My Math Forum relationship between tangent, sine and cosine

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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 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 Math Focus: Mainly analysis and algebra 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. Thanks from matisolla Last edited by v8archie; June 3rd, 2015 at 06:33 PM.
June 4th, 2015, 07:24 AM   #3
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Quote:
 Originally Posted by v8archie 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.

 June 4th, 2015, 09:17 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,932 Thanks: 1127 Math Focus: Elementary mathematics and beyond $\displaystyle \cos(x)=\frac{\sqrt3}{2},\sin(x)=\frac12$
 June 4th, 2015, 10:14 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,932 Thanks: 1127 Math Focus: Elementary mathematics and beyond $\displaystyle \dfrac{1+\tan(x)}{1-\tan(x)}=\tan(x+45^\circ)=2+\sqrt3$ $\displaystyle x+45^\circ=\tan^{-1}(2+\sqrt3)$ $\displaystyle x=\tan^{-1}(2+\sqrt3)-45^\circ$ $\displaystyle \sin(x)=\sin(\tan^{-1}(2+\sqrt3)-45^\circ)$ $\displaystyle \sin(x)=\frac{\sqrt2}{2}\left(\dfrac{2+\sqrt3}{ \sqrt{8+4\sqrt3}}-\dfrac{1}{\sqrt{8+4\sqrt3}}\right)=\frac{\sqrt2}{2 }\left(\frac{1+\sqrt3}{2\sqrt{2+\sqrt3}}\right)= \frac{\sqrt2}{2}\cdot\frac12\sqrt{\frac{4+2\sqrt3} {2+ \sqrt3}}=\frac{\sqrt2}{2}\cdot\frac{\sqrt2}{2}= \frac12$ $\displaystyle \Rightarrow\sin(x)=\frac12,\cos(x)=\frac{\sqrt3}{2 }$. Last edited by greg1313; June 4th, 2015 at 10:21 AM.
 June 4th, 2015, 12:43 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,475 Thanks: 2039 Note that tan(45° + 30°) = (1 + 1/√3)/(1 - 1/√3) = 2 + √3.
June 4th, 2015, 02:29 PM   #7
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Quote:
 Originally Posted by matisolla 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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