help with expressions

**paulaa** · January 5th 2011, 05:40 AM

values of two of the three expressions cos (x), sin (x), and tan (x), has not been set. Decide these values.

sin(x) = -((2)/(5^0.5)) , 3n/2<x<2n
cos(x)^2 + sin(x)^2 = 1

cos(x) = (1-sin(x)^2)^0.5 = (1-(-2/(5^0.5))^2)^0.5 = (9/5)^0.5

tan(x) = ((-2/(5^0.5))/((9/5)^0.5) = (-10^0.5)/5

When I leave it to my teacher I get error, does anyone know what the problem is? He says that you should show 3n/2<x<2n in sin, cos and tan in any way.

By n I mean pi

**Soroban** · January 5th 2011, 06:08 AM

Hello, paulaa!

Your arithmetic is off . . .

Values of two of the three expressions $\cos\theta,\:\sin\theta,\:\tan\theta$ have not been set.
Decide these values.

. . $\sin\theta \,=\,\text{-}\dfrac{2}{\sqrt{5}},\;\;\frac{3\pi}{2} < x < 2\pi$

$\sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \cos\theta \:=\:\pm\sqrt{1-\sin^2\!\theta}$

$\cos\theta \:=\:\pm\sqrt{1 - \left(\text{-}\frac{2}{\sqrt{5}}\right)^2} \:=\:\pm\sqrt{1 - \frac{4}{5}} \:=\:\pm\sqrt{\frac{1}{5}} \;=\;\pm\dfrac{1}{\sqrt{5}}$

Since $\,\theta$ is in Quadrant 4, $\cos\theta$ is positive: . $\boxed{\cos\theta \:=\:\frac{1}{\sqrt{5}}}$

Then: . $\tan\theta \:=\:\dfrac{\sin\theta}{\cos\theta} \:=\:\dfrac{\text{-}\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} \quad\Rightarrow\quad \boxed{\tan\theta \:=\:-2}$

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