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Mathematics 7 Online
OpenStudy (anonymous):

cot2x + csc2x = 2csc2x - 1 my answer > cot^2+csc^4*x^3

OpenStudy (anonymous):

Teacher:You cannot take terms across the = when doing a proof. You must show that the LHS=RHS. See if you can use cot^2(x)+1 = csc^2(x) to help you do this.??

OpenStudy (anonymous):

@freckles

OpenStudy (freckles):

so those things in your problem are not double angles right?

OpenStudy (anonymous):

idont thinks so no there not

OpenStudy (xapproachesinfinity):

looks like square not double angle that way the left hand side is correct

OpenStudy (anonymous):

yes

OpenStudy (freckles):

so the problem is: \[\cot^2(x)+\csc^2(x)=2\csc^2(x)-1\]

OpenStudy (anonymous):

yes

OpenStudy (xapproachesinfinity):

this problem wants you to use the id 1+cot^2=csc^2

OpenStudy (freckles):

well let's play with this: \[\cot^2(x)+1=\csc^2(x)\] subtract 1 on both sides

OpenStudy (freckles):

this will isolate the cot^2(x) term

OpenStudy (freckles):

in then you will be able to write left hand side in terms of csc

OpenStudy (anonymous):

umm okkk

OpenStudy (freckles):

do this one step: subtract 1 on both sides of cot^2(x)+1=csc^2(x)

OpenStudy (anonymous):

cot^2(x)csc^2 right

OpenStudy (freckles):

what happen to the equation

OpenStudy (freckles):

if I asked you to subtract 1 on both sides of x+1=y what would you get after just doing that one step

OpenStudy (anonymous):

you would get x=y

OpenStudy (freckles):

i'm asking you to subtract one on both sides not one side remember whatever you do to one side of the equation you have to do to the other

OpenStudy (freckles):

x+1=y subtract 1 on both sides (x+1)-1=y-1 x+1-1=y-1 x=y-1

OpenStudy (anonymous):

ohh ok

OpenStudy (freckles):

\[\cot^2(x)+1=\csc^2(x) \\ \text{ subtract 1 on both sides here }\]

OpenStudy (anonymous):

cot^2(X)=CSC^2(X)-1

OpenStudy (freckles):

right \[\cot^2(x)+\csc^2(x) \text{ was your left hand side } \\ \text{ replace} \cot^2(x) \text{ with } \csc^2(x)-1\]

OpenStudy (freckles):

\[(\csc^2(x)-1)+\csc^2(x) \\ \text{ you can drop ( ) } \\ \csc^2(x)-1+\csc^2(x)\] do you think you can finish here:

OpenStudy (freckles):

you know how to combine like terms right?

OpenStudy (anonymous):

no

OpenStudy (freckles):

examples of combining like terms: 5+5=2(5)=10 a+a=2a x^2+x^2=2x^2 log(x)+log(x)=2log(x) sin(x)+sin(x)=2sin(x) sin^2(x)+sin^2(x)=2sin^2(x) sin^2(x)+4sin^2(x)=5sin^2(x) can you find this: csc^2(x)+csc^2(x)=?

OpenStudy (freckles):

or not find but simplify

OpenStudy (anonymous):

(2*x)*csc^2

OpenStudy (freckles):

what happen to csc^2 's angle part?

OpenStudy (anonymous):

??

OpenStudy (freckles):

csc^2 has no meaning csc^2(x) has meaning

OpenStudy (freckles):

something has to be plugged into the csc^2 function to actually have meaning csc^2 alone means nothing

OpenStudy (anonymous):

so (2*x)*csc^2(X)

OpenStudy (freckles):

so I think you meant to write 2csc^2(x) this does not mean csc^2 times x it means csc^2 of x like means x is being plugged into the csc^2( ) function

OpenStudy (freckles):

where does the extra x come from

OpenStudy (freckles):

\[\csc^2(x)+\csc^2(x)=2 \csc^2(x) \]

OpenStudy (anonymous):

ohh i see what you mena venr mind that last statment lol

OpenStudy (freckles):

anyways do you have any questions on this one before I take off for a bit

OpenStudy (anonymous):

id thnx alot

OpenStudy (freckles):

np

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