# New trigo-system

#### dhiab

Solve :
$$\displaystyle sin(x)cos(y)=\frac{1}{4}$$
$$\displaystyle 3Tan(x)=Tan(y)$$

#### dwsmith

MHF Hall of Honor
Solve :
$$\displaystyle sin(x)cos(y)=\frac{1}{4}$$
$$\displaystyle 3Tan(x)=Tan(y)$$
Infinitely many solutions.

#### simplependulum

MHF Hall of Honor
$$\displaystyle \sin(x) \cos(y) = \frac{1}{4}$$

$$\displaystyle 3\tan(x) = \tan(y)$$ or

$$\displaystyle 3\sin(x)\cos(y) = \sin(y)\cos(x) = \frac{3}{4}$$

we have

$$\displaystyle \sin(y)\cos(x) = \frac{3}{4}$$ and

$$\displaystyle \sin(x) \cos(y) = \frac{1}{4}$$

$$\displaystyle \sin(x+y) = \sin(x) \cos(y) + sin(y)\cos(x) = 1$$

$$\displaystyle \sin(y-x) = sin(y)\cos(x) - \sin(x) \cos(y) = \frac{1}{2}$$

$$\displaystyle x+y = \frac{4n+1}{2} \pi$$

$$\displaystyle y-x = m\pi + (-1)^m \frac{\pi}{6}$$

Therefore

$$\displaystyle x = \frac{1}{2}(\frac{4n+1}{2} \pi - m\pi - (-1)^m \frac{\pi}{6} )$$

$$\displaystyle y = \frac{1}{2}(\frac{4n+1}{2} \pi + m\pi + (-1)^m \frac{\pi}{6} )$$

• dhiab

#### Soroban

MHF Hall of Honor
Hello, dhiab!

I will assume that $$\displaystyle x$$ and $$\displaystyle y$$ are acute angles.
You can generalize them later.

Solve: . $$\displaystyle \begin{array}{cccc} \sin x\cos y &=& \frac{1}{4} &  \\ 3\tan x&=& \tan y &  \end{array}$$

From , we have: .$$\displaystyle 3\frac{\sin x}{\cos x} \:=\:\frac{\sin y}{\cos y} \quad\Rightarrow\quad \cos x\sin y \:=\:3\sin x\cos x$$

Substitute : . $$\displaystyle \cos x\sin y \:=\:3\left(\tfrac{1}{4}\right)$$

. . $$\displaystyle \begin{array}{cccccc}\text{We have:} & \cos x\sin y &=& \frac{3}{4} \\ \\[-3mm] \text{Add :} & \sin x\cos y &=& \frac{1}{4} \end{array}$$

And we have: .$$\displaystyle \sin x\cos y + \cos x \sin y \:=\:\frac{3}{4} + \frac{1}{4} \quad\Rightarrow\quad \sin(x+y) \:=\:1 \quad\Rightarrow\quad x + y \:=\:\frac{\pi}{2}$$

. . $$\displaystyle x$$ and $$\displaystyle y$$ are complementary . .$$\displaystyle \Rightarrow\quad \cos y \:=\:\sin x$$

Substitute into : . $$\displaystyle \sin x\cdot\sin x \:=\:\frac{1}{4} \quad\Rightarrow\quad \sin x \:=\:\frac{1}{2}$$

Therefore: . $$\displaystyle x \:=\:\frac{\pi}{6},\;\;y \:=\:\frac{\pi}{3}$$