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June 3rd, 2015, 06:05 PM  #1 
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  relationship between tangent, sine and cosine
Hi, how can I approach this problem? Also, could you tell me what results do you obtain?

June 3rd, 2015, 06:28 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,355 Thanks: 2469 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 rightangled triangle to determine $\sin x$ and $\cos x$. My answers aren't pretty at all. Last edited by v8archie; June 3rd, 2015 at 06:33 PM. 
June 4th, 2015, 07:24 AM  #3  
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  Quote:
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: 7,851 Thanks: 1075 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: 7,851 Thanks: 1075 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: 19,501 Thanks: 1739 
Note that tan(45° + 30°) = (1 + 1/√3)/(1  1/√3) = 2 + √3.

June 4th, 2015, 02:29 PM  #7 
Senior Member Joined: Apr 2014 From: Europa Posts: 571 Thanks: 176  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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cosine, relationship, sine, tangent 
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