The Attempt at a Solution

sin_x_pi.jpg
 
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sin(pi)=0, cos(pi)=(-1). So, of course, sin(∏)cos(ω∏)-cos(∏)sin(ω∏) = sin(ω∏). I'm not sure what that has to do with what follows.
 
Dick said:
sin(pi)=0, cos(pi)=(-1). So, of course, sin(∏)cos(ω∏)-cos(∏)sin(ω∏) = sin(ω∏). I'm not sure what that has to do with what follows.

Ohhh :shy: Just been thinking about using trig identity, thank you Dick !
 

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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