# Finding the value of the angle theta in radians, knowing sec theta and cot theta

• Mar 29th 2010, 06:50 AM
cozza
Finding the value of the angle theta in radians, knowing sec theta and cot theta
Hi,

The question I have is:

Given that $\sec \theta = -2$, $\cot \theta = -1/3\sqrt{3}$ and $-\pi < 0 < \pi$, find the exact value of the angle $\theta$ in radians. Justify your answer.

Is the following along the right lines?

$\sec\theta=1/cos\theta$ therefore $\cos\theta=1/\sec \theta=-1/2$
$\cot\theta=1/\tan\theta$ therefore $\tan\theta=1/\cot\theta=-3 \sqrt{3}/3$

$\sin\theta =\tan\theta\cos\theta=-3\sqrt{3}/3x-1/2=1/2sqrt{3}$

Since both $\tan\theta$ and $\cos\theta$ are negative and lie in the range $-\pi < 0 < \pi$, then $\theta$ must lie in the 2nd quadrant so is positive.

• Mar 29th 2010, 07:18 AM
HallsofIvy
Quote:

Originally Posted by cozza
Hi,

The question I have is:

Given that $\sec \theta = -2$, $\cot \theta = -1/3\sqrt{3}$ and $-\pi < 0 < \pi$, find the exact value of the angle $\theta$ in radians. Justify your answer.

Is the following along the right lines?

$\sec\theta=1/cos\theta$ therefore $\cos\theta=1/\sec \theta=-1/2$
$\cot\theta=1/\tan\theta$ therefore $\tan\theta=1/\cot\theta=-3 \sqrt{3}/3$

$\sin\theta =\tan\theta\cos\theta=-3\sqrt{3}/3x-1/2=1/2sqrt{3}$

Since both $\tan\theta$ and $\cos\theta$ are negative and lie in the range $-\pi < 0 < \pi$, then $\theta$ must lie in the 2nd quadrant so is positive.

Good. Now you also need to know some basic angles. All three angles in an equilateral triangle have measure $\pi/3$ radians. If you drop a perpendicular from one vertex to the opposite side, you divide the triangle into two congruent right triangle with angles of $\pi/3$ and [/tex]\pi/6[/tex]. If you take the equilateral triangle to have sides of length 1, the hypotenuse of the right triangles is 1 and the leg opposite the $\pi/6$ angle is 1/2.
Good. Now you also need to know some basic angles. All three angles in an equilateral triangle have measure $\pi/3$ radians. If you drop a perpendicular from one vertex to the opposite side, you divide the triangle into two congruent right triangle with angles of $\pi/3$ and [/tex]\pi/6[/tex]. If you take the equilateral triangle to have sides of length 1, the hypotenuse of the right triangles is 1 and the leg opposite the $\pi/6$ angle is 1/2.
Thank you for such a quick reply! So does the angle $\theta=1/2sqrt{3}$? Can I then justify this answer by explaining what you have written? Why do I need to know the lenths of the sides in the right angle traingle to find the exact value of the angle $\theta$ in rtadians? Sorry if this seems like a really stupid question.