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Mathematics 12 Online
OpenStudy (anonymous):

Verify the identity: [sin(3t)-sin(t)]/[cos(3t)+cos(t)]=tan(t)

OpenStudy (anonymous):

\[\Huge \frac{[\sin(3t)-\sin(t)]}{[\cos(3t)+\cos(t)]}~=\tan(t)\]??????

OpenStudy (anonymous):

yes. i just had the brackets to make it look a little easier to read. i am stumped on this one

OpenStudy (anonymous):

Ok, sorry, do you know about De Moivre's theorem?

OpenStudy (anonymous):

No, i have not gotten to that in the class yet. this is just supposed to be verified using indentities

OpenStudy (anonymous):

Ok, well Iw ould just say that \[\Large \sin(3t)= 3\cos^2t \sin t−\sin^3t\] and \[\Large \cos(3t)=\cos ^3 θ − 3 \cosθ \sin^2 θ\]

OpenStudy (anonymous):

What we want to end up with is sin(t)/cos(t) becaues that equals tan(t)

OpenStudy (anonymous):

\[\Large \frac {3\cos^2t \sin t−\sin^3t-\sin t}{\cos ^3 θ − 3 \cosθ \sin^2 θ+ \cos t}\]

OpenStudy (anonymous):

\[\Large \frac {\sin t (3\cos^2t −\sin^2t-1)}{\cos t (\cos ^2 t −3 \sin^2 t+1)}\]

OpenStudy (anonymous):

\[\Large \tan t \times \frac { (−4\sin^2t+2)}{ (−4 \sin^2 t+2)}\]

OpenStudy (anonymous):

Sorry I did a lot of simplifying in this last step

OpenStudy (anonymous):

lol yeah im a little lost at that point but thank you i think i can go from the last step and get to where your at! thank you!

OpenStudy (anonymous):

No, thats my b, i just i went through it lol, do you still want the full tutorial?

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