ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Nomvalt on April 26, 2010, 08:51:22 pm

Title: exact values and symmetry
Post by: Nomvalt on April 26, 2010, 08:51:22 pm: If $sin(x)= \frac{-\sqrt{3}}{2}$ , and $\frac{3\pi }{2}<x<2\pi$ , find $cos(x)$ and $tan(x)$ .

When I tried to find the value for the third side using pythagora's theorem I got the number √-1, which when evaluated appeared as 'error 2' on my calculator so is not a real value (as you cannot take the square of a negative number). Anyway, how would you find cos(x) and tan(x) in this situation? I have a feeling that this question does not make sense and was just a mistake, but I just want to make sure. :-\
Title: Re: exact values and symmetry
Post by: the.watchman on April 26, 2010, 08:55:22 pm: $cos^2 (x)=1-sin^2(x) = 1-(-\frac{\sqrt{3}}{2})^2 = \frac{1}{4}$

$\cos (x)=\pm \frac{1}{2}$

BUT in the fourth quadrant, cos is positive

So $\cos (x)=\frac{1}{2}$

sin/cos for tan :)
Title: Re: exact values and symmetry
Post by: vea on April 27, 2010, 04:07:20 pm: Just for fun
$\frac{sinx}{cosx}=\frac{\frac{-\sqrt{3}}{2}}{\frac{1}{2}}$

$tanx=(\frac{-\sqrt{3}}{2}}) (\frac{2}{1})$

$tanx=-\sqrt{3}$

Hope the tex comes out right.