matrices: give exact value for cos(-13(pi)/12)

M

mathwiz

New member
Joined
May 12, 2008
Messages
4
I'm utterly confused.. i skipped trig and my last math class didn't go through sin and cos enough, so the matrix transformations are realllly confusing

i guess that would technically be formulas for cos(a+b) and sin(a+b) but not a & b... hha

my main problem is giving the exact values for the given... one is cos(-13(pi)/12)
 
Re: matrices...ew

You'll need the value of \(\displaystyle cos(\pi/12)\).
 
Re: matrices...ew

\(\displaystyle cos(\frac{-13 \pi}{12}) = cos(\frac{13\pi}{4}-\frac{13\pi}{3})\)

Then just apply the formula for cos(a-b).
 
Re: matrices...ew

Hello, mathwiz!

\(\displaystyle \text{Find the exact value: }\;\cos\left(\text{-}\frac{13\pi}{12}\right)\)
Click to expand...

\(\displaystyle \text{With some experimenting, we find that: }\:-\frac{13\pi}{12} \;=\;\frac{5\pi}{3} - \frac{11\pi}{4}\)

. . \(\displaystyle \text{And we will use: }\;\cos(A - B) \;=\;\cos A\cos B + \sin A\sin B\)


\(\displaystyle \text{We have: }\;\cos\left(\frac{5\pi}{3} - \frac{11\pi}{4}\right) \;=\;\cos\left(\frac{5\pi}{3}\right)\cos\left(\frac{11\pi}{4}\right) + \sin\left(\frac{5\pi}{3}\right)\sin\left(\frac{11\pi}{4}\right)\)

. . Got it?

 
\(\displaystyle \frac{\pi}{12}\, = \, \frac{\pi}{4}\, - \frac{\pi}{6}\,\)

\(\displaystyle cos(-\frac{13\pi}{12})\, = cos(\frac{13\pi}{12})\, = \, cos(\pi\, + \frac{\pi}{12})\, =-\, cos(\frac{\pi}{12})\,=- \, cos(\frac{\pi}{4}\, - \frac{\pi}{6})\,\)

edit - fixed the sign problem
 
\(\displaystyle \cos\left(-\frac{13\pi}{12}\right) =\)

\(\displaystyle -\cos\left(\frac{\pi}{12}\right) =\)

\(\displaystyle -\cos\left(\frac{\pi}{3}- \frac{\pi}{4}\right) =\)

\(\displaystyle -\left[\cos\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{4}\right)\right] =\)

\(\displaystyle -\left(\frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}\right)=\)

\(\displaystyle -\frac{\sqrt{2} + \sqrt{6}}{4}\)
 
so i still don't get HOW to do that
it's not the problem itself it's the whole...concept
 
mathwiz said:
I'm utterly confused.. i skipped trig and....

it's not the problem itself it's the whole...concept
Click to expand...
It sounds like skipping trig might not have been the best decision.... :shock:

Since you're taking trig-based calculus, you clearly now need the trig. Obviously, a semester's-worth of material cannot be conveyed in a short forum posting -- there's a reason the trig books were hundreds of pages long! :wink:

At a guess, you're either going to need to get a "dummies" type book and do some self-study, or else think about hiring a qualified local tutor and setting aside a few hours a week for concentrated teaching sessions. :idea:

Good luck! :D

Eliz.
 
This is an identity:
\(\displaystyle \cos (A - B) = \cos A \cos B + \sin A \sin B\)

What we want to do is split \(\displaystyle -13\pi / 12\) into the subtraction of two numbers A and B whose exact cosine and sine values are known. So after playing with a few combinations, one combination that works is \(\displaystyle 5\pi /3\) and \(\displaystyle 11\pi / 4\) where:

\(\displaystyle -\frac{13\pi}{12} = \underbrace{\frac{5\pi}{3}}_{A} - \underbrace{\frac{11\pi}{4}}_{B}\)

as suggested by Soroban. These values are chosen because you know the exact sine and cosine values of both angles (at least hopefully you know them):

\(\displaystyle \begin{array}{lll} \cos A = \cos \left(\frac{5\pi}{3}\right) = \frac{1}{2} & \quad \quad & \sin A = \sin \left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2} \\ \cos B =\cos \left(\frac{11\pi}{4}\right) = \cos \left(\frac{3\pi}{4} + 2\pi\right) = -\frac{\sqrt{2}}{2} & \quad \quad & \sin B = \sin \left(\frac{11\pi}{4}\right) = \frac{\sqrt{2}}{2}\)

So now you can apply the formula for \(\displaystyle \cos (A - B)\)
 
mathwiz said:
it's just this chapter...
the rest of it isn't very trig-based
Click to expand...

Calculus never leaves trig. If you don't feel comfortable with trig - you'll have whole host of problems as you advance - specially if you major in engineering. You'll need to do "trig - substitution", "cartesian-polar transformation","vector analysis" ... just to name a few.

If you are planning to do any more calculus - or go into engineering - take a rigorous trig class. It will make your life liveable....
 
You must log in or register to reply here.
Top