Melody

Lets see..

\({sin(2x)-cos(x) \over 4sin^2x - 1} = {sin^2 x (cos x) + cos^2x \over 2 sin(x)+1}\\ LHS\\ ={sin(2x)-cos(x) \over 4sin^2x - 1} \\ =\frac{2sinxcosx-cos(x) } {(2sinx - 1)(2sinx+1)} \\ =\frac{cosx(2sinx-1) } {(2sinx - 1)(2sinx+1)} \\ =\frac{cosx } {2sinx+1} \\ \)

\(RHS\\ ={sin^2 x (cos x) + cos^2x \over 2 sin(x)+1}\\ \ne LHS\)

I suspect this is an error on our part on on the part of your teacher or text.

\((6cos^2 24°-6sin^2 24°)tan48°\)

I used the double angle identity \(cos2A = cos^2 A-sin^2 A \)

and I got \(6cos2(24)°(tan48°)\)

I could change the tan to sin/cos but then I'm not sure what to do next?

--------

\(6cos[\color{red}{2(24)}\color{black}{]°(tan48°)}\\ =6cos[\color{red}{48}\color{black}{]°(tan48°)}\\ =6cos[\color{red}{48}\color{black}{]°\cdot\frac{sin48}{cos48}}\\ =6sin48\)

Umm

The asyptotes are true for both graphs so they do not effect the identity.

The holes are only present for the right hand side so they are the ones that should be stated as non-permissable for the identity.

It is fine to say it is incorrect but please give the reason that you think so. 

No e is not a coefficient. e is the constant.

Coefficients are the numbers in front of the pronumerals

(if the pronumeral is 1 then it is invisable but it is still there - that is not relevant here though.)

Yea,

Set notation is the absolute worst.

Mathematicians like to have their own private  language. 

I often think that 'mathematics' is my second language.  I am just not all that literate in it.   LOL.

Thanks Chris and Rom,

I always get confused by this notation too.    

sin ^ 2x-9cos ^ 2x If cos = -0.9

\(find\;\;\;sin ^ 2x-9cos ^ 2x\;\;\;\;\; If \;\;\;\;cosx = -0.9\\~\\ sin ^ 2x-9cos ^ 2x\\ =sin ^ 2x-9cos ^ 2x+cos^2x-cos^2x\\ =(sin ^ 2x+cos^2x)-10cos ^ 2x\\ =1-10cos ^ 2x\\ =1-10*0.81\\ =1-8.1\\ =-7.1\)

I am not sure what you problems is. I have graphed them here