# Conjecturing sinusoids using a CAS?

• Aug 18th 2010, 08:54 AM
treetop282
Conjecturing sinusoids using a CAS?
I really have no idea what any of this means, and I think I just need a translation...
Let $\displaystyle y=sin(ax) + cos(ax)$.
Use the symbolic manipulator of a computer algebra system (CAS) to conjecture the following:
a) Express y as a sinusoid for a=2,3,4 and 5
b) Conjecture another formula for y for a equal to any positive integer n.
c) Check your conjecture with a CAS.
d) Use the formula for the sine of the sum of two angles to confirm your conjecture.

Now, I have no idea what a symbolic manipulator, computer algebra system, or sinusoid is. I have a TI-84 and I really have no idea what just happened... help?
• Aug 18th 2010, 09:54 AM
Vlasev
Do you know what the functions sin and cos are? You probably should but if you don't, here is something that may help Trigonometric functions - Wikipedia, the free encyclopedia. The stuff inside the functions is just a times x, where a is a constant, usually a real number.

Symbolic manipulator is a program like Mathematica, which lets you plug in equations without number (such as the above equation) and you can literally manipulate the expressions. In Mathematica for example, there are powerful algorithms that let you factor and simplify expressions and also transform expression using known identities. A CAS is a computer program with which you can do numerical math calculations. Nowadays a CAS and symbolic manipulator come hand in hand. Mathematica and Maple are the one I use.

In part a), you need to use trig identities for the double angles and for the sums of angles to express y as a sum of products of sine and cosine functions that involve only a single x. Once you do this for the first couple of values of a, you should see a clear pattern. Then you need to conjecture a formula.

In part c) you need to use a CAS to manipulate either the conjectured formula or simplify y.
• Aug 18th 2010, 11:27 AM
treetop282
I understand it a little more, but I still can't figure anything out.
I do understand the trig functions, but I can't figure out how to get the x alone using them. This is what I have, and it's only a few lines:
a) $\displaystyle y=sin(2x)+cos(2x)$
$\displaystyle y=2sin(x)cos(x)+(cos(x))^2-(sin(x))^2$
$\displaystyle y=2sin(x)cos(x)+1-2(sin(x))^2$
And yeah, I hope that part's right. I downloaded that Mathematica thingy, but holy cow is it complicated. It keeps telling me I'm wrong, and I can't help but to agree with it. Where should I go from here...?
• Aug 18th 2010, 12:23 PM
Vlasev
You were on the right track. Just leave it as it is in the 2nd line. Do the 3x case and you'll see the pattern.
• Aug 18th 2010, 01:42 PM
treetop282
This is what I've collected so far, with help from the website: Summary of Trigonometric Identities.
(I omitted the x's.)
$\displaystyle 1a) y=sin(2cos)-1+2cos^2$
$\displaystyle 2a) y=sin(-1+4cos^2)-3cos+4cos^3$
$\displaystyle 3a) y=sin(-4cos+8cos^3)+1-8cos^2+8cos^4$
$\displaystyle 4a) y=sin(1-12cos^2+16cos^4)+5cos-20cos^3+16cos^5$
$\displaystyle b) y=2sin((n-1)x)cos(x)-sin((n-2)x)+2cos((n-1)x)cos(x)-cos((n-2)x)$

Now, my mind is like blown because I tried to get the triple and quadruple angle ones by like putting 2x into the previous one in place of x, but everything got like ten million times more confusing so I just trusted the site's triple and quadruple angle formulas. And I can see the rising exponents thing, but everything else just looks like random gibberish to me and I cannot see how anybody can draw a formula from that.
Anyways, I have no way of checking this with a CAS and since the answer in the back of my book for part (c) is "It works," I think I'm okay. The only problem is that the answer in my book is this:
$\displaystyle sqrt(2)sin(ax+(pi/4)$
I don't even understand where any of that comes from... and the proof they show for part (d) is just as overwhelming.
Can someone explain where they got their answer wrong, or why my answer is so long and possibly not right...?
• Aug 19th 2010, 11:38 PM
Vlasev
Here is something to think about.

Let $\displaystyle \displaystyle \cos(x) = a$ and $\displaystyle \sin(x) = b$. Then

$\displaystyle \displaystyle \cos(2x)+\sin(2x) = a^2+ 2ab - b^2$

$\displaystyle \displaystyle (a-b)^2 = a^2 -2ab +b^2$

Next

$\displaystyle \displaystyle \cos(3x)+\sin(3x) = a^3+3a^2b-3ab^2+b^3$

$\displaystyle \displaystyle (a-b)^3 = a^3 - 3a^2b + 3ab^2 -b^3$

This is a definite hint on the general form of the identity. I believe the general identity is not far from this.