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anonymous
 one year ago
Can anyone help me to understand trig identities? Thanks!
anonymous
 one year ago
Can anyone help me to understand trig identities? Thanks!

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freckles
 one year ago
Best ResponseYou've already chosen the best response.3well the most basic one is the Pythagorean identities This circle I drew has radius 1 and center (0,0). dw:1437618239963:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.3dw:1437618291353:dw Say I choose a random point on there (x,y) such that x^2+y^2=1

freckles
 one year ago
Best ResponseYou've already chosen the best response.3dw:1437618328438:dw And I draw a right triangle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you this is helping!

freckles
 one year ago
Best ResponseYou've already chosen the best response.3and remember this circle has radius 1 dw:1437618385419:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.3cos(theta)=x/1=x sin(theta)=y/1=y

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[x^2+y^2=1 \\ (\cos(\theta))^2+(\sin(\theta))^2=1 \\ \text{ replaced } x \text{ with } \cos(\theta) \text{ since } x=\cos(\theta) \\ \text{ replaced } y \text{ with } \sin(\theta) \text{ since } y=\sin(\theta)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\cos^2(\theta)+\sin^2(\theta)=1 \text{ is a Pythagorean identitiy }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I 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
 one year ago
Best ResponseYou've already chosen the best response.0Right, 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
 one year ago
Best ResponseYou've already chosen the best response.3dividing 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
 one year ago
Best ResponseYou've already chosen the best response.3dividing 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
 one year ago
Best ResponseYou've already chosen the best response.0Fantastic, thank you! Could you explain how identities might look on a graph?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3identities 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
 one year ago
Best ResponseYou've already chosen the best response.0oh 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
 one year ago
Best ResponseYou've already chosen the best response.3Say 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
 one year ago
Best ResponseYou've already chosen the best response.0right 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
 one year ago
Best ResponseYou've already chosen the best response.3you can use other identities to prove it for example: Say we want to disprove or prove the following is an identity: \[\frac{x^21}{x+1}=x1 \] We could use the identity called difference of squares on the x^21 (x^21)=(x1)(x+1) \[\frac{(x1)(x+1)}{x+1} \text{ but } \frac{x+1}{x+1}=1 \text{ as long as } x \neq 1 \\ \text{ so } \frac{x^21}{x+1}=\frac{(x1)(x+1)}{x+1}=\frac{(x1) \cancel{(x+1)}}{\cancel{x+1}} ; x \neq 1 \\ \text{ so } \frac{x^21}{x+1}=x1 \text{ is an identity as long as } x \neq 1 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so so much sir!!!!

freckles
 one year ago
Best ResponseYou've already chosen the best response.3we 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
 one year ago
Best ResponseYou've already chosen the best response.3There are other trig identities though besides the Pythagorean identity but that is a very lengthy discussion

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea i understand, i think ill be better at figuring those out now that i have at least the basic knowledge! Thanks again!

freckles
 one year ago
Best ResponseYou've already chosen the best response.3Say 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... \[d_1=\sqrt{(\cos(\alpha)\cos(b))^2+(\sin(\alpha)\sin(b))^2} \\ d_1^2=(\cos(\alpha)\cos(b))^2+(\sin(\alpha)\sin(b))^2 \\ \text{ now recall } (xy)^2=x^2+y^22xy \\ d_1^2=\cos^2(\alpha)+\cos^2(b)2 \cos(\alpha) \cos(b)+\sin^2(\alpha)+\sin^2(b)2 \sin(\alpha)\sin(b) \\ \text{ recall the Pythagorean identity } \cos^2(\alpha)+\sin^2(\alpha)=1 \text{ and you can also say } \\ \cos^2(b)+\sin^2(b)=1 \\ d_1^2=22\cos(\alpha)\cos(b)2 \sin(a) \sin(b) \\ d_1^2=22(\cos(\alpha) \cos(b)+ \sin(a) \sin(b)) \\ \text{ but notice } \\ \\ \text{ the distance \between the points } (\cos(b\alpha),\sin(b\alpha)) \text{ and } (1,0) \text{ should provide the}\] \[\text{ same distance } \\ d_1^2= (\cos(b\alpha) 1)^2+(\sin(b\alpha)0)^2 \\ d_1^2=\cos^2(ba) 2\cos(b\alpha)+1 +\sin^2(ba) \\ d_1^2=2\cos(b\alpha)+2 \] By the way I did use \[\cos^2(b\alpha)+\sin^2(b\alpha)=1 \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3anyways we have then that: \[22(\cos(\alpha)\cos(b)+\sin(\alpha)\sin(b))=22 \cos(b\alpha) \\ \text{ which means } \cos(\alpha)\cos(b)+\sin(\alpha)\sin(b)=\cos(b\alpha)\] this is the difference identity for cosine

freckles
 one year ago
Best ResponseYou've already chosen the best response.3dw:1437620598502:dw
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