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June 3rd, 2015, 06:05 PM   #1
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relationship between tangent, sine and cosine

Hi, how can I approach this problem? Also, could you tell me what results do you obtain?
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June 3rd, 2015, 06:28 PM   #2
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Find $\tan x$ - perhaps you'll find it easier to write $t = \tan x$ and then solve for $t$.

Once you have $\tan x$, you can picture a right-angled triangle to determine $\sin x$ and $\cos x$.

My answers aren't pretty at all.
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Last edited by v8archie; June 3rd, 2015 at 06:33 PM.
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June 4th, 2015, 07:24 AM   #3
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Quote:
Originally Posted by v8archie View Post
Find $\tan x$ - perhaps you'll find it easier to write $t = \tan x$ and then solve for $t$.

Once you have $\tan x$, you can picture a right-angled triangle to determine $\sin x$ and $\cos x$.

My answers aren't pretty at all.

Thanx! I got sinx= 2sqrt(39)/13 and cosx=sqrt(13)/13.
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June 4th, 2015, 09:17 AM   #4
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$\displaystyle \cos(x)=\frac{\sqrt3}{2},\sin(x)=\frac12$
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June 4th, 2015, 10:14 AM   #5
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$\displaystyle \dfrac{1+\tan(x)}{1-\tan(x)}=\tan(x+45^\circ)=2+\sqrt3$

$\displaystyle x+45^\circ=\tan^{-1}(2+\sqrt3)$

$\displaystyle x=\tan^{-1}(2+\sqrt3)-45^\circ$

$\displaystyle \sin(x)=\sin(\tan^{-1}(2+\sqrt3)-45^\circ)$

$\displaystyle \sin(x)=\frac{\sqrt2}{2}\left(\dfrac{2+\sqrt3}{ \sqrt{8+4\sqrt3}}-\dfrac{1}{\sqrt{8+4\sqrt3}}\right)=\frac{\sqrt2}{2 }\left(\frac{1+\sqrt3}{2\sqrt{2+\sqrt3}}\right)= \frac{\sqrt2}{2}\cdot\frac12\sqrt{\frac{4+2\sqrt3} {2+ \sqrt3}}=\frac{\sqrt2}{2}\cdot\frac{\sqrt2}{2}= \frac12$

$\displaystyle \Rightarrow\sin(x)=\frac12,\cos(x)=\frac{\sqrt3}{2 }$.

Last edited by greg1313; June 4th, 2015 at 10:21 AM.
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June 4th, 2015, 12:43 PM   #6
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Note that tan(45° + 30°) = (1 + 1/√3)/(1 - 1/√3) = 2 + √3.
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June 4th, 2015, 02:29 PM   #7
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Quote:
Originally Posted by matisolla View Post
I got sinx= 2sqrt(39)/13 and cosx=sqrt(13)/13.
Note $\displaystyle t = \tan x$ and solve for $\displaystyle t$

I got $\displaystyle t = \dfrac{\sqrt3}{3}
\\\;\\
So,\ \ \tan x = \dfrac{\sqrt3}{3}
\\\;\\
x=30^o
\\\;\\
\sin 30^o = \dfrac{1}{2}
\\\;\\
\cos 30^o = \dfrac{\sqrt3}{2} $
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