Trig indentity

• May 2nd 2006, 08:24 AM
Trig indentity
Hi All,

How about do I go to answer this trig. identity question. I have attached it as a picture as wasn't to sure on how to insert the symbols.

Also if it is not possible to plot the sketch just explain it in words what I have to do, i'm sure I would be able to understand. :o

Thanks.
• May 2nd 2006, 09:08 AM
topsquark
Quote:

Hi All,

How about do I go to answer this trig. identity question. I have attached it as a picture as wasn't to sure on how to insert the symbols.

Also if it is not possible to plot the sketch just explain it in words what I have to do, i'm sure I would be able to understand. :o

Thanks.

I can't show the graph, but I can help with the rest.

Via the identity:
$\displaystyle R cos(\theta - \beta)=R cos\theta cos\beta + R sin\theta sin\beta$

By comparison then:
$\displaystyle a cos\theta + b sin\theta = R cos\theta cos\beta + R sin\theta sin\beta$
So
$\displaystyle a = R cos\beta$
$\displaystyle b = R sin\beta$

Dividing these two equations gives us:
$\displaystyle \frac{R sin\beta}{R cos\beta}=tan\beta = \frac{b}{a}$
or
$\displaystyle \beta = tan^{-1} \left ( \frac{b}{a} \right )$

and
$\displaystyle R^2 sin^2 \beta + R^2 cos^2 \beta = R^2 (1) = a^2+b^2$
so
$\displaystyle R = \sqrt{a^2+b^2}$

Now, in order to graph $\displaystyle 2 cos\theta + 3 sin\theta$ you are essentially graphing:
$\displaystyle \sqrt{2^2+3^2}cos \left ( \theta - \left [ tan^{-1} \left ( \frac{3}{2} \right ) \right ] \right )$

or roughly $\displaystyle \sqrt{13} \, cos(\theta - 0.983)$ vs. $\displaystyle \theta$.
This graph is simply a cosine graph, with amplitude $\displaystyle \sqrt{13}$ and translated 0.983 radians (or so) to the right.

-Dan
• May 2nd 2006, 09:19 AM
CaptainBlack
Quote:

Hi All,

How about do I go to answer this trig. identity question. I have attached it as a picture as wasn't to sure on how to insert the symbols.

Also if it is not possible to plot the sketch just explain it in words what I have to do, i'm sure I would be able to understand. :o

Thanks.

Here is the plot

RonL
• May 2nd 2006, 10:01 AM
re:
Hey Cheers for that guys! :)

I get want you have done throughout but how do you get $\displaystyle a = R cos\beta$ and $\displaystyle b = R sin\beta$
• May 2nd 2006, 01:23 PM
topsquark
Quote:

Originally Posted by topsquark
$\displaystyle a cos\theta + b sin\theta = R cos\theta cos\beta + R sin\theta sin\beta$

You get them by comparing forms. Let me write it out a bit more explicitly:
$\displaystyle a cos\theta + b sin\theta = R cos\theta cos\beta + R sin\theta sin\beta$

Now, take a look at the $\displaystyle cos\theta$ term. They have to be the same on both sides of the equation so we can write:
$\displaystyle a cos\theta = R cos\theta cos\beta$
Dividing both sides by $\displaystyle cos\theta$ we obtain:
$\displaystyle a = R cos\beta$

A similar argument with the sine function gives the second relation.

Technically, I am assuming that $\displaystyle cos\theta \neq 0$ so some care needs to be taken as far as interpretation for all values of $\displaystyle \theta$. However, since $\displaystyle \theta$ is being taken as an independent variable and not a constant I feel we can more or less disregard this problem. (Theoretically I suppose we should be plotting an open circle at any point that $\displaystyle cos\theta = 0$.)

-Dan
• May 2nd 2006, 01:30 PM