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

[mathwiz]

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)

 

[royhaas]

Re: matrices...ew

You'll need the value of $\displaystyle cos(\pi/12)$.

 

[daon]

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).

 

[soroban]

Re: matrices...ew

Hello, mathwiz!

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

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

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


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

. . Got it?

 

[Deleted member 4993]

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

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

edit - fixed the sign problem

 

[mathwiz]

well somehow it's supposed to be like
-sq rt of 6-sq rt of 2/2?

 

[skeeter]

$\displaystyle \cos\left(-\frac{13\pi}{12}\right) =$

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

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

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

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

$\displaystyle -\frac{\sqrt{2} + \sqrt{6}}{4}$

 

[mathwiz]

so i still don't get HOW to do that
it's not the problem itself it's the whole...concept

 

[stapel]

I'm utterly confused.. i skipped trig and....

it's not the problem itself it's the whole...concept

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!

Eliz.

 

[mathwiz]

it's just this chapter...
the rest of it isn't very trig-based

 

[o_O]

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 $\displaystyle \cos (A - B)$

 

[Deleted member 4993]

it's just this chapter...
the rest of it isn't very trig-based

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....