1. ## Urgent Question

Hi guys,

How do I do arithematic operations with functions like:

2cos(2t-45(degrees)) - 3cos(4t-45) + 5sin(3t+45)

Any help would be greatly appreciated.

Thanks

2. Originally Posted by thebearot
Hi guys,

How do I do arithematic operations with functions like:

2cos(2t-45(degrees)) - 3cos(4t-45) + 5sin(3t+45)

Any help would be greatly appreciated.

Thanks
The basic rules are:
$\displaystyle sin(A \pm B) = sin(A)cos(B) \pm sin(B)cos(A)$
$\displaystyle cos(A \pm B) = cos(A)cos(B) \mp sin(A)sin(B)$

We get the "double angle" formulae from these:
$\displaystyle sin(2A) = 2sin(A)cos(A)$
$\displaystyle cos(2A) = cos^2(A) - sin^2(B) = 2cos^2(A) - 1 = 1 - 2sin^2(A)$

Hint: It takes a bit of work, but you can simplify sin(3t) as sin(2t+t) and cos(4t) as cos(2t+2t).

-Dan

3. Thanks for the quick reply. I'll try to solve it with the formulas.

4. Originally Posted by topsquark
Hint: It takes a bit of work, but you can simplify sin(3t) as sin(2t+t) and cos(4t) as cos(2t+2t).
There is an elegant way to derive a general formula for,
$\displaystyle \sin nx \mbox{ and }\cos nx$

5. Hello, ThePerfectHacker!

There is an elegant way to derive a general formula for: $\displaystyle \sin nx \mbox{ and }\cos nx$

I "discovered" a method while in college . . . with my very own ritual.

Suppose we want the trig formulas for $\displaystyle 6\theta$

(1) Expand $\displaystyle (x + 1)^6:\;\;x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1$

(2) Construct a fraction with these terms,
Write the terms alternately in the denominator and numerator.

. . $\displaystyle \frac{6x^5\qquad 20x^3\qquad 6x}{x^6\qquad 15x^4\qquad 15x^2\qquad 1}$

(3) Insert alternating signs in the numerator and the denominator.

. . $\displaystyle \frac{6x^5 -20x^3 + 6x}{x^6 - 15x^4 + 15x^2 -1}$

(4) Replace $\displaystyle x$ with $\displaystyle \tan\theta$ and we have:

. . $\displaystyle \boxed{\tan6\theta \;= \;\frac{6\tan^2\!\theta -20\tan^3\!\theta + 6\tan\theta}{\tan^6\!\theta - 15\tan^4\!\theta + 15\tan^2\!\theta -1}}$ $\displaystyle \bf{(1)}$

(5) Multiply top and bottom by $\displaystyle \cos^6\!\theta$ and we get two formulas:

. . $\displaystyle \tan6\theta \;=$$\displaystyle \;\frac{\boxed{\sin6\theta}}{\boxed{\cos6\theta}} \;= \;\frac{\boxed{6\sin^2\!\theta\cos^4\!\theta - 20\sin^3\!\theta\cos^3\!\theta + 6\sin\theta\cos^5\!\theta}}{\boxed{\sin^6\!\theta - 15\sin^4\!\theta\cos^2\!\theta + 15\sin^2\!\theta\cos^4\!\theta - \cos^6\!\theta}}$ $\displaystyle \begin{array}{ccc}\bf{(2)} \\ \\ \bf{(3)}\end{array}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This is all possible because of DeMoivre's Theorem.

Since $\displaystyle \cos6\theta + i\sin6\theta\;= \;(\cos\theta + i\sin\theta)^6$, we can expand the right side,

then equate real and imaginary components, giving us (2) and (3) above.

6. Originally Posted by Soroban
Hello, ThePerfectHacker!

I "discovered" a method while in college . . . with my very own ritual.
That must have been 80 years ago. Anyway you have a good method. The one I was speaking about also uses de Moiver's theorem. Then you just use binomial expansion.