Question

sin2A + sin4A = cosA + cos3A

Answer 1

Answer on Question #70274 – Math – Trigonometry

Question

$$\sin(2A) + \sin(4A) = \cos(A) + \cos(3A)$$

Solution

Method 1

$$\sin(2A) - \cos(A) = \cos(3A) - \sin(4A)$$ $$-2\sin(-2A/2 - A/2 + \pi/4) \sin(-2A/2 + A/2 + \pi/4) = -(-2\sin(-4A/2 - 3A/2 + \pi/4) \sin(-4A/2 + 3A/2 + \pi/4))$$ $$\sin(-3A/2 + \pi/4) \sin(-A/2 + \pi/4) = -\sin(-7A/2 + \pi/4) \sin(-A/2 + \pi/4)$$ $$\sin(-A/2 + \pi/4) (\sin(-3A/2 + \pi/4) + \sin(-7A/2 + \pi/4)) = 0$$

1) $\sin(-A/2 + \pi/4) = 0$

$$-A/2 + \pi/4 = \pi n$$ $$A = 2(\pi/4 - \pi n) = \pi/2 - 2\pi n = (1 - 4n)\pi/2$$

2) $\sin(-3A/2 + \pi/4) + \sin(-7A/2 + \pi/4) = 0$

$$2\sin((-3A/2 + \pi/4 - 7A/2 + \pi/4)/2) \cos((-3A/2 + \pi/4 + 7A/2 - \pi/4)/2) = 0$$ $$2\sin(-5A/2 + \pi/4) \cos(A) = 0$$ $$-5A/2 + \pi/4 = n\pi \text{ or } A = \pi/2 + n\pi$$ $$A = 2/5(\pi/4 - n\pi) \text{ or } A = \pi/2 + n\pi$$ $$A = \pi/10 - 2n\pi/5 \text{ or } A = \pi/2 + n\pi$$ $$A = (1 - 4n)\pi/10 \text{ or } A = (2n+1)\pi/2$$

Method 2

$$y(A) = \sin(2A) + \sin(4A)$$$$y (A) = \cos (A) + \cos (3 A)$$

LHS:

$$\begin{array}{l} \sin (2 A) + \sin (4 A) = \sin (2 A) + \sin 2 (2 A) = \\ = \sin (2 A) + 2 \sin (2 A) \cos (2 A) = \sin (2 A) (1 + 2 \cos (2 A)) = 2 \sin A \cos A (1 + 2 \cos (2 A)) = \\ = 2 \cos A (\sin A + 2 \cos (2 A)) = 2 \cos A (\sin A + 2 (1 - 2 \sin^ {2} A) \sin A) = \\ = 2 \cos A (\sin A + 2 \sin A - 4 (\sin A) ^ {3}) = 2 \cos A \sin (3 A) \end{array}$$

RHS:

$$\begin{array}{l} \cos (A) + \cos (3 A) = \cos A + 4 \cos^ {3} A - 3 \cos A = \\ = 4 \cos^ {3} A - 2 \cos A = 2 \cos A (2 \cos^ {2} A - 1) = 2 \cos A \cos (2 A) \end{array}$$

LHS=RHS, so

$$2 \cos A \sin (3 A) = 2 \cos A \cos (2 A)$$ $$\text{If } \cos A = 0 \text{ then } A = - \frac {\pi}{2} + n \pi$$

If $\cos A \neq 0$ then we can divide by $\cos A$

$$\sin (3 A) = \cos (2 A) = \sin (\pi / 2 - 2 A)$$

Let's find the corresponding A:

$$\sin (2 A + A) = \cos (2 A)$$ $$\sin (2 A) \cos A + \cos (2 A) \sin A = \cos (2 A)$$ $$\begin{array}{l} 2 \sin A \cos^ {2} A + (1 - 2 \sin^ {2} A) \sin A = \cos (2 A) \\ 2 \sin A (1 - \sin^ {2} A) + \sin A - 2 \sin^ {3} A = \cos (2 A) \\ 2 \sin A - 2 \sin^ {3} A + \sin A - 2 \sin^ {3} A = \cos (2 A) \\ \end{array}$$

As you can see

$$\sin (3 A) = \cos (2 A)$$ $$3 \sin A - 4 \sin^ {3} A = 1 - 2 \sin^ {2} A$$ $$\begin{array}{l} - 4 \sin^ {3} A + 2 \sin^ {2} A + 3 \sin A - 1 = 0 \\ \sin A = t \\ - (t - 1) (4 t ^ {2} + 2 t - 1) = 0 \\ \end{array}$$ $$\begin{array}{l} D = 2 ^ {2} - 4 * 4 * (- 1) = 2 0 \\ t = \frac {- 2 \pm 2 \sqrt {5}}{2 * 4} = \frac {- 1 \pm \sqrt {5}}{4} \\ \sin A = \frac {- 1 \pm \sqrt {5}}{4} \\ \sin A = \frac {- 1 - \sqrt {5}}{4} \Rightarrow A = - \left(- \frac {3 \pi}{1 0}\right) + n \pi \text{ or } A = \left(- \frac {3 \pi}{1 0}\right) + n \pi \\ \sin A = \frac {- 1 + \sqrt {5}}{4} \Rightarrow A = - \frac {\pi}{1 0} + n \pi \text{ or } A = \frac {\pi}{1 0} + n \pi \\ \end{array}$$

Answer: $A = \frac{3\pi}{10} + n\pi$, $A = -\frac{3\pi}{10} + n\pi$, $A = -\frac{\pi}{10} + n\pi$, $A = \frac{\pi}{10} + n\pi$, $A = -\frac{\pi}{2} + n\pi$

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