i have been trying to crack this but i have had no luck. hope someone can help. $\displaystyle (sin4x + sin2x)/(cos4x+cos2x) = tan3x$ the X varibale is actually (pie), i just could not type it into the window...
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Originally Posted by sigma1 i have been trying to crack this but i have had no luck. hope someone can help. $\displaystyle (sin4x + sin2x)/(cos4x+cos2x) = tan3x$ the X varibale is actually (pie), i just could not type it into the window... What exactly are you trying to crack? $\displaystyle \sin(n \pi) = 0$ for $\displaystyle n = \pm 1$,$\displaystyle \pm 2$, ... $\displaystyle \tan(n \pi) = 0$ for $\displaystyle n = \pm 1$,$\displaystyle \pm 2$, ... Hence both sides = 0.
Oh yeah and the Latex for pi (not pie lol) is... \pi $\displaystyle \pi$
the actual question is prove that $\displaystyle (sin4A + sin2A)/(cos4A+cos2A) = tan3A $
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