cot2x + csc2x = 2csc2x - 1 my answer > cot^2+csc^4*x^3
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.??
@freckles
so those things in your problem are not double angles right?
idont thinks so no there not
looks like square not double angle that way the left hand side is correct
yes
so the problem is: \[\cot^2(x)+\csc^2(x)=2\csc^2(x)-1\]
yes
this problem wants you to use the id 1+cot^2=csc^2
well let's play with this: \[\cot^2(x)+1=\csc^2(x)\] subtract 1 on both sides
this will isolate the cot^2(x) term
in then you will be able to write left hand side in terms of csc
umm okkk
do this one step: subtract 1 on both sides of cot^2(x)+1=csc^2(x)
cot^2(x)csc^2 right
what happen to the equation
if I asked you to subtract 1 on both sides of x+1=y what would you get after just doing that one step
you would get x=y
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
x+1=y subtract 1 on both sides (x+1)-1=y-1 x+1-1=y-1 x=y-1
ohh ok
\[\cot^2(x)+1=\csc^2(x) \\ \text{ subtract 1 on both sides here }\]
cot^2(X)=CSC^2(X)-1
right \[\cot^2(x)+\csc^2(x) \text{ was your left hand side } \\ \text{ replace} \cot^2(x) \text{ with } \csc^2(x)-1\]
\[(\csc^2(x)-1)+\csc^2(x) \\ \text{ you can drop ( ) } \\ \csc^2(x)-1+\csc^2(x)\] do you think you can finish here:
you know how to combine like terms right?
no
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)=?
or not find but simplify
(2*x)*csc^2
what happen to csc^2 's angle part?
??
csc^2 has no meaning csc^2(x) has meaning
something has to be plugged into the csc^2 function to actually have meaning csc^2 alone means nothing
so (2*x)*csc^2(X)
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
where does the extra x come from
\[\csc^2(x)+\csc^2(x)=2 \csc^2(x) \]
ohh i see what you mena venr mind that last statment lol
anyways do you have any questions on this one before I take off for a bit
id thnx alot
np
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