Question

cos3tita+3costitasin2tita=x‚sin3tita+3sintitacos2tita=y

Answer 1

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Answer on Question #80410 – Math – Trigonometry

Question

If $\cos^3\theta + 3\cos\theta\sin^2\theta = x, \sin^3\theta + 3\sin\theta\cos^2\theta = y$, then prove

$$(x + y)^{2/3} + (x - y)^{2/3} = 2$$

Solution

$$\begin{aligned} x + y &= \cos^3\theta + 3\cos\theta\sin^2\theta + \sin^3\theta + 3\sin\theta\cos^2\theta \\ &= \cos^3\theta + \sin^3\theta + 3\sin\theta\cos\theta(\cos\theta + \sin\theta) \\ &= (\cos\theta + \sin\theta)(\cos^2\theta - \sin\theta\cos\theta + \sin^2\theta) + 3\sin\theta\cos\theta(\cos\theta + \sin\theta) \\ &= (\cos\theta + \sin\theta)(\cos^2\theta + 2\sin\theta\cos\theta + \sin^2\theta) = (\cos\theta + \sin\theta)^3 \end{aligned}$$ $$\begin{aligned} x - y &= \cos^3\theta + 3\cos\theta\sin^2\theta - \sin^3\theta - 3\sin\theta\cos^2\theta \\ &= \cos^3\theta - \sin^3\theta + 3\sin\theta\cos\theta(\sin\theta - \cos\theta) \\ &= (\cos\theta - \sin\theta)(\cos^2\theta + \sin\theta\cos\theta + \sin^2\theta) - 3\sin\theta\cos\theta(\cos\theta - \sin\theta) \\ &= (\cos\theta - \sin\theta)(\cos^2\theta - 2\sin\theta\cos\theta + \sin^2\theta) = (\cos\theta - \sin\theta)^3 \end{aligned}$$ $$\begin{aligned} (x + y)^{2/3} + (x - y)^{2/3} &= ((\cos\theta + \sin\theta)^3)^{2/3} + ((\cos\theta - \sin\theta)^3)^{2/3} \\ &= (\cos\theta + \sin\theta)^2 + (\cos\theta - \sin\theta)^2 \\ &= \cos^2\theta + 2\sin\theta\cos\theta + \sin^2\theta + \cos^2\theta - 2\sin\theta\cos\theta + \sin^2\theta \\ &= 2(\cos^2\theta + \sin^2\theta) = 2(1) = 2, \theta \in R. \end{aligned}$$