Mathematics
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OpenStudy (anonymous):
Simplify the trig expression. cos(x)(sec(x)-cos(x))
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OpenStudy (anonymous):
start with \[\cos(x)\left(\frac{1}{\cos(x)}-\cos(x)\right)\]
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
i still have no idea where to go.
OpenStudy (unklerhaukus):
expand the brackets!
OpenStudy (anonymous):
use the distributive property to multiply
then use the most famous trig identity of them all
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OpenStudy (anonymous):
i got 1-cos^2x
OpenStudy (unklerhaukus):
do you remember the most famous trig identity?
OpenStudy (anonymous):
no
OpenStudy (anonymous):
nvm is it sin^2x???
OpenStudy (unklerhaukus):
something^2+something^2 = 1
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OpenStudy (anonymous):
yeah sin^2x+cos^2x=1
OpenStudy (unklerhaukus):
cool,
can you rearrange this to be cos^2x = ....
OpenStudy (anonymous):
=1-sin^2x
OpenStudy (unklerhaukus):
you had
1-cos^2x
now sub in cos^2x = 1-sin^2x
OpenStudy (anonymous):
wait what?
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OpenStudy (anonymous):
im getting sin^2x
OpenStudy (unklerhaukus):
Yep, that's right
OpenStudy (anonymous):
oh ok cool
OpenStudy (anonymous):
what about (tanx+cotx) cotx
OpenStudy (unklerhaukus):
use cot = 1 / tan
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OpenStudy (anonymous):
im actually splittting the expression into 2 fractions. tanx/cotx + cotx/cotx
OpenStudy (anonymous):
thats what i did first anyway
OpenStudy (anonymous):
then cot over cot is just 1.
OpenStudy (anonymous):
so (tanx/cotx)+1
OpenStudy (anonymous):
then what u said cot=1/tan... so my result is = tan^2x+1 which is sec^2x
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OpenStudy (unklerhaukus):
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