My teacher gave my class review questions to do over the winter break. I did almost everything except these three:

How do you draw and calculate these?

1) cos x = -(sqrt(2))/2

2) csc x = 2

3) sec x = -1

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- Jan 1st 2009, 06:57 PMAlphaRockHow To Do This Trigonometry
My teacher gave my class review questions to do over the winter break. I did almost everything except these three:

How do you draw and calculate these?

1) cos x = -(sqrt(2))/2

2) csc x = 2

3) sec x = -1 - Jan 1st 2009, 10:53 PMGrandadTrigonometry
Hello AlphaRock

I'm not sure whether you're used to working in degrees or radians, so I'll keep it simple and work in degrees.

1) Look first at $\displaystyle \cos y = +\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

This comes from an isosceles $\displaystyle 45^o - 45^o - 90^o$ right-angled triangle with sides, using Pythagoras, $\displaystyle 1 - 1- \sqrt{2}$. You should see then that $\displaystyle \cos 45^o= \frac{1}{\sqrt{2}}$ , and so $\displaystyle y = 45^o$.

To solve $\displaystyle \cos x = -\frac{\sqrt{2}}{2}$, simply take this value of $\displaystyle y$ from $\displaystyle 180^o$, because $\displaystyle \cos x = -\cos(180 -x)$.

A more advanced answer, involving radians is the general solution $\displaystyle x=(2n+1)\pi \pm \frac{\pi}{4}$, but if you haven't covered that sort of thing in your course, you can ignore this!

2) $\displaystyle \csc x = 2 \implies \sin x = \frac{1}{2}$. Draw a 30 - 60 - 90 triangle, with a hypotenuse 2 units long, and figure out the lengths of the shortest side using the fact that this triangle is half of an equilateral triangle. Then you should see that the sine of one of the angle is $\displaystyle \frac{1}{2}$

3) $\displaystyle \sec x = -1 \implies \cos x = -1$. Use the same formula as in (1), and the fact that $\displaystyle \cos 0^o = 1$.

Can you complete them now?

Grandad

- Jan 3rd 2009, 01:55 AMPn0yS0ld13r
I don't know if you covered this yet in your class, but an alternative way to solve these is to use the inverse of these trig functions to solve for x.

Example:

$\displaystyle \cos x=-\dfrac{\sqrt{2}}{2}$.

We need to restrict our domain to make the cosine function one-to-one: $\displaystyle [0,\pi]$.

Taking the arccosine of both sides,

$\displaystyle \arccos(\cos x))=\arccos \left(-\dfrac{\sqrt{2}}{2}\right) $

$\displaystyle x = \arccos \left(-\dfrac{\sqrt{2}}{2}\right)=\dfrac{3\pi}{4}$.

We know that $\displaystyle \cos x = \cos -x$.

So solving $\displaystyle \cos (-x) = -\dfrac{\sqrt{2}}{2}$,

$\displaystyle \arccos(\cos (-x)) = \arccos \left(-\dfrac{\sqrt{2}}{2}\right) $

$\displaystyle -x = \arccos \left(-\dfrac{\sqrt{2}}{2}\right) = \dfrac{3\pi}{4} \Rightarrow x =-\dfrac{3\pi}{4}$.

Since the period of cos x is $\displaystyle 2\pi$, we have the general solution:

$\displaystyle x=\pm \dfrac{3\pi}{4} \pm 2\pi n$ where $\displaystyle n \in \mathbb{Z}$