Can anyone help me to understand trig identities? Thanks!

- anonymous

Can anyone help me to understand trig identities? Thanks!

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- chestercat

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- freckles

well the most basic one is the Pythagorean identities
This circle I drew has radius 1 and center (0,0).
|dw:1437618239963:dw|

- freckles

|dw:1437618291353:dw|
Say I choose a random point on there (x,y)
such that x^2+y^2=1

- freckles

|dw:1437618328438:dw|
And I draw a right triangle

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## More answers

- anonymous

Thank you this is helping!

- freckles

and remember this circle has radius 1
|dw:1437618385419:dw|

- freckles

cos(theta)=x/1=x
sin(theta)=y/1=y

- freckles

\[x^2+y^2=1 \\ (\cos(\theta))^2+(\sin(\theta))^2=1 \\ \text{ replaced } x \text{ with } \cos(\theta) \text{ since } x=\cos(\theta) \\ \text{ replaced } y \text{ with } \sin(\theta) \text{ since } y=\sin(\theta)\]

- freckles

\[\cos^2(\theta)+\sin^2(\theta)=1 \text{ is a Pythagorean identitiy }\]

- anonymous

Oh! I see!

- freckles

I could divide both sides by cos^2(theta)
or sin^2(theta) to make the identity look a bit different
but they are still Pythagorean identities

- anonymous

Right, that's what i saw on the chart i looked up of basic identities.. i saw the identities but had no clue what they meant.

- freckles

dividing both sides by cos^2(theta) gives:
\[\frac{\cos^2(\theta)+\sin^2(\theta)}{\cos^2(\theta)}=\frac{1}{\cos^2(\theta)} \\ \frac{\cos^2(\theta)}{\cos^2(\theta)}+\frac{\sin^2(\theta)}{\cos^2(\theta)}=\frac{1}{\cos^2(\theta)} \\ 1+\tan^2(\theta)=\sec^2(\theta)\]

- freckles

dividing both sides by sin^2(theta) gives:
\[\frac{\cos^2(\theta)+\sin^2(\theta)}{\sin^2(\theta)}=\frac{1}{\sin^2(\theta)} \\ \frac{\cos^2(\theta)}{\sin^2(\theta)}+\frac{\sin^2(\theta)}{\sin^2(\theta)}=\frac{1}{\sin^2(\theta)} \\ \cot^2(\theta)+1=\csc^2(\theta)\]

- anonymous

Fantastic, thank you! Could you explain how identities might look on a graph?

- freckles

identities on a graph?
Like you mean you want me to graph
f(x)=sin^2(x)+cos^2(x)
though this is really saying
f(x)=1
so you would be graphing the constant function y=1.

- anonymous

oh ok, Sometimes we are asked "which of these id's are the same" and the teacher says "think about it on a graph" if that helps make my question

- freckles

Say you have f(x)=g(x) is an identity
if you graph y=f(x) and y=g(x) they should be lying on top of each other .

- anonymous

right i know thats the idea but is there a way to tell what id's are the same without looking on a graph every time?

- freckles

you can use other identities to prove it
for example:
Say we want to disprove or prove the following is an identity:
\[\frac{x^2-1}{x+1}=x-1 \]
We could use the identity called difference of squares on the x^2-1
(x^2-1)=(x-1)(x+1)
\[\frac{(x-1)(x+1)}{x+1} \text{ but } \frac{x+1}{x+1}=1 \text{ as long as } x \neq -1 \\ \text{ so } \frac{x^2-1}{x+1}=\frac{(x-1)(x+1)}{x+1}=\frac{(x-1) \cancel{(x+1)}}{\cancel{x+1}} ; x \neq -1 \\ \text{ so } \frac{x^2-1}{x+1}=x-1 \text{ is an identity as long as } x \neq -1 \]

- anonymous

Thank you so so much sir!!!!

- freckles

we can do the same thing with a trig expression:
\[\frac{\sin^2(\theta)-1}{\sin(\theta)+1} =\sin(\theta)-1 \\ \text{ say we want \to prove this is or isn't an identity } \\ \text{ well } \sin^2(\theta)-1=(\sin(\theta)-1)(\sin(\theta)+1) \text{ by difference of squares } \\ \frac{\sin^2(\theta)-1}{\sin(\theta)+1}=\sin(\theta)-1 \text{ as long as } \sin(\theta) \neq -1 \]
You could actually figure out for what theta sin(theta) is -1
and exclude them
\[\frac{\sin^2(\theta)-1}{\sin(\theta)+1}=\sin(\theta)-1 \text{ is an identity as long as } \\ \theta \neq \frac{3 \pi}{2}+2\pi n \text{ where } n \text{ is an integer }\]

- freckles

There are other trig identities though besides the Pythagorean identity but that is a very lengthy discussion

- anonymous

Yea i understand, i think ill be better at figuring those out now that i have at least the basic knowledge!
Thanks again!

- freckles

Say we have two angles with the coordinates given:
|dw:1437619665480:dw|
|dw:1437619706210:dw|
Say I put both of these on the same picture:
|dw:1437619765363:dw|
And that broken line symbolizing I'm going to find the distance between the two points...
\[d_1=\sqrt{(\cos(\alpha)-\cos(b))^2+(\sin(\alpha)-\sin(b))^2} \\ d_1^2=(\cos(\alpha)-\cos(b))^2+(\sin(\alpha)-\sin(b))^2 \\ \text{ now recall } (x-y)^2=x^2+y^2-2xy \\ d_1^2=\cos^2(\alpha)+\cos^2(b)-2 \cos(\alpha) \cos(b)+\sin^2(\alpha)+\sin^2(b)-2 \sin(\alpha)\sin(b) \\ \text{ recall the Pythagorean identity } \cos^2(\alpha)+\sin^2(\alpha)=1 \text{ and you can also say } \\ \cos^2(b)+\sin^2(b)=1 \\ d_1^2=2-2\cos(\alpha)\cos(b)-2 \sin(a) \sin(b) \\ d_1^2=2-2(\cos(\alpha) \cos(b)+ \sin(a) \sin(b)) \\ \text{ but notice } \\ \\ \text{ the distance \between the points } (\cos(b-\alpha),\sin(b-\alpha)) \text{ and } (1,0) \text{ should provide the}\]
\[\text{ same distance } \\ d_1^2= (\cos(b-\alpha) -1)^2+(\sin(b-\alpha)-0)^2 \\ d_1^2=\cos^2(b-a) -2\cos(b-\alpha)+1 +\sin^2(b-a) \\ d_1^2=-2\cos(b-\alpha)+2 \]
By the way I did use
\[\cos^2(b-\alpha)+\sin^2(b-\alpha)=1 \]

- freckles

anyways we have then that:
\[2-2(\cos(\alpha)\cos(b)+\sin(\alpha)\sin(b))=2-2 \cos(b-\alpha) \\ \text{ which means } \cos(\alpha)\cos(b)+\sin(\alpha)\sin(b)=\cos(b-\alpha)\]
this is the difference identity for cosine

- freckles

|dw:1437620598502:dw|

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