Trigonometric Identities

GHKNJH-
After wrestling with Blogger for an hour I finally figured out where my invitation email went. So uh, hi guys, I don't want to do this but I'm being forced, I guess. This is Casey, that quiet girl in the back of the classroom with the kinda citrus-colored hair.

I'm going to cut straight to the chase and do a quick rundown of what we've learned about trigonometric identities recently. I got maybe three hours of sleep last night and I'm too lazy right now to include a bunch of pretty images and graphs, so have a billion colors instead.
 

What is a trigonometric identity exactly? 

A trigonometric identity is essentially a couple of trig functions in disguise.

Tanx is the trigonometric identity of sinx/cosx, therefore tanx = sinx/cosx. Anywhere you see tanx used in an equation, you can substitute sinx/cosx in for it, should that prove helpful in solving it. That's the general basis for a trigonometric identity.

Other standard trigonometric identities can be boiled down to the three seen on our formula sheet:

sin^2(x) + cos^2(x) = 1

tan^2(x) + 1 = sec^2(x)

1 + cot^2(x) = csc^2(x)

And a few derived by logic:

cscx = 1/sinx

secx = 1/cosx

cotx = 1/tanx

tanx = sinx/cosx

How can I use trigonometric identities?

During this course, you will have to solve a multitude of daunting looking trigonometric equations.
Ones such as:
a) cos^2(x) - sin^2(x) = 0,
Or even simplifying equations like these:
b) (cotx)/(cscx*cosx).

In order to solve a), we have to change that cos^2(x) into a sine function so we can simplify, or vice-versa. In order to do that, we can take the Pythagorean identity formula, and rearrange it to solve for cos^2(x).
After doing so, we will have something that looks like this: cos^2(x) = 1 - sin^2(x). 
Now we can substitute 1 - sin^2(x) in for cos^2(x), and simplify the equation:

 cos^2(x) - sin^2(x) = 0
1 - sin^2(x) - sin^2(x) = 0
-sin^2(x) - sin^2(x) + 1 = 0
sin^2(x) + sin^2(x) - 1 = 0
x^2 + x^2 - 1 = 0
2x^2 - 1 = 0
x^2 = 1/2
x = square-root(1/2)
sinx = square-root(1/2)
x reference = 45 degrees

The rest is history because I am lazy.


In order to solve b), we have to bring up something called:

Non-permissible values

Oh gosh what is all this nonsense now?
It's simply a range of values an expression can't use for x, nothing too fancy. If you have an expression, 1/x, x obviously cannot equal 0, because dividing by zero yields an undefined value.

If we look at (cotx)/(cscx*cosx), we can see three trig functions here. A quick expansion/investigation of them shows us:

cotx = cosx/sinx
csc = 1/sinx
cosx = part of the product of a denominator

Judging by this, we can see that cosx =/= 0, and sinx =/= 0, because with a value of 0, we'd be dividing by 0 in more than one place, and we can't have that, no sir.

If we turn our attention towards the unit circle, the only places we see where cosx equaling 0 are at pi/2 + pi*n, where nEI.
If we also look again, we see the only places where sinx equals 0 are at pi*n, where nEI.

We can condense this information into a single equation to describe these non-permissible values.
x =/= (pi/2)*n, where nEI.
This equation must be coupled with the final simplified version of the equation, (cotx)/(cscx*cosx), for the final answer to be considered correct.

Now, in order to simplify this gross, ugly thing, we have to change all trig functions in the expression to either sin or cos functions, because this makes it nice and easy to handle. We've already partially done this while calculating the non-permissible values above, so we simply substitute the appropriate functions in for cot and csc.

(cotx)/(cscx*cosx)
(cosx/sinx)/((1/sinx)*cosx)
(cosx/sinx)/((1/sinx)*cosx/1)
(cosx/sinx)/(cosx/sinx)
= 1

As it turns out, that whole expression simplifies down to 1.
Toss the non-permissible values in there anyway, just for good measure.

(cotx)/(cscx*cosx) = 1
x =/= (pi/2)n, where nEI.



HHNGH I HAVEN'T EVEN GOTTEN TO DOUBLE ANGLE IDENTITIES YET.

Using double angle identities

Double angle identities are pretty much the exact same thing as above, except used for, but not limited to, when we have stuff to solve or prove or calculate that involves more than one angle crammed into an equation.
Like this scary stuff:
a) sin48 * cos17 = cos48 * sin17
This even scarier stuff also:
b) sin(7pi/12) = x

Thankfully we get a ton of formulas on our formula sheets about this. Nearly the entire left side of the page is dedicated to this Alpha Beta nonsense. 

I should probably list them here but there's a lot. Sorry people who don't have our formula sheet.
Pretend a=Alpha and B=Beta for the purposes of this blog post.

RIGHTO.
I have to sleep soon so I better get powering through this.

a)
(All angles are in degrees)
sin48 * cos17 - cos48 * sin17
Is the thing we got, 
sina * cosB - cosa * sinB = sin(a - B)
Is the formula we can use. 

Lining up the formulas like this, we can see that a should equal 48, and B should equal 17. We plug these values into our equation and do a quick calculation.
sin48 * cos17 - cos48 * sin17 = sin(48 - 17)
= sin(48 - 17)
= sin(31) is our expression written as a single trigonometric equation.


b)
sin(7pi/12) = x
This angle is clearly not one of our special triangles.
But you know what? We can make it out to be the sum of two special triangles.

What we do, is we take 7/12, omitting the pi for now, and break the 7 up into two values that add together to equal back to 7.
There are multiple sets of numbers like this:
2+5
4+3
1+6
But what we are looking for is a magical set of numbers that when put as the numerator, both divide cleanly by the denominator. Only 4+3 accomplishes this.

4/12 + 3/12 
Simplify.
1/3 + 1/4
Now we reattach pi to the numerator.
pi/3 + pi/4

We now see how sin(7pi/12) = sin(pi/3 + pi/4)
We can use these two new angles as our Alpha and Beta respectively. We select the appropriate formula: sin(a + B) = sina * cosB + cosa * sinB and plug these new values in:

sin(pi/3) * cos(pi/4) + cos(pi/3) * sin(pi/4)
Now, according to the special triangles associated with each angle, we can pick out the exact values of each trig function.
[sqroot(3)/2 * 1/sqroot(2)] + [1/2 * 1/sqroot(2)]
Simplify...
[sqroot(3)/2sqroot(2)] + [1/2sqroot(2)]
Simplify again...

[sqroot(3)+1]/[2sqroot(2)]
This is the exact value of sin(7pi/12)





Oops haha now I have no time left for the rest of my homework. 
Hope this helped someone. I'm proud of my big messy colored number explosion. 

Hope I didn't make any stupid errors like I keep doing now for some reason.