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 February 26th, 2013, 05:00 AM #1 Senior Member   Joined: Sep 2012 Posts: 201 Thanks: 1 Trig identities So I have this question fairly simple, I would say, but my issue is I can't actual visualise what it is asking so to speak; I will explain. Question: Given that x is an acute, find tan x if $\sin^2x=\frac{3}{4}$. So I first went back to the unit circle to try and give a better understanding. From this I get the 2 identities $\tan x\equiv\frac{\sin x}{\cos x}$ and $cos^2x\equiv1-\sin^2x$. So this is where I am sort of winging it so to speak. Because I have a $\sin^2x$ in the equation my tan would look like the $(\tan x)^2\equiv(\frac{\sin x}{\cos x})^2\rightarrow \tan^2x\equiv\frac{\sin^2x}{\cos^2x}$, I then sub the identities in and solve; have checked the answer in back of the book, which gives the correct answer, my concern is do I have the right method for working the question out, or have I gone wrong and by some luck got the right answer? Big thanks in advance.
 February 26th, 2013, 05:44 AM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Trig identities Using your reasoning, you should have that cos^2 = 1 - 3/4 = 1/4. So tan^2 = sin^2/cos^2 = (3/4)/(1/4) = 3. Tan(x) is therefore the square root of 3. Here is where the assumption that x is acute comes in. Of course, knowing that x is acute is enough to conclude that sin(x) = sqrt(3)/2, whence tan(x) = sqrt(3) immediately.
 February 26th, 2013, 05:54 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,169 Thanks: 1640 If x is acute, tan(x) is positive. tan²(x) = sin²(x)/cos²(x) = sin²(x)/(1 - sin²(x)) = (3/4)/(1 - (3/4)) = 3, so tan(x) = ?3. Alternatively, cot²(x) = csc²(x) - 1 = 1/sin²(x) - 1 = 4/3 - 1 = 1/3, so cot(x) = 1/?3. Hence tan(x) = ?3. For a third method, x must be the angle in a right-angled triangle with leg opposite angle x of length ?3, and with hypotenuse of length 2. By Pythagoras, the leg adjacent to the angle x has length 1. Hence tan(x) = ?3.
February 26th, 2013, 08:05 AM   #4
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Re: Trig identities

Hello, taylor_1989_2012!

Quote:
 $\text{Given that }x\text{ is acute and }\,\sin^2x\,=\,\frac{3}{4},\,\text{ find }\tan x.$

$\text{W\!e have: }\:\sin x \:=\:\frac{\sqrt{3}}{2} \:=\:\frac{opp}{hyp}$

$x\text{ is an angle in a right triangle with: }\,opp= \sqrt{3},\:hyp = 2$

Code:
                  *
* *
*   *  _
2 *     * /3
*       *
* x       *
*  *  *  *  *
adj
$\text{Pythagoras says: }\,adj= 1.$

$\text{Therefore: }\:\tan x \:=\:\frac{opp}{adj} \:=\:\frac{\sqrt{3}}{1} \:=\:\sqrt{3}$

 March 4th, 2013, 02:05 AM #5 Senior Member   Joined: Sep 2012 Posts: 201 Thanks: 1 Re: Trig identities Thanks for all the replies, once again big help, and thanks again.

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