Question

sin(x+3y) + sin(3x+y) ÷ sin2x + sin2y = 2cos(x+y)

Answer 1

Answer on Question #62166 – Math – Trigonometry

Question

$$(\sin(x+3y) + \sin(3x+y)) \div (\sin 2x + \sin 2y) = 2\cos(x+y)$$

Solution

We shall use the following formulas:

$$\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cdot \cos \frac{\alpha - \beta}{2}, \quad (1)$$ $$\sin(2\alpha) = 2 \sin(\alpha) \cos(\alpha), \quad (2)$$ $$\cos(-\alpha) = \cos(\alpha). \quad (3)$$

Then

$$\begin{aligned} & (\sin(x+3y) + \sin(3x+y)) = |(1)| = 2 \sin \frac{x+3y+3x+y}{2} \cdot \cos \frac{x+3y-3x-y}{2} = \\ & = 2 \sin(2x+2y) \cdot \cos(-x+y) = |(3)| = 2 \sin(2x+2y) \cdot \cos(x-y) = |(2)| = \\ & = 4 \sin(x+y) \cos(x+y) \cos(x-y); \end{aligned}$$ $$\sin(2x) + \sin(2y) = |(1)| = 2 \sin \frac{2x+2y}{2} \cdot \cos \frac{2x-2y}{2} = 2 \sin(x+y) \cdot \cos(x-y).$$

Thus,

$$\frac{\sin(x+3y) + \sin(3x+y)}{\sin(2x) + \sin(2y)} = \frac{4 \sin(x+y) \cos(x+y) \cos(x-y)}{2 \sin(x+y) \cdot \cos(x-y)} = 2 \cos(x+y).$$

Q.E.D.

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