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

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1. freckles

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

2. freckles

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

3. freckles

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

4. anonymous

Thank you this is helping!

5. freckles

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

6. freckles

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

7. freckles

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

8. freckles

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

9. anonymous

Oh! I see!

10. 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

11. 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.

12. freckles

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

13. freckles

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

14. anonymous

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

15. 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.

16. 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

17. 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 .

18. 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?

19. 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) $\frac{(x-1)(x+1)}{x+1} \text{ but } \frac{x+1}{x+1}=1 \text{ as long as } x \neq -1 \\ \text{ so } \frac{x^2-1}{x+1}=\frac{(x-1)(x+1)}{x+1}=\frac{(x-1) \cancel{(x+1)}}{\cancel{x+1}} ; x \neq -1 \\ \text{ so } \frac{x^2-1}{x+1}=x-1 \text{ is an identity as long as } x \neq -1$

20. anonymous

Thank you so so much sir!!!!

21. 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 $\frac{\sin^2(\theta)-1}{\sin(\theta)+1}=\sin(\theta)-1 \text{ is an identity as long as } \\ \theta \neq \frac{3 \pi}{2}+2\pi n \text{ where } n \text{ is an integer }$

22. freckles

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

23. anonymous

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

24. 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 $\cos^2(b-\alpha)+\sin^2(b-\alpha)=1$

25. 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

26. freckles

|dw:1437620598502:dw|

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