Solving trigonometric equations

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as we know that (sin(x))^2 + (cos(x))^2=1
how about (sin(3x))^2 + (cos(3x))^2 and 5sin(3x)^2 + 6cos(3x)^2??

how can we solve these problems??
thanx
 
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What are you trying to solve? You could substitute numbers for the x's and compute the value.

I don't believe there's a nifty identity for (sin(3x))^2 + (cos(3x))^2 even though it does look somewhat similar to (sin(x))^2 + (cos(x))^2.
 
Actually, aslong as the arguments match it equals one. So (sin(3x))^2 + (cos(3x))^2 does equal one.
 
Indeed, if we apply the identity \sin^{2}\theta +\cos^{2}\theta = 1,to the above expression and simply use the substitution 3x = \theta... :wink:
 
For 5sin(3x)^2 + 6cos(3x)^2, write it as 5sin^3(3x)+ 5cos^2(3x)+ cos^2(3x)= 5(sin^3(3x)+ cos^2(3x))+ cos^2(3x)= 5+ cos^2(3x).
 
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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