## ajprincess Note to the viewers:This is not a question but short notes on trigonometric identities that may be of some use. one year ago one year ago

1. ajprincess

|dw:1339728896748:dw| a-opposite b-adjacent c-hypotenuse. $\sin \theta=a/c$ $\cos \theta=b/c$ $\tan \theta=a/b$ $\csc \theta=1/\sin \theta=1/a/c=c/a$ $\sec \theta=1/\cos \theta=1/b/c=c/b$ $\cot \theta=1/\tan \theta=1/a/b=b/a$ $\sin ^{2}\theta+ \cos ^{2}\theta=1$ $\tan ^{2}\theta+1=\sec ^{2}\theta$ $1+\cot^{2}\theta=\csc^{2}\theta$ $tan\theta*\cot\theta=1$ $sin(A+B)=sinAcosB+sinBcosA$ $cos(A+B)=cosAcosB+sinAsinB$ $tan(A+B)=(tanA+tanB)/(1-tanA*tanB)$ $sin(A-B)=sinAcosB-sinBcosA$ $cos(A-B)=cosAcosB-sinAsinB$ $tan(A-B)=(tanA-tanB)/(1+tanA*tanB)$

2. ajprincess

$sin(A+B)+sin(A-B)=2sinAcosB$ $cos(A+B)+cos(A-B)=2cosAcosB$ $sin(A+B)-sin(A-B)=2cosAsinB$ $cos(A-B)-cos(A+B)=2sinAsinB$ $sin2\theta=2sin\theta\cos\theta$ $cos2\theta=cos^2\theta-sin^2\theta$ $cos2\theta=1-2sin^2\theta$ $cos2\theta=2cos^2\theta-1$ $tan2\theta=2tan\theta/(1-tan^2\theta)$ $sin3\theta=3sin\theta-4sin^3\theta$ $cos3\theta=4cos^3\theta-3cos\theta$ $sin(-\theta)=-sin\theta$ $cos(-\theta)=cos\theta$ $tan(-theta)=-tan\theta$

3. ajprincess

$sin(90-\theta)=cos\theta$ $cos(90-\theta)=sin\theta$ $tan(90-\theta)=cot\theta$ $sin(180-\theta)=sin\theta$ $cos(180-\theta)=-cos\theta$ $tan(180-\theta)=-tan\theta$ $sin(180+\theta)=-sin\theta$ $cos(180+\theta)=-cos\theta$ $tan180+\theta)=tan\theta$ $sin(90+\theta)=cos\theta$ $cos(90+\theta)=-sin\theta$ $tan(90+\theta)=-cot\theta$

4. ajprincess

5. maheshmeghwal9

Nice & thanx @ajprincess :)

6. ajprincess

U r welcome nd thanxx. I thank all who gav me medals for givng them.:)

7. alexwee123

nice :)

8. maheshmeghwal9

very nice:)

9. ajprincess

Thanxxx a lot for the comments.:)

10. experimentX

seems like you used in few ,,,

11. lgbasallote

this is a true tutorial indeed $\Huge \color{maroon}{\mathtt{\text{<tips hat>}}}$

12. ajprincess

Thanx a lot @lgbasallote

13. WONDEMU

how should I memorize these?

14. Loujoelou

I liked it :) Good Job making it :)

15. ajprincess

Thanx a lot @Loujoelou and as for question @wondemu memorise them group by group. For example (1)sin(A+B)=sinAcosB+sinBcosA cos(A+B)=cosAcosB+sinAsinB tan(A+B)=(tanA+tanB)/(1−tanA∗tanB) (2)sin(A−B)=sinAcosB−sinBcosA cos(A−B)=cosAcosB−sinAsinB tan(A−B)=(tanA−tanB)/(1+tanA∗tanB) You'll find them useful when u hav finishd memorising them.

16. apoorvk

Simple, clear, lucid, and great! :P

17. ajprincess

Thanx a lot @apoorvk

18. maheshmeghwal9

Wow reaching the same target as lgba {30 medals} But hope u would get more than that ^_^

19. ajprincess

Thanx a lot @maheshmeghwal9

20. maheshmeghwal9

:)

21. mathslover

good tutorial must appreciate ...

22. ajprincess

Thanx a lot @mathslover.

23. mathslover

no thanks @ajprincess actually many of students present here needed this

24. ajprincess

:)

25. Calcmathlete

Good job :) I actually didn't know the negative and triple identities.

26. ajprincess

thanx a lot @calcmathlete

27. phani09

that was a good revision.....

28. ajprincess

Thanx a lot @phani09

29. waterineyes

@ajprincess , I want to tell you something if you will not feel bad.. May I tell??

30. eliassaab
31. ajprincess

ya sure. @waterineyes.

32. waterineyes

Can you tell me the formulas for $$cos(A + B) \quad and \quad cos(A-B)$$ ??

33. waterineyes

@ajprincess, do you think that you have given the right identities for $$cos(A + B) \quad and \quad cos(A - B)$$ ??

34. Calcmathlete

@waterineyes is correct. Those two identities are incorrect. $cos(A + B) = \cos A\cos B - \sin A\sin B$$cos(A - B) = \cos A\cos B + \sin A\sin B$

35. waterineyes

See, the difference between $$sin(A \pm B) \quad And \quad cos(A \pm B)$$: $\Large \color{green}{\sin(A \pm B) = sinA.cosB \pm cosA.sinB}$ $\Large \color{green}{\cos(A \pm B) = cosA.cosB \mp cosA.cosB}$

36. waterineyes

@ajprincess , don't misunderstand me and don't think I always look for mistakes but everywhere you used this identity wrong.. So, I think your mind says it is correct but sorry to say it is incorrect... In sin(A + B) there is respective + in the middle, In sin(A - B) there is respective - in the middle.. But for cos(A + B), there is - in the middle, for cos(A - B), there is + in the middle... Getting it??

37. ajprincess

oh no it's k. Thanx for telling t. i copied it from a note of mine. i must hav written it wrong when I wrote. thanxx a lot for mentioning it. I welcome this sorts of comments. @waterineyes.

38. waterineyes

26 people gave you medals and I think you deserve more than that.. But how will you let it know to all the 26 people that you have written slight wrong there so that they can know the real formula... Think about it.. @ajprincess ..

39. ajprincess

it's my mistake nd I am greatly sorry for t. Let me post t again in a new post so that they can get . I am nt sure of any other way other than this.