6sin2(theta)cos2(theta)
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Originally Posted by dno 6sin2(theta)cos2(theta) You should know that 2 sin A cos A = sin(2A) ....
Originally Posted by mr fantastic You should know that 2 sin A cos A = sin(2A) .... yeah I know you've got to use that but I dont know how to apply it to that equation
Originally Posted by dno yeah I know you've got to use that but I dont know how to apply it to that equation Just to make it clearer: $\displaystyle \sin{(2n\theta)} = 2\sin{n\theta}\cos{n\theta}$ where n is a real number. For your case: $\displaystyle 6\sin{2\theta}\cos{2\theta} = 3\cdot \underbrace{2\sin{2\theta}\cos{2\theta}}_{= ?} = \ldots$
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