anonymous
  • anonymous
ANYBODY HELP !!!!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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whpalmer4
  • whpalmer4
Which problem do you need help solving?
anonymous
  • anonymous
any

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More answers

whpalmer4
  • whpalmer4
Okay, let's do the first one. |dw:1371431326484:dw| a. Find the length of side b to two decimal places. b. Find the length of side a to two decimal places in three different ways. Any idea how to tackle this?
anonymous
  • anonymous
yeah im sorry i knew that one. Can you help with the second one?
whpalmer4
  • whpalmer4
Okay, fine. Can you find the circumference of that circle? The angle can be found by this proportion: \[\frac{15}{C} = \frac{\theta}{360}\] if \(C\) is the circumference.
anonymous
  • anonymous
5400/Ctheta?
anonymous
  • anonymous
i have till ten to get points!AHHHH
Mertsj
  • Mertsj
C = 12 pi Replace C with 12 pi. Multiply 15x360 and then divide by 12 pi. That will be the angle theta
anonymous
  • anonymous
theta = 450
whpalmer4
  • whpalmer4
Isn't all the way around the circle only 360 degrees? :-)
Mertsj
  • Mertsj
Seems too big since the entire circle would only be 360 degrees.
whpalmer4
  • whpalmer4
You forgot to divide by pi = 3.14...
anonymous
  • anonymous
oops sorry forgot the pi sign ooops
whpalmer4
  • whpalmer4
always look at your results with a questioning eye and ask yourself if they make sense.
anonymous
  • anonymous
143.3121019
anonymous
  • anonymous
which problem are you doing?
whpalmer4
  • whpalmer4
That's more like it...although not exactly what I got, and the problem does ask for the nearest 100th...
whpalmer4
  • whpalmer4
this is the 2nd problem
anonymous
  • anonymous
oh
anonymous
  • anonymous
143.31
whpalmer4
  • whpalmer4
What are you using for the value of pi?
anonymous
  • anonymous
calculator pi
whpalmer4
  • whpalmer4
Interesting. I get 143.239 using 3.1415926535
anonymous
  • anonymous
but 15 times 360 = 5400 12 times 3.14 = 37.68 5400/37.68=143.31
whpalmer4
  • whpalmer4
http://www.wolframalpha.com/input/?i=15*360%2F(12*pi) Ah, but you are using 3.14, and those extra digits can make a difference, obviously.
anonymous
  • anonymous
okay well my calculator must stop at 3.14 cuz i get the same thing using calculator pi
whpalmer4
  • whpalmer4
Maybe it does. But here, punch in some extra digits and have a better chance of getting the right answer: pi = 3.141592
whpalmer4
  • whpalmer4
I'd hate to see you do the work correctly but get marked off just because your answer didn't round to the right digit in the 100ths place!
anonymous
  • anonymous
143.2394786
whpalmer4
  • whpalmer4
Right, now round that to the nearest 100th of a degree, as requested by the problem statement.
anonymous
  • anonymous
143.24
whpalmer4
  • whpalmer4
Yep, I believe that to be a true and correct answer for problem #2. Shall we try #3?
anonymous
  • anonymous
yes!
whpalmer4
  • whpalmer4
Great! I like sine curves :-)
whpalmer4
  • whpalmer4
You can draw nearly anything with the right combination of them! \[y=-\sin(x-\frac{\pi}{4}) + 2\]
whpalmer4
  • whpalmer4
Let's do the easy part first. What does adding +2 to the function do to the graph?
whpalmer4
  • whpalmer4
adding +2 to the result of the function, that is
whpalmer4
  • whpalmer4
if we've got \(y = \sin x\) and we change it to \(y = \sin x + 2\) what happens?
anonymous
  • anonymous
my calculator is being stupid. should it move 2 spaces to the left?
anonymous
  • anonymous
@whpalmer4
whpalmer4
  • whpalmer4
Here, I'll be your human calculator :-) x y = sin x y = sin x + 2 0 0 2 pi/2 1 3 pi 0 2 3pi/2 -1 1 2pi 0 2 etc.
anonymous
  • anonymous
so it moves by 2 each time
whpalmer4
  • whpalmer4
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anonymous
  • anonymous
ok i got it now
whpalmer4
  • whpalmer4
gives us a vertical translation of 2
whpalmer4
  • whpalmer4
if we had subtracted 3 instead of adding 2, we'd have a vertical translation of -3.
anonymous
  • anonymous
okay!
whpalmer4
  • whpalmer4
Next, let's tackle amplitude. Amplitude is just the vertical scale of the function. Normally, the amplitude of the sine function is 1, because starting at y = 0, it goes to y = 1 and y = -1 as its most extreme values. written more precisely, if you have a function \(y = a \sin x\), the amplitude is given by \(|a|\).
whpalmer4
  • whpalmer4
So what is the amplitude of \(y = -\sin(x-\frac{\pi}{4})+2\) ?
whpalmer4
  • whpalmer4
It might be helpful to remember that \( -\sin x = -1\sin x\)
anonymous
  • anonymous
1
whpalmer4
  • whpalmer4
right!
anonymous
  • anonymous
:)
whpalmer4
  • whpalmer4
okay, we've done amplitude, we've done vert. translation, that leaves period and phase shift, which are trickier to grasp for many people (myself included). maybe you'll be an exception :-)
whpalmer4
  • whpalmer4
you know, before we do that, let's do the vertical translation and amplitude for the other half of the problem. \(y = 2 \cos 2\pi x\) what is the amplitude, and what is the vertical translation?
anonymous
  • anonymous
amplitude is 1 and vertical translation is 2
whpalmer4
  • whpalmer4
How did you arrive at those answers?
whpalmer4
  • whpalmer4
Oh, those are correct for the 1st equation, yes. How about the second equation, \(y = 2 \cos 2\pi x\) ?
anonymous
  • anonymous
oh oops sorry i though t you were asking what we already did above! Sorry!
whpalmer4
  • whpalmer4
No problem.
whpalmer4
  • whpalmer4
It would be a problem if you thought those were the answers for the second equation, however :-)
anonymous
  • anonymous
lol
anonymous
  • anonymous
would it be zero?
whpalmer4
  • whpalmer4
So, what do you think the corresponding answers for the equation \(y = 2\cos 2 \pi x\) are?
whpalmer4
  • whpalmer4
for amplitude, or vertical translation?
anonymous
  • anonymous
vertical translation?
whpalmer4
  • whpalmer4
is that an answer, or a prayer? :-)
anonymous
  • anonymous
a prayer lol
whpalmer4
  • whpalmer4
(my old math teacher loved that line, and so do I!)
whpalmer4
  • whpalmer4
vertical translation of 0 is correct.
anonymous
  • anonymous
lol its awesome! and im right?
whpalmer4
  • whpalmer4
FWIW, my goal is always to make anyone I help understand well enough that they answer with confidence.
whpalmer4
  • whpalmer4
yes, you are correct, vertical translation is 0. how about amplitude?
anonymous
  • anonymous
okay um hold on
anonymous
  • anonymous
is it 2?
whpalmer4
  • whpalmer4
yes? ;-)
anonymous
  • anonymous
what?
whpalmer4
  • whpalmer4
yes, the answer is 2. I was hoping you were certain about it :-)
anonymous
  • anonymous
lol :) yay im right!
whpalmer4
  • whpalmer4
great. now for the tricky stuff. sorry, the really, really, really easy stuff :-)
anonymous
  • anonymous
lol
whpalmer4
  • whpalmer4
Do you remember translations of functions, shifting things back and forth along the x-axis?
whpalmer4
  • whpalmer4
or how you move the center of a circle from the origin to some other point?
whpalmer4
  • whpalmer4
("no" is a perfectly acceptable answer here)
anonymous
  • anonymous
sorry my mom wanted me. Anyways maybe. but for the sake of learning im gonna sayno just in case i be wrong.
whpalmer4
  • whpalmer4
okay, we'll leave that for a bit and do the period part. a periodic function is one that repeats. the period of the periodic function is just the amount of space along the x-axis that it takes to get to the identical spot on the function the next time around. \(y = \sin x\) has a period of \(2\pi\) because \(y=\sin x = \sin (x+2\pi)\) for any value of \(x\).
anonymous
  • anonymous
ok
whpalmer4
  • whpalmer4
If you carefully plotted \(y = \sin x\) and \(y = \sin(x+2\pi)\) on the same graph, you'd find that they are identical.
anonymous
  • anonymous
woo hoo!
anonymous
  • anonymous
lol
whpalmer4
  • whpalmer4
So, how do we know that the period of \(y = \sin x\) is \(2\pi\)? Well, if you look at the graph of the function, it starts at 0 at x = 0, goes up to 1 at x = pi/2, back down to 0 at x = pi, down to -1 at x = 3pi/2, and back to 0 at x=2pi, whereupon it does the same damn thing again. If you look at the distance along the x-axis to go from one spot anywhere on the curve to the corresponding spot in the next period, the distance along the x-axis is 2pi. So, our period is 2pi. Fair enough?
anonymous
  • anonymous
yes
whpalmer4
  • whpalmer4
That means you can identify the period any way you like — from one peak to the next identical peak, from the spot where it crosses the x-axis going up to the spot where it next crosses the x-axis going up, etc.
whpalmer4
  • whpalmer4
Okay, what happens if we change our function from \(y = \sin x\) to \(y = \sin 2x\), besides wearing out our pencil ever so slightly sooner? Care to venture a guess?
anonymous
  • anonymous
moves up 2 spaces?
whpalmer4
  • whpalmer4
nope. that would be the effect of adding to the result of the sine function...
anonymous
  • anonymous
oh were multiplying?
whpalmer4
  • whpalmer4
yes, \(y = \sin 2x\) means that we multiply \(x\) by 2 before feeding it into the sine computing machinery...
anonymous
  • anonymous
moves to the right 2 places?
whpalmer4
  • whpalmer4
nope. I'll stop tormenting you now, I can tell you're dying to know what happens :-) to go around one full cycle of the graph, x goes from 0 to 2 pi, right? what if we multiplied x by 2 before computing the sine, but still took x from 0 to 2 pi? We'd get two copies of the sine function, squashed into the width normally occupied by 1.
whpalmer4
  • whpalmer4
Here I've plotted \(y = \sin x\) and \(y = \sin 2x + 2\) — the latter is shifted up by 2 just so you can see each one clearly...
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anonymous
  • anonymous
thats cool!
anonymous
  • anonymous
can we finish this later, i need your help on stuff that closes at 12 if that is okay with you
whpalmer4
  • whpalmer4
here, for completeness, let's multiply by a number smaller than 1 as well. here I've added \(y = \sin \frac{1}{2}x\) (shifted down 2 to keep it clear of the others)
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whpalmer4
  • whpalmer4
Okay, what's the stuff that needs to get done first?
anonymous
  • anonymous
f(x)=-3cos(2x+pi/4) i need the amplitude and period i say the amplitude is 3 but i dont know what the period is
anonymous
  • anonymous
providing im right about the amplitude
whpalmer4
  • whpalmer4
Okay, amplitude is correct, give yourself a pat on the back. period is what we were discussing. if we have a multiplier in front of x *in the arguments to the sine function* then that either makes the thing wiggle faster, or slower. a number > 1 means wiggle more times in a fixed length of the x-axis, and a number < 1 means wiggle fewer times. (I'm referring to the absolute value of the multiplier here)
anonymous
  • anonymous
okay
whpalmer4
  • whpalmer4
looking at the stunning graphs I just provided, what happened to the period of \(y = \sin x\) when I changed it to \(y = \sin 2x\) ?
anonymous
  • anonymous
wiggled 2 times as much in the short distance?
whpalmer4
  • whpalmer4
wiggled 2 times in the same distance, yes. what does that mean for the period? a) period is 2 times as big, and equals 4 pi b) period is 1/2 as big, and equals 1 pi
whpalmer4
  • whpalmer4
(1 wiggle = 1 period for purposes of this discussion)
anonymous
  • anonymous
ok
whpalmer4
  • whpalmer4
So, to find the period, you look at the argument to the sine or cosine function, take only the part involving multiplication, and compare it with \(x\). let's call whatever is multiplying with \(x \) \(Z\). the period changes by a factor of \(1/Z\). So, if we have a \(y = \sin(2x+3)\), we take the \(2x\) part, say "Z = 2", so our period becomes \[P = 1/Z * 2\pi = \frac{1}{2}*2\pi = \pi\]
whpalmer4
  • whpalmer4
If we had \( y = \sin(47x)\), our period would become \[P = \frac{1}{47}*2\pi = \frac{2\pi}{47}\]
whpalmer4
  • whpalmer4
Here's the slightly confusing one: what if we had \(y = \sin(\pi x)\) ? Well, \(Z = \pi\) and our period becomes \[P = \frac{1}{Z}*2\pi = \frac{1}{\pi}*2\pi = 2\]
anonymous
  • anonymous
ok
whpalmer4
  • whpalmer4
That's a little different because the pi disappears from the period, unlike the others we've had:
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whpalmer4
  • whpalmer4
but you can see that the period is clearly 2, right?
anonymous
  • anonymous
yes
whpalmer4
  • whpalmer4
So, it's really just a proportion: if you write the sine function as \(y = \sin (Bx)\) where \(B\) is some constant, then the period is \[\frac{1}{B} = \frac{P}{2 \pi}\]
whpalmer4
  • whpalmer4
double the number in front of \(x\), and the period is cut in half. an equivalent way of saying that is that \(B*P = 2\pi\)
whpalmer4
  • whpalmer4
yet another way of thinking of it: a value of B such that |B|> 1 makes the sine function live its life faster :-)
whpalmer4
  • whpalmer4
So, back to your function \(f(x)=-3\cos(2x+\pi/4) \): the amplitude is |-3| = 3 the period is (1/2)*2pi = pi there is no vertical translation
anonymous
  • anonymous
f(x)=tan(3x-pi/2) asymptotes and period.
whpalmer4
  • whpalmer4
hey, that's not a sin or cos :-)
whpalmer4
  • whpalmer4
here the period of the tan function is \pi, but we have a multplier of 3 in front of it, so what does that do to the period? a) pi/3 b) 3pi (Jeopardy theme plays in background)
anonymous
  • anonymous
tripples the period?
whpalmer4
  • whpalmer4
Bzzt! Wrong, but thanks for playing! :-) No, the same transformation as for sin and cos — a multiplier in front of x means we divide the period by the multiplier. tan(3x) has 3 times as many cycles in a given stretch as tan(x), so the period must be 1/3 as long, right?
anonymous
  • anonymous
oh ok
whpalmer4
  • whpalmer4
now we still have to find those pesky asymptotes, or "retriceytotes" as my old calc teacher used to say
whpalmer4
  • whpalmer4
ah, the openstudy censor has made that much less funny! a s s m y t o t e s
whpalmer4
  • whpalmer4
So, the tan function is equivalent to sin / cos with the same arguments. in other words, \[\tan x = \frac{\sin x}{\cos x}\] Any place where \(cos x = 0\) is going to be an asymptote, because we can't divide by 0
whpalmer4
  • whpalmer4
So, we can find the asymptotes by finding the places where \(\cos(2x+\pi/4) = 0\)
whpalmer4
  • whpalmer4
Normally, \(\cos x = 0\) for \(x = \pm k\pi+ (\frac{\pi}{2}), k = 0,1,2,3...\)
anonymous
  • anonymous
it kicked me out sorry lol
whpalmer4
  • whpalmer4
With that \(+\pi/4\) thrown in the mix, we need to shift the values by \(-\pi/4\). If some feature on the graph happened at \(x = 0\), it now happens at \(x = -\pi/4\).
whpalmer4
  • whpalmer4
That means our asymptotes are at \[x = \pm k \pi + (\frac{\pi}{2}-\frac{\pi}{4}) = \pm k\pi +(\frac{\pi}{4}), k = 0,1,2,3...\] if I did everything correctly :-)
whpalmer4
  • whpalmer4
Did the problem specify the domain of the tan function?
anonymous
  • anonymous
so it would be period pi/3 and asymptote at 0?
whpalmer4
  • whpalmer4
Oh, nuts, I didn't account for the 2...grrr... Maybe because it was spread out too far along my screen, that's the ticket! Let's try again: \[f(x)=\tan(3x-pi/2)\] Amplitude is |1| = 1 Period is \(\pi/3\)
whpalmer4
  • whpalmer4
Asymptotes occur wherever \(\cos(3x-\pi/2) = 0\)
whpalmer4
  • whpalmer4
Normally \(\cos(x) = 0\) at \(\pm\pi/2, \pm3\pi/2, \pm5\pi/2\) etc. With the multiplier of 3 in there, the asymptotes happen every \(\pi/3\). Here's a picture of \(y = \cos(3x-\pi/2)\) and \(y = \tan(3x-\pi/2)\)
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whpalmer4
  • whpalmer4
You can see wherever cos crosses the x-axis, the tan function has a vertical asymptote
whpalmer4
  • whpalmer4
Have your eyes glazed over permanently yet? :-)
anonymous
  • anonymous
lol probably
whpalmer4
  • whpalmer4
Were there any more you wanted to go through?
anonymous
  • anonymous
which are identities \[tanxcosxcscx=1\] \[\frac{ secx-cosx }{ secx } = \sin^2x\] \[1-tanxtany=\frac{ \cos(x+4) }{ cosxcosy }\] \[4cosxsinx=2cosx+1-2sinx\]
anonymous
  • anonymous
\[\cos^22x + \sin^22x = 1 \] \[\sec^2x + \csc^2x = 1 \] \[\sec^2x - \tan2x = 1 \] \[\tan^2x - \sec^2x = 1 \]
whpalmer4
  • whpalmer4
Well, one quick and dirty way to check if something is an identity is to try it with a few points. I typically use 0, pi/2, pi, and pi/4 if the other 3 have all worked. For example, let's try \(\sec^2x + \csc^2x = 1\) \[\sec 0 = 1\]\[\csc 0 = \infty\]Guess that one isn't an identity!
whpalmer4
  • whpalmer4
For others, substituting basic trig definitions works: \[\tan x\cos x\csc x=1\]\[\frac{\sin x}{\cos x}\cos x\frac{1}{\sin x} = 1\]As you can see, everything cancels out nicely, leaving 1=1, so it is an identity.
whpalmer4
  • whpalmer4
If you can do valid substitutions and get to a spot where you have something that obviously is NOT equal (1 = 0, for example), then that is not an identity.
whpalmer4
  • whpalmer4
Also handy in this stuff is a decent table of trig identities, such as http://www.sosmath.com/trig/Trig5/trig5/trig5.html
anonymous
  • anonymous
thanks!
whpalmer4
  • whpalmer4
Try to do those yourself, but if you get stuck, let me know and I can make suggestions
anonymous
  • anonymous
okie dokie!
anonymous
  • anonymous
as far as the question about the asymptotes and the period question, i forgot to write down the answer
whpalmer4
  • whpalmer4
it should still be there if you scroll back far enough... period is pi/3 asymptotes at -2pi, -5pi/3, -4pi/3, -pi, -2pi/3, -pi/3, 0, pi/3, 2pi/3, pi, 4pi/3, 5pi/3, 2pi, etc.
whpalmer4
  • whpalmer4
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anonymous
  • anonymous
ok thx. do you know if\[Sin \frac{ a }{ 2 } = \pm \frac{ \sqrt{1-cosa} }{ 2 }\] for all values of a?
whpalmer4
  • whpalmer4
it works for 0: \[\sin 0/2 = \pm \frac{\sqrt{1-\cos 0}}{2} \rightarrow 0=0 \]But at \(a=\pi/2\) it doesn't. \[\sin ((\pi/2)/2) = \pm\frac{\sqrt{1-\cos \pi/2}}{2}\]\[\frac{1}{\sqrt{2}} = \pm\frac{1}{2}\]which isn't true.
anonymous
  • anonymous
okay thanks!
anonymous
  • anonymous
we can finish the original paper we were working on
whpalmer4
  • whpalmer4
Pick the most interesting/important question, I reserve the right to call it a night after each question :-)
anonymous
  • anonymous
okay how bout number 7?
whpalmer4
  • whpalmer4
#7 it is! I would start by digging out that trig identity page and looking up the identities for \(\sin 2x\) and \(\cos 2x\)
whpalmer4
  • whpalmer4
They should give you expansions in just terms of sin and cos x
whpalmer4
  • whpalmer4
For example, the first term \[\frac{\sin(2x)}{\sin(x)} \]becomes \[\frac{2\sin(x)\cos(x)}{\sin(x)}\]which cancels to make \(2\cos(x)\)
anonymous
  • anonymous
so what would go in the first line?
whpalmer4
  • whpalmer4
Then I would use \(\cos(2x)=2\cos^2(x)-1\) for the second term. \[\frac{\sin(2x)}{\sin(x)}-\frac{\cos(2x)}{\cos(x)}=\sec(x)\] \[\frac{2\cancel{\sin(x)}\cos(x)}{\cancel{\sin(x)}}-\frac{2\cos^2(x)-1 }{\cos(x) }=\sec(x)\]Now multiply the first term by \(\frac{(\cos(x))}{(\cos(x))}=1\) to get a common denominator \[\frac{2\cos^2(x)}{\cos(x)}-\frac{(2\cos^2(x)-1)}{\cos(x)}=\sec(x)\]
whpalmer4
  • whpalmer4
the left hand side simplifies to \[\frac{1}{\cos(x)}\] which is the definition of \(\sec(x)\)...
whpalmer4
  • whpalmer4
See how that worked?
anonymous
  • anonymous
yes!
whpalmer4
  • whpalmer4
Here's some of the chickenscratch I have to type to make the equations look all pretty like that: \frac{2\cos^2(x)}{\cos(x)}-\frac{(2\cos^2(x)-1)}{\cos(x)}=\sec(x) (that's the last line of the work) those flippin' { ( ) } look an awful lot alike on my screen, and if they aren't matched up exactly right, the computer makes rude raspberries in my general direction, but provides no help in finding the missing or extra character in there :-)
anonymous
  • anonymous
lol thank you!
whpalmer4
  • whpalmer4
I think the result is worth it, mind you...
anonymous
  • anonymous
lol very true!
whpalmer4
  • whpalmer4
but I could scribble it out on paper faster, for sure!
anonymous
  • anonymous
\[\frac{2\cos^2(x)}{\cos(x)}-\frac{(2\cos^2(x)-1)}{\cos(x)}=\sec(x)\]
whpalmer4
  • whpalmer4
How does that simplify? I'll just do the numerator: \[2\cos^2(x)-(2\cos^2(x)-1)) = 2\cos^2(x)-2\cos^2(x)-(-1)) = 1\]
anonymous
  • anonymous
wouldnt the denominator be 0?
whpalmer4
  • whpalmer4
no...we just subtract the numerators, the denominator stays the same \[\frac{2}{5}-\frac{(2-1)}{5} = \frac{2}{5}-\frac{2}{5}-\frac{(-1)}{5} = \frac{1}{5}\]right? (5 has nothing to do with this problem)
anonymous
  • anonymous
so the denominator would be cosx - cosx?
whpalmer4
  • whpalmer4
no, that's two separate fractions...
anonymous
  • anonymous
uh! im tired lol! cosx?
whpalmer4
  • whpalmer4
yes. both fractions have cos x as the denominator ( I multiplied the first one by cos x/cos x to get it that way) the left side of the equation simplifies down to just 1/cos x, and the right side is sec x, which is defined as 1/cos x
anonymous
  • anonymous
thank you! i don't think i'm normally this stupid but i'm not sure lol.
whpalmer4
  • whpalmer4
hey, it's nearly 2 AM, right? an occasional minor mistake isn't unexpected. dump a bucket of ice cold water over your head, and you'll be fine for another 15 minutes :-)
anonymous
  • anonymous
lol thanks. and ya its 2am ill be up till like 9 ha FUN!
whpalmer4
  • whpalmer4
next question?
anonymous
  • anonymous
how about cos^2(x)csc(x)-csc(x)
anonymous
  • anonymous
=___ x
whpalmer4
  • whpalmer4
Okay, \[\cos^2(x)\csc(x)-\csc(x) = \cos^2(x)*\frac{1}{\sin(x)}-\frac{1}{\sin(x)}=\frac{\cos^2(x)-1}{\sin(x)}\]\[\sin^2(x)+\cos^2(x)=1\]\[\cos^2(x)-1=-\sin^2(x)\]So left hand side becomes\[\frac{\cos^2(x)-1}{\sin(x)} =\frac{-\sin^2(x)}{\sin(x)}=-\sin(x) \]
whpalmer4
  • whpalmer4
I guess maybe a bit simpler is \[\cos^2(x)\csc(x)-\csc(x) = \csc(x)(\cos^2(x)-1) = \csc(x)(-\sin^2(x)) \]\[= \frac{-\sin^2(x)}{\sin(x)}=-\sin(x)\]
anonymous
  • anonymous
and what if it is ----^2 x = (csc x-1)(csc x+1)
whpalmer4
  • whpalmer4
\[(\csc x -1)(\csc x+1) = (\frac{1}{\sin x}-1)(\frac{1}{\sin x}+1) \]\[= -1+\frac{1}{\sin^2x} = -1+\csc^2x = \cot^2x\]because \[1+\cot^2x=\csc^2x\]
anonymous
  • anonymous
and how bout pi/4cospi/6=1/2(___pi/12+cos5pi/12)
whpalmer4
  • whpalmer4
Are you sure you have that written correctly? \[\frac{1}{4} \pi \cos \left(\frac{\pi }{6}\right)=\frac{1}{2} \left(\frac{\pi n}{12}+\cos \left(\frac{5 \pi }{12}\right)\right)\] (I'm using \(n\) as the blank)

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