New trigo-system

May 2009
596
31
ALGERIA
Solve :
\(\displaystyle
sin(x)cos(y)=\frac{1}{4}
\)
\(\displaystyle
3Tan(x)=Tan(y)
\)
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
\(\displaystyle \sin(x) \cos(y) = \frac{1}{4} \)

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

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

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

MHF Hall of Honor
May 2006
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6,341
Lexington, MA (USA)
Hello, dhiab!

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


Solve: . \(\displaystyle \begin{array}{cccc}
\sin x\cos y &=& \frac{1}{4} & [1] \\ 3\tan x&=& \tan y & [2] \end{array}\)

From [2], 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 [1]: . \(\displaystyle \cos x\sin y \:=\:3\left(\tfrac{1}{4}\right)\)


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


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 [1]: . \(\displaystyle \sin x\cdot\sin x \:=\:\frac{1}{4} \quad\Rightarrow\quad \sin x \:=\:\frac{1}{2}\)


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