Trig indentity

**dadon** · May 2nd 2006, 08:24 AM

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.

Thanks.

**topsquark** · May 2nd 2006, 09:08 AM

Originally Posted by dadon

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.

Thanks.

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

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

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

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

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

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

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

-Dan

**CaptainBlack** · May 2nd 2006, 09:19 AM

Originally Posted by dadon

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.

Thanks.

Here is the plot

RonL

**dadon** · May 2nd 2006, 10:01 AM

Hey Cheers for that guys!

I get want you have done throughout but how do you get $<br /> a = R cos\beta<br />$ and $<br /> b = R sin\beta<br />$

**topsquark** · May 2nd 2006, 01:23 PM

Originally Posted by topsquark

$a cos\theta + b sin\theta = R cos\theta cos\beta + R sin\theta sin\beta$

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

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

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

Technically, I am assuming that $cos\theta \neq 0$ so some care needs to be taken as far as interpretation for all values of $\theta$ . However, since $\theta$ 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 $cos\theta = 0$ .)

-Dan

**dadon** · May 2nd 2006, 01:30 PM

Thanks for sharing you time to solve my problem and explainly in a way that I understand fully.

Cheers

dadon

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