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Yes, they are the main ones.

A lot can be derived just from these.

Your teacher may push you to remember a whole heap more - I mean ones that you can derive.

It depends how quick you can get as to whether you memorize more or not.

You probably haven't done any calculus yet but I am just youwing you the bit where you will need to rearrage trig.

In calculus you may be asked to integrate cos^2x

You can't do this unless you change it.

\(cos(2x)=sin^2x-cos^2x\\ cos(2x)=1-cos^2x-cos^2x\\ cos(2x)=1-2cos^2x\\ 1-2cos^2x=cos(2x)\\ 1-cos(2x)=2cos^2x\\ cos^2x=\frac{1-cos(2x)}{2}\\ cos^2x=\frac{1}{2}-\frac{cos(2x)}{2}\\\)

The RHS you can integrate.  

Now you will probably be told to memorize this resultant rearrangement.  

I work it out EVERY time.  But my method is slow.

Ah. So are the 'double' and 'half' angle identities both derived from the angle addition/difference identities?

So the most important identities to remember would be the sum/difference identities.

sin(a + b) = sin(a)cos(b) + cos(a)sin(b) 
sin(a – b) = sin(a)cos(b) – cos(a)sin(b) 
cos(a + b) = cos(a)cos(b) – sin(a)sin(b) 
cos(a – b) = cos(a)cos(b) + sin(a)sin(b)

As well as:

sin^2(x) + cos^2(x) = 1 (the other identities are easily derived from this).

So most functions with some trig function can be solved using these 2 sets of identities?

This function popped up towards the end of my derivatives chapter, and the book on trig barely covered those identities at all! :( (it mentioned the sin(a+b) identity but never used it).

Thanks!

sin(A+B)=sinACosB-SinBcosA        (This is an identity that you need to memorize.)

therefore

sin(2x)=2cos(x)sin(x) 

therefore

-sin(2x)=-2cos(x)sin(x) 

Do you want a proof for the original identity - I mean for the first line?

h(t) = 2,5(1-0,035t¨)^0,4 m

when is 9 cm/sec

\(h = 2.5(1-0.035t)^{0.4}\;metres\\ h = 250(1-0.035t)^{0.4}\;cm\\ \frac{dh}{dt} = 250*0.4(1-0.035t)^{-0.6}*(-0.035)\\ \frac{dh}{dt} = -250*0.4*0.035(1-0.035t)^{-0.6}\\ \frac{dh}{dt} = -3.5(1-0.035t)^{-0.6}\\ \mbox{Find time when velocity =9cm/sec}\\ 9=-3.5(1-0.035t)^{-0.6}\\ \left(\frac{-90}{35}\right)^{-1/0.6}=((1-0.035t)^{-0.6})^{-1/0.6}\\ \left(\frac{-18}{7}\right)^{-5/3}=1-0.035t\\ 1-0.035t=\left(\frac{-18}{7}\right)^{-5/3}\\ 1-0.035t=-\left(\frac{7}{18}\right)^{5/3}\\ 0.035t=1+\left(\frac{7}{18}\right)^{5/3}\\ t=\frac{1+\left(\frac{7}{18}\right)^{5/3}}{0.035}\\ \)

(1+(7/18)^(5/3))/0.035 = 34.4912396506397926

So if I haven't made a mistake, the speed will be 9cm/sec after  34.5 seconds

It is still infinity :)

Thanks EP  :))

Calculate the time needed for a drip to deliver 65 drops/minute, given a 1.6L solution has a drip rate of 15 drops/mL

I have my own pet method for doing rates like these  :))

\(you\;\;have\\~\\ \frac{65drops}{1minute},\qquad 1600ml, \qquad \frac{15drops}{1ml} \qquad want \;\; minutes?\\~\\ \frac{1minute}{65drops}\times \frac{15drops}{1ml}\times \frac{1600ml}{1}\\~\\ \mbox{drops cancel and ml cancel}\\~\\ =\frac{1minute\times 15\times 1600}{65\times 1 \times 1}\\~\\ =\frac{ 15\times 1600\;}{65}\;minutes\\~\\ \approx369\;minutes\\~\\ \approx\;6hours\;\; and \;\;9\;minutes\\~\\\)

Any questions anyone?

a^2 + a^x = a^10 

If a is a constant and x is a variable then:

\(\begin{align} a^2 + a^x& = a^{10}\\ a^x &= a^{10}-a^2\\ log( a^x)& =log( a^{10}-a^2)\\ xlog( a)& =log( a^{10}-a^2)\\ x& =\frac{log( a^{10}-a^2)}{log(a)}\\ \end{align} \)

if (1 + (1/m))^m = e What is m

\(\displaystyle\lim_{m\rightarrow \infty}\;\; (1 + \frac{1}{m})^m = e \\ and\\ \displaystyle\lim_{m\rightarrow -\infty}\;\; (1 + \frac{1}{m})^m = e \\~\\ (1 + \frac{1}{m})^m \ne e\qquad \mbox{where }m\in R \\\)

Here is the graph

Here's another approach:

.

As \(\sqrt{27}=\sqrt{3^3}\rightarrow3\sqrt{3}\)  then \(\sqrt{3}+\sqrt{27}\rightarrow\sqrt{3}+3\sqrt{3}\rightarrow4\sqrt{3}\)

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CONGRATULATIONS   BRITTANY  !!!!

        That is great...    

You are not bad at math, according to http://www.slate.com/articles/life/classes/2015/09/why_am_i_bad_at_math_take_a_math_class_in_college_and_learn_to_reason_abstractly.html and me.

If u are bad, don't stick it in your head. try to get better, and ESPECIALLY dont share in a math forum.

"3*9=27+4=31 that over 1 gets"

Huh?   Where did you get    ' over 1 '     your question has it   'over SEVEN'    ....read your question!!!!!   AND the '4' is not included in the fraction the way you have it written.....

turns out the answer was actually 31. I BELIEVED IN YOU AND YOU LET ME DOWN lol. It makes complete sense now though. 3*9=27+4=31 that over 1 gets you 31 thanks for the help though

Multiply through to get

3x +4x -8 = 13    add 8 to both sides of the eq   and combine the 'x' terms

7x = 21               divide both sides by  7

x =3

This one equals:

3*(2)  + 12/3 *3^2         |5-7|  =  |-2| = 2

6       +    4 *  9

6 + 36

42

This equals:

4 +3(9)/7 -6

4 + 27/7 -6

-2 + 27/7

-2 + 3 6/7

1 6/7

Do it this way   ( -12) ^2 =144   (use the parentheses ...with the neg sign INSIDE )

You can set it up as a ratio:

.41/x = 1/64

.41 (64) = x = 26.24 "

|3k-2|=2|k+2|

Formula:

\(\begin{array}{|rcll|} \hline |a|^2 &=& a^2 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline |3k-2| &=& 2|k+2| \qquad | \qquad \text{square both sides}\\ (3k-2)^2 &=& 2^2\cdot (k+2)^2 \\ 9k^2-12k+4 &=& 4\cdot (k^2+4k+4) \\ 9k^2-12k+4 &=& 4k^2+16k+16 \\ \dots \\ 5k^2-28k-12 &=& 0 \\\\ k_{1,2} &=& \frac{28 \pm \sqrt{28^2-4\cdot 5\cdot (-12) } }{2\cdot 5} \\ k_{1,2} &=& \frac{28 \pm \sqrt{1024 } }{10} \\ k_{1,2} &=& \frac{28 \pm 32 } {10} \\\\ k_1 &=& \frac{28 + 32 } {10} \\ k_1 &=& \frac{ 60 } {10} \\ \mathbf{ k_1 }& \mathbf{=} & \mathbf{6} \\\\ k_2 &=& \frac{28 - 32 } {10} \\ k_2 &=& \frac{ -4 } {10} \\ \mathbf{ k_2 } & \mathbf{=} & \mathbf{-0.4} \\ \hline \end{array}\)

OK.....but I think you messed up your question....

I will assume    65 drops/ min  and  15 drops / ml    how much time to 'drip out '  1.6 liter

65 d/min   /   15 d/ ml   =  4.3333   ml/min

1600 ml /4.3333 ml/min = 369.23 min  = 6 hrs 9.23 min

Area = pi  r^2

Circ = pi  d          I would say you would use the pi key and the 'squared' key the most....maybe the = or x  key (haha)