Pre-Calculus 40S Winter 2013 Section CB: Proving Trigonometric Identities 101

Hey guys! John B. here, and in case any of you were still wondering, or just plain didn't pay attention when I explained, I'm Malkuthe on here because my gmail account is still linked to my old Blogger account which I used for all of my literary work. If you didn't know, apart from being an answer key, I'm also a high-fantasy writer. If you follow this link, it should take you to one of the stories in my new blog. Have fun!

Anyway, on to more pressing matters: Trigonometric Identities.

In the Philippines, when I took my Trigonometry and Advanced Algebra course, this was one of the most frustrating units for me just because it was so tedious. So, to help you guys navigate this article, I'll give you a table of contents!

The Identities

Reciprocal Identities
Sum and Difference Identities
Double Angle Identities

Non-Permissible Values
Proving Identities

The Identities

To be able to prove trig identities, you have to start by knowing the basic identities. From the last unit that we did on the unit circle, you have to remember

$\sin^2\theta + \cos^2\theta = 1$

Which is otherwise known as the Pythagorean Identity.

The Reciprocal Identities

We know about the Sine, Cosine and Tangent identities, but you also have to take note of the reciprocal identities. There are three:

$\csc\theta=\frac{1}{\sin\theta}\;\;\;\sec\theta=\frac{1}{\cos\theta}\;\;\;\cot\theta=\frac{1}{\tan\theta}$

The Sum and Difference Identities

The sum and difference identities are pretty straightforward. Normally, what I would do is show you how to derive them, but the sum and difference identities for at least Sine and Cosine rely on concepts that we haven't yet discussed in class. In any case, if you wanted to know what is used to derive them, it's the Euler Formula:

$\cos x + \imath\sin x=e^{\imath x}$

I can show you how to derive the Tangent identities, however. Anywho, on to the sum identities:

$\cos(\alpha+\beta)=\cos\alpha\cos\beta - \sin\alpha\sin\beta$ $\sin(\alpha+\beta)=\sin\alpha\cos\beta + \cos\alpha\sin\beta$

And then the difference identities:

$\cos(\alpha-\beta)=\cos\alpha\cos\beta + \sin\alpha\sin\beta$ $\sin(\alpha-\beta)=\sin\alpha\cos\beta - \cos\alpha\sin\beta$

How about Tangent? Well, let's take a look at the sum identity of Tan. Knowing that $\tan\theta=\frac{\sin\theta}{\cos\theta}$ we can say that $\tan(\alpha+\beta)=\frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}$

So, let's expand this a bit.

$\tan(\alpha+\beta)=\frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta}$

Well, wait a minute, we can't do anything about this! There's no way to simplify things out and cancel! However, you're missing something very important. Going back, we know that Tan is equal to Sin over Cos. Is there any way to get that to show up in this formula? Well, let me show you!

$\tan(\alpha+\beta)=\frac{\sin\alpha{\color{red}\cos\beta + \cos\alpha}\sin\beta}{{\color{red}\cos\alpha\cos\beta} - \sin\alpha\sin\beta}$
$\tan(\alpha+\beta)=\frac{\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta}}{\frac{\cos\alpha\cos\beta+\sin\alpha\sin\beta}{\cos\alpha\cos\beta}}$

The Cosines in the numerator cancel out, and you're left with:

$\tan(\alpha+\beta)=\frac{\frac{\sin\alpha}{\cos\alpha}+\frac{\sin\beta}{\cos\beta}}{1-\frac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}}$

And would you look at that, we can substitute in Tan using the identity. Therefore, the Tan sum and difference identities are:

$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$
$\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$

The Double Angle Identities

For these identities, I will not show you how to derive them, because we already did this in class. These are the double angle identities:

$\sin2\theta=2\sin\theta\cos\theta$ $\cos2\theta=\cos^2\theta-\sin^2\theta$ $\cos2\theta=1-2\sin^2\theta$ $\cos2\theta=2\cos^2\theta-1$ $\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$
Return to Table of Contents -->

Non-Permissible Values

When it comes to proving Trig Identities, it's important to note that this is only true for all permissible values meaning that the same doesn't hold true for non-permissible values. Which brings up a good question: what are non-permissible values and how do we find them?

Non-permissible values are, by definition, just values that will make a denominator zero. So in the case of $\frac{1}{\cos^2\theta-1}$ , all you have to do is "equate" the denominator to zero.

$\cos^2\theta-1\neq0$
$\cos^2\theta\neq1$
$\cos\theta\neq\sqrt{1}$
$\cos\theta\neq\pm1$
$\theta\neq\pi n, n\in\mathbb{I}$

Tips and Tricks!

If you have $\csc\theta\;or\;\sec\theta$ in your denominator, you don't have to find the non-permissible values for those. Remember this: reciprocal functions can never be zero.

If you have $\cot\theta$ in your denominator, all you have to do is look for the non-permissible values for $\cos\theta$ .

If you have $\csc\theta\;or\;\sec\theta$ in your numerator, find the non-permissble values for $\sin\theta\;or\;\cos\theta$ respectively.

If you have $\tan\theta\;or\;\cot\theta$ in your numerator, then look for the non-permissible values for $\cos\theta\;or\;\sin\theta$ respectively.
Return to Table of Contents -->

Proving Identities

Now, after that almost unnecessarily lengthy discussion, let's move on to the meat of the matter, proving Trigonometric Identities. As long as you can remember everything that was said before this part, you will do just fine. It can seem quite daunting at first glance, but once you practice and get to know the identities, proving them won't be so hard.

To begin with, let's start with a simple identity. $\csc\theta\sin\theta=1$

We substitute csc with the proper identity: $\frac{1}{\sin\theta}\cdot\sin\theta=1$

Sin cancels out and we're left with $1=1$ , $LHS=RHS$

Now that we've got the easy out of the way, let's move on to a more challenging identity to prove. And as we go along, I will point out things that you have to recognize once you look at a problem. Prove the following identity:

$\cos{\color{red}4\theta} + \sin^2\theta = 4\sin^4\theta - 7\sin^2\theta + 4\cos^4\theta - 8\cos^2\theta + 5$

First of all, there isn't a denominator, so all values are permissible. Second, Mr. Piatek said that you want to work on the more complicated side first. For this case, you might say almost immediately that it's the right hand side that's difficult to work with. However, you'll be wrong. Highlighted in red is $4\theta$ , which is something we don't have an identity for. This is why we have to work on the left hand side.

That being said, you have to recognize that $4\theta=2\theta+2\theta$ which brings me to my first tip: when you see an angle other than $\theta\;or\;2\theta$ , you have to know that that angle can be expressed as the sum or difference of two other angles. So, let's work with what we've got. For convenience's sake, let's set the right hand part of the identity aside for a moment.

$\cos(2\theta+2\theta) + \sin^2\theta$

Using the sum identity for cosine, we get:

${\color{red}\cos^22\theta - sin^22\theta} + \sin^2\theta$

Oh no! But we don't have an identity for $\cos^22\theta\;OR\;\sin^22\theta$ ! Well, don't despair yet, we can rewrite them as this:

${\color{Blue}\cos2\theta\cdot\cos2\theta} - {\color{DarkOrange}\sin2\theta\cdot\sin2\theta} + \sin^2\theta$

Let's use the double angle identities and we get the following:

${\color{Blue}(2\cos^2\theta-1)(2\cos^2\theta-1)} - {\color{DarkOrange}(2\sin\theta\cos\theta)(2\sin\theta\cos\theta)} + \sin^2\theta$

Okay, we seem to be getting closer! Let's FOIL everything.

${\color{Blue}(4\cos^4\theta-4\cos^2\theta+1)} - {\color{DarkOrange}(4\sin^2\theta\cos^2\theta)} + \sin^2\theta$

Oh boy. We've got $\cos^4\theta$ . Don't worry, we can rewrite this as: $\cos^2\theta\cdot\cos^2\theta$ . However, at this point, I think it will be a wise decision to go back to the problem --> and look at what we should have. We have two sin terms and two cos terms. But wait! Notice how the sin terms are on the left and the cos terms are on the right? Why don't we try to make one of the BLUE cos terms into sin terms and all the ORANGE sin terms into cos terms? At this point, you just have to experiment!

This brings me to my second tip: whenever you see a trig ratio squared, always remember the Pythagorean Identity.

${\color{Blue}(4(1-\sin^2\theta)(1-\sin^2\theta)-4\cos^2\theta+1)} - {\color{DarkOrange}(4(1-\cos^2\theta)\cos^2\theta)} + \sin^2\theta$
${\color{Blue}(4(1-2\sin^2\theta+\sin^4\theta)-4\cos^2\theta+1)} - {\color{DarkOrange}(4(\cos^2\theta-\cos^4\theta))} + \sin^2\theta$
${\color{Blue}(4 - 8\sin^2\theta + 4\sin^4\theta - 4\cos^2\theta+1)} - {\color{DarkOrange}(4\cos^2\theta - 4\cos^4\theta)} + \sin^2\theta$
${\color{Blue}-8\sin^2\theta + 4\sin^4\theta - 4\cos^2\theta+5} - {\color{DarkOrange}4\cos^2\theta + 4\cos^4\theta} + \sin^2\theta$

Whew. That was a lot, but we're close now! All we have to do is combine like terms and what do we get?

$4\sin^4\theta - 7\sin^2\theta + 4\cos^4\theta - 8\cos^2\theta + 5$

Look familiar? --> We can now say $LHS=RHS$ Return to Table of Contents -->

Anyway, I hope you guys learned something! That's it for my scribing. Happy Math-ing ~~yeah, riiight~~ to all of you!

Pre-Calculus 40S Winter 2013 Section CB

Thursday, May 2, 2013

Proving Trigonometric Identities 101