## ajprincess Group Title Note to the viewers:This is not a question but short notes on trigonometric identities that may be of some use. 2 years ago 2 years ago

1. ajprincess Group Title

|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 Group Title

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

3. ajprincess Group Title

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

4. ajprincess Group Title

5. maheshmeghwal9 Group Title

Nice & thanx @ajprincess :)

6. ajprincess Group Title

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

7. alexwee123 Group Title

nice :)

8. maheshmeghwal9 Group Title

very nice:)

9. ajprincess Group Title

Thanxxx a lot for the comments.:)

10. experimentX Group Title

seems like you used in few ,,,

11. lgbasallote Group Title

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

12. ajprincess Group Title

Thanx a lot @lgbasallote

13. WONDEMU Group Title

how should I memorize these?

14. Loujoelou Group Title

I liked it :) Good Job making it :)

15. ajprincess Group Title

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 Group Title

Simple, clear, lucid, and great! :P

17. ajprincess Group Title

Thanx a lot @apoorvk

18. maheshmeghwal9 Group Title

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

19. ajprincess Group Title

Thanx a lot @maheshmeghwal9

20. maheshmeghwal9 Group Title

:)

21. mathslover Group Title

good tutorial must appreciate ...

22. ajprincess Group Title

Thanx a lot @mathslover.

23. mathslover Group Title

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

24. ajprincess Group Title

:)

25. Calcmathlete Group Title

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

26. ajprincess Group Title

thanx a lot @calcmathlete

27. phani09 Group Title

that was a good revision.....

28. ajprincess Group Title

Thanx a lot @phani09

29. waterineyes Group Title

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

30. eliassaab Group Title
31. ajprincess Group Title

ya sure. @waterineyes.

32. waterineyes Group Title

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

33. waterineyes Group Title

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

34. Calcmathlete Group Title

@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 Group Title

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 Group Title

@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 Group Title

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 Group Title

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 Group Title

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.