Mathematics
22 Online
OpenStudy (anonymous):
Use basic identities to simplify the expression:
cos(x)-cos^3(x)
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OpenStudy (anonymous):
I have no clue which equations to even look at. I know the answer but not how to get to it.
OpenStudy (anonymous):
factorise cos x
OpenStudy (anonymous):
I dont know how :/
OpenStudy (anonymous):
can you factor this \[a-a^3\]
OpenStudy (anonymous):
factor a
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OpenStudy (anonymous):
no...
OpenStudy (anonymous):
I dont understand...its in its lowest form?
OpenStudy (anonymous):
\[a-a^3=a(1-a^2)\]
OpenStudy (anonymous):
do you understand this
OpenStudy (anonymous):
wouldnt it be a(1-a)? cause youre gonna factor?
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OpenStudy (anonymous):
but it is a^2 cos when we multiply out that it has to give us the original expression
ie a-a^3
OpenStudy (anonymous):
but a(1-a)=a-a*a=a-a^2 not a-a^3
OpenStudy (anonymous):
I dont follow :(
OpenStudy (anonymous):
can you factor this
\[x^3+x^4\]
OpenStudy (anonymous):
is that a superscript 3 or 2
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OpenStudy (anonymous):
x^3(1+x)
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
ok...
OpenStudy (anonymous):
so \[x-x^3\]
OpenStudy (anonymous):
factor this
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OpenStudy (anonymous):
cosx(1-cos^2x)
OpenStudy (anonymous):
yes so use this identity
\[\cos ^2x+\sin^2x=1\]...hint take \[\cos ^2x\] to the other side
OpenStudy (anonymous):
sin^2x-1=cos^2x?
OpenStudy (anonymous):
or do i leave the 1
OpenStudy (anonymous):
yes leave the 1
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OpenStudy (anonymous):
sin^2x=cos^2x+1
OpenStudy (anonymous):
cos went to the other side so it is negetive
OpenStudy (anonymous):
oh ok
sin^2x=1-cos^2x
OpenStudy (anonymous):
yes now go back to your question and look if it is similar to this
OpenStudy (anonymous):
i dont understand how you decided to use that identity...
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OpenStudy (anonymous):
and thats not the right answer..
OpenStudy (anonymous):
\[\cos x(\color{red}{1-\cos ^2x})=\cos x \color{red}{\sin^2x}\]
red part from identity
OpenStudy (anonymous):
This isnt easy =..erg.. alright
OpenStudy (anonymous):
is this the answer or not
OpenStudy (anonymous):
no
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OpenStudy (anonymous):
cos(x)sin^2(x) is the answer
OpenStudy (anonymous):
yes thats the one i wrote
OpenStudy (anonymous):
i dont see it...
OpenStudy (primeralph):
Still need help?
OpenStudy (anonymous):
yes
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OpenStudy (anonymous):
please
OpenStudy (primeralph):
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