Question

How to solve equations of the form 4sinAcosA=1

Answer 1

Answer on Question #59781 - Math - Trigonometry

Question

How to solve equations of the form 4sinAcosA=1

Solution

It is known that

2sinAcosA = sin2A;

$$\sin \left(\frac {\pi}{6}\right) = \frac {1}{2};$$ $$\sin \left(\frac {5 \pi}{6}\right) = \frac {1}{2};$$ $$\arcsin \left(\frac {1}{2}\right) = \frac {\pi}{6},$$

where $\arcsin(x)$ is the inverse sine function.

Besides, the period of the sine function is $2\pi$, hence $\sin(x) = \sin(x + 2\pi) = \sin(x + 2\pi n)$, where $n$ is integer.

Next,

4sinAcosA = 1;

2 · (2sinAcosA) = 1;

2sin2A = 1;

sin2A = 1/2;

$2A = \frac{\pi}{6} + 2\pi n$ and $2A = \frac{5\pi}{6} + 2\pi n$, where $n$ is integer;

$A = \frac{\pi}{12} + \pi n$ and $A = \frac{5\pi}{12} + \pi n$, where $n$ is integer.

More general form of solution is

$2A = (-1)^{k}\arcsin\left(\frac{1}{2}\right) + k\pi$, where $k$ is integer;

$2A = (-1)^{k} \frac{\pi}{6} + k\pi$, where $k$ is integer;

$A = (-1)^{k} \frac{\pi}{12} + \frac{k\pi}{2}$, where $k$ is integer.

The equation 4sinAcosA=1 has only four roots in the range $[0; 2\pi]$:

$\frac{\pi}{12}$ (or $15{}^{\circ}$), $\frac{5\pi}{12}$ (or $75{}^{\circ}$), $\frac{13\pi}{12}$ (or $195{}^{\circ}$), $\frac{17\pi}{12}$ (or $255{}^{\circ}$).

Answer: $A = (-1)^{k} \frac{\pi}{12} + \frac{k\pi}{2}$, where $k$ is integer.

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