Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
|dw:1339728896748:dw| a-opposite b-adjacent c-hypotenuse. \[\sin \theta=a/c\] \[\cos \theta=b/c\] \[\tan \theta=a/b\] \[\csc \theta=1/\sin \theta=1/a/c=c/a\] \[\sec \theta=1/\cos \theta=1/b/c=c/b\] \[\cot \theta=1/\tan \theta=1/a/b=b/a\] \[\sin ^{2}\theta+ \cos ^{2}\theta=1\] \[\tan ^{2}\theta+1=\sec ^{2}\theta\] \[1+\cot^{2}\theta=\csc^{2}\theta\] \[tan\theta*\cot\theta=1\] \[sin(A+B)=sinAcosB+sinBcosA\] \[cos(A+B)=cosAcosB+sinAsinB\] \[tan(A+B)=(tanA+tanB)/(1-tanA*tanB)\] \[sin(A-B)=sinAcosB-sinBcosA\] \[cos(A-B)=cosAcosB-sinAsinB\] \[tan(A-B)=(tanA-tanB)/(1+tanA*tanB)\]
\[sin(A+B)+sin(A-B)=2sinAcosB\] \[cos(A+B)+cos(A-B)=2cosAcosB\] \[sin(A+B)-sin(A-B)=2cosAsinB\] \[cos(A-B)-cos(A+B)=2sinAsinB\] \[sin2\theta=2sin\theta\cos\theta\] \[cos2\theta=cos^2\theta-sin^2\theta\] \[cos2\theta=1-2sin^2\theta\] \[cos2\theta=2cos^2\theta-1\] \[tan2\theta=2tan\theta/(1-tan^2\theta)\] \[sin3\theta=3sin\theta-4sin^3\theta\] \[cos3\theta=4cos^3\theta-3cos\theta\] \[sin(-\theta)=-sin\theta\] \[cos(-\theta)=cos\theta\] \[tan(-theta)=-tan\theta\]
\[sin(90-\theta)=cos\theta\] \[cos(90-\theta)=sin\theta\] \[tan(90-\theta)=cot\theta\] \[sin(180-\theta)=sin\theta\] \[cos(180-\theta)=-cos\theta\] \[tan(180-\theta)=-tan\theta\] \[sin(180+\theta)=-sin\theta\] \[cos(180+\theta)=-cos\theta\] \[tan180+\theta)=tan\theta\] \[sin(90+\theta)=cos\theta\] \[cos(90+\theta)=-sin\theta\] \[tan(90+\theta)=-cot\theta\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

1 Attachment
Nice & thanx @ajprincess :)
U r welcome nd thanxx. I thank all who gav me medals for givng them.:)
nice :)
very nice:)
Thanxxx a lot for the comments.:)
seems like you used in few ,,,
this is a true tutorial indeed \[\Huge \color{maroon}{\mathtt{\text{}}}\]
Thanx a lot @lgbasallote
how should I memorize these?
I liked it :) Good Job making it :)
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.
Simple, clear, lucid, and great! :P
Thanx a lot @apoorvk
Wow reaching the same target as lgba {30 medals} But hope u would get more than that ^_^
Thanx a lot @maheshmeghwal9
:)
good tutorial must appreciate ...
Thanx a lot @mathslover.
no thanks @ajprincess actually many of students present here needed this
:)
Good job :) I actually didn't know the negative and triple identities.
thanx a lot @calcmathlete
that was a good revision.....
Thanx a lot @phani09
@ajprincess , I want to tell you something if you will not feel bad.. May I tell??
http://www.sosmath.com/trig/Trig5/trig5/trig5.html
ya sure. @waterineyes.
Can you tell me the formulas for \(cos(A + B) \quad and \quad cos(A-B)\) ??
@ajprincess, do you think that you have given the right identities for \(cos(A + B) \quad and \quad cos(A - B)\) ??
@waterineyes is correct. Those two identities are incorrect. \[cos(A + B) = \cos A\cos B - \sin A\sin B\]\[cos(A - B) = \cos A\cos B + \sin A\sin B\]
See, the difference between \(sin(A \pm B) \quad And \quad cos(A \pm B)\): \[\Large \color{green}{\sin(A \pm B) = sinA.cosB \pm cosA.sinB}\] \[\Large \color{green}{\cos(A \pm B) = cosA.cosB \mp cosA.cosB}\]
@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??
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.
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 ..
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question