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Trigonometry 14 Online
OpenStudy (anonymous):

sin2x/cosx

OpenStudy (shaik0124):

sin2x=2sinxcosx therforesin2x/cos2x= 2sinxcosc/cosx=2sinx is answer

OpenStudy (zehanz):

Hey @Davidrrp, you know now that sin2x=2sinxcosx. Put that in the place of sin2x!

OpenStudy (anonymous):

i know sin2x = 2sinxcosx

OpenStudy (zehanz):

@shaik0124: good to be helping, but better not give the complete answer away...

OpenStudy (shaik0124):

ok

OpenStudy (zehanz):

So you have: \[\frac{ 2\sin x \cos x }{ \cos x }\]Remember what the last step was in the other problem?

OpenStudy (anonymous):

cancel out on both sides

OpenStudy (zehanz):

Yes! So now you know the answer...

OpenStudy (anonymous):

i also know cosx = 1/secx

OpenStudy (zehanz):

That is also true, but because you've just cancelled out cos x, there is no use for it now. Save it for another occasion!

OpenStudy (anonymous):

so then cosx cancels up and bottom?

OpenStudy (anonymous):

oooh i see

OpenStudy (zehanz):

Just like the other one... That is what you have learned, dude! Next time you can do it yourself, no excuse as: "I don't know what to do" ;)

OpenStudy (anonymous):

lol i'll try my best. i still got more though lol

OpenStudy (zehanz):

Just keep on practising!

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