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Trigonometry
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OpenStudy (anonymous):
sin2x/cosx
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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
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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!
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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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