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Find an angle x where sin x = cos x

Guest Mar 18, 2015

Best Answer 

 #4
avatar+92623 
+5

WOW heureka, you gave yourself something to chew on there!    

My answer is no where near as impressive.   

 

sinx=cosx          When cosx=0,    sinx ≠0     so    cosx ≠ 0  therefore I can divide by it.

 

$$\\\frac{sinx}{cosx}=1\\\\
tanx=1\\\\
$tan is pos in the 1st an 3rd quad$\\\\
x=\frac{\pi}{4},\;\;\frac{5\pi}{4}\;\;etc\\\\
x=n\pi+\frac{\pi}{4}\qquad n\in Z\qquad$ (n is an integer)$$

Melody  Mar 18, 2015
 #1
avatar
+5

sinx = cos x @45 degrees

Guest Mar 18, 2015
 #2
avatar+86859 
+5

This will occur at 45° ± n180°  where n is an integer

 

  

CPhill  Mar 18, 2015
 #3
avatar+19480 
+5

Find an angle x where sin x = cos x

 

$$\cos(x)-\sin(x)=0 \\\\
a\cdot \cos(x) + b\cdot \sin(x) = c \qquad | \qquad a=1 \qquad b=-1 \qquad c= 0\\
a\cdot \cos(x) + b\cdot \sin(x) = c \quad | \quad : a \\
\cos(x) + \frac{b}{a} \cdot \sin(x) = \frac{c}{a} \quad | \quad c = 0 \\\\
\cos(x) + \frac{b}{a} \cdot \sin(x) = 0 \\
\small{\text{
$
\text{We set } \tan{(\varepsilon)} = \frac{b}{a} \quad \text{ we have } a = 1 \text{ and } b = -1 \text{ so } \varepsilon = \arctan{(-1)} = -\frac{\pi}{4}
$
}}\\\\
\cos(x) + \frac{b}{a} \cdot \sin(x) = 0 \quad | \quad \tan{(\varepsilon)} = \frac{b}{a}\\\\
\cos(x) + \tan{(\varepsilon)}\cdot \sin(x) = 0 \\
\cos(x) + \frac{ \sin{(\varepsilon)} } { \cos{(\varepsilon)} } \cdot \sin(x) = 0 \quad | \quad \cdot \cos{(\varepsilon)} \\
\small{\text{
$
\cos{ (\varepsilon)} \cdot \cos(x) + \sin{(\varepsilon)} \cdot \sin(x) = 0 \quad | \quad \cos{ (x-\varepsilon )} = \cos{ (\varepsilon)} \cdot \cos(x) + \sin{(\varepsilon)} \cdot \sin(x)
$}}\\
\cos{ (x-\varepsilon )} =0 \quad | \quad \pm \arccos{}\\
x-\varepsilon = \pm \arccos{(0)} = \pm \frac{\pi}{2}\\
x-\varepsilon = \pm \frac{\pi}{2} \\
x= \varepsilon \pm \frac{\pi}{2} \quad | \quad \varepsilon = -\frac{\pi}{4}\\
x= -\frac{\pi}{4} \pm \frac{\pi}{2} \\\\
x_1= -\frac{\pi}{4} +\frac{\pi}{2} = \frac{\pi}{4} \\
x_1= 45\ensurement{^{\circ}} \pm k\cdot 360\ensurement{^{\circ}}$$

$$\\x_2= -\frac{\pi}{4} -\frac{\pi}{2}
= -\frac{3}{4} \cdot \pi
= -\frac{3}{4} \cdot \pi + 2\pi
= \frac{5}{4} \cdot\pi \\
x_2 = 225\ensurement{^{\circ}} \pm k\cdot 360\ensurement{^{\circ}}$$

$$k=0,1,2\cdots$$

heureka  Mar 18, 2015
 #4
avatar+92623 
+5
Best Answer

WOW heureka, you gave yourself something to chew on there!    

My answer is no where near as impressive.   

 

sinx=cosx          When cosx=0,    sinx ≠0     so    cosx ≠ 0  therefore I can divide by it.

 

$$\\\frac{sinx}{cosx}=1\\\\
tanx=1\\\\
$tan is pos in the 1st an 3rd quad$\\\\
x=\frac{\pi}{4},\;\;\frac{5\pi}{4}\;\;etc\\\\
x=n\pi+\frac{\pi}{4}\qquad n\in Z\qquad$ (n is an integer)$$

Melody  Mar 18, 2015

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