ANYBODY HELP !!!!!!

- anonymous

ANYBODY HELP !!!!!!

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

##### 1 Attachment

- whpalmer4

Which problem do you need help solving?

- anonymous

any

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

yeah im sorry i knew that one. Can you help with the second one?

- 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

5400/Ctheta?

- anonymous

i have till ten to get points!AHHHH

- Mertsj

C = 12 pi
Replace C with 12 pi.
Multiply 15x360 and then divide by 12 pi. That will be the angle theta

- anonymous

theta = 450

- whpalmer4

Isn't all the way around the circle only 360 degrees? :-)

- Mertsj

Seems too big since the entire circle would only be 360 degrees.

- whpalmer4

You forgot to divide by pi = 3.14...

- anonymous

oops sorry forgot the pi sign ooops

- whpalmer4

always look at your results with a questioning eye and ask yourself if they make sense.

- anonymous

143.3121019

- anonymous

which problem are you doing?

- whpalmer4

That's more like it...although not exactly what I got, and the problem does ask for the nearest 100th...

- whpalmer4

this is the 2nd problem

- anonymous

oh

- anonymous

143.31

- whpalmer4

What are you using for the value of pi?

- anonymous

calculator pi

- whpalmer4

Interesting. I get 143.239 using 3.1415926535

- anonymous

but 15 times 360 = 5400
12 times 3.14 = 37.68
5400/37.68=143.31

- 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

okay well my calculator must stop at 3.14 cuz i get the same thing using calculator pi

- 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

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

143.2394786

- whpalmer4

Right, now round that to the nearest 100th of a degree, as requested by the problem statement.

- anonymous

143.24

- whpalmer4

Yep, I believe that to be a true and correct answer for problem #2. Shall we try #3?

- anonymous

yes!

- whpalmer4

Great! I like sine curves :-)

- whpalmer4

You can draw nearly anything with the right combination of them!
\[y=-\sin(x-\frac{\pi}{4}) + 2\]

- whpalmer4

Let's do the easy part first. What does adding +2 to the function do to the graph?

- whpalmer4

adding +2 to the result of the function, that is

- whpalmer4

if we've got \(y = \sin x\) and we change it to \(y = \sin x + 2\) what happens?

- anonymous

my calculator is being stupid. should it move 2 spaces to the left?

- anonymous

@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

so it moves by 2 each time

- whpalmer4

##### 1 Attachment

- anonymous

ok i got it now

- whpalmer4

gives us a vertical translation of 2

- whpalmer4

if we had subtracted 3 instead of adding 2, we'd have a vertical translation of -3.

- anonymous

okay!

- 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

So what is the amplitude of \(y = -\sin(x-\frac{\pi}{4})+2\) ?

- whpalmer4

It might be helpful to remember that \( -\sin x = -1\sin x\)

- anonymous

1

- whpalmer4

right!

- anonymous

:)

- 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

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

amplitude is 1 and vertical translation is 2

- whpalmer4

How did you arrive at those answers?

- whpalmer4

Oh, those are correct for the 1st equation, yes. How about the second equation, \(y = 2 \cos 2\pi x\) ?

- anonymous

oh oops sorry i though t you were asking what we already did above! Sorry!

- whpalmer4

No problem.

- whpalmer4

It would be a problem if you thought those were the answers for the second equation, however :-)

- anonymous

lol

- anonymous

would it be zero?

- whpalmer4

So, what do you think the corresponding answers for the equation \(y = 2\cos 2 \pi x\) are?

- whpalmer4

for amplitude, or vertical translation?

- anonymous

vertical translation?

- whpalmer4

is that an answer, or a prayer? :-)

- anonymous

a prayer lol

- whpalmer4

(my old math teacher loved that line, and so do I!)

- whpalmer4

vertical translation of 0 is correct.

- anonymous

lol its awesome! and im right?

- whpalmer4

FWIW, my goal is always to make anyone I help understand well enough that they answer with confidence.

- whpalmer4

yes, you are correct, vertical translation is 0. how about amplitude?

- anonymous

okay um hold on

- anonymous

is it 2?

- whpalmer4

yes? ;-)

- anonymous

what?

- whpalmer4

yes, the answer is 2. I was hoping you were certain about it :-)

- anonymous

lol :) yay im right!

- whpalmer4

great. now for the tricky stuff. sorry, the really, really, really easy stuff :-)

- anonymous

lol

- whpalmer4

Do you remember translations of functions, shifting things back and forth along the x-axis?

- whpalmer4

or how you move the center of a circle from the origin to some other point?

- whpalmer4

("no" is a perfectly acceptable answer here)

- anonymous

sorry my mom wanted me. Anyways maybe. but for the sake of learning im gonna sayno just in case i be wrong.

- 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

ok

- 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

woo hoo!

- anonymous

lol

- 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

yes

- whpalmer4

That means you can identify the period any way you like â€” 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

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

moves up 2 spaces?

- whpalmer4

nope. that would be the effect of adding to the result of the sine function...

- anonymous

oh were multiplying?

- whpalmer4

yes, \(y = \sin 2x\) means that we multiply \(x\) by 2 before feeding it into the sine computing machinery...

- anonymous

moves to the right 2 places?

- 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

Here I've plotted \(y = \sin x\) and \(y = \sin 2x + 2\) â€” the latter is shifted up by 2 just so you can see each one clearly...

##### 1 Attachment

- anonymous

thats cool!

- anonymous

can we finish this later, i need your help on stuff that closes at 12 if that is okay with you

- 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)

##### 1 Attachment

- whpalmer4

Okay, what's the stuff that needs to get done first?

- 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

providing im right about the amplitude

- 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

okay

- 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

wiggled 2 times as much in the short distance?

- 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

(1 wiggle = 1 period for purposes of this discussion)

- anonymous

ok

- 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

If we had \( y = \sin(47x)\), our period would become \[P = \frac{1}{47}*2\pi = \frac{2\pi}{47}\]

- 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

ok

- whpalmer4

That's a little different because the pi disappears from the period, unlike the others we've had:

##### 1 Attachment

- whpalmer4

but you can see that the period is clearly 2, right?

- anonymous

yes

- 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

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

yet another way of thinking of it: a value of B such that |B|> 1 makes the sine function live its life faster :-)

- 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

f(x)=tan(3x-pi/2) asymptotes and period.

- whpalmer4

hey, that's not a sin or cos :-)

- 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

tripples the period?

- whpalmer4

Bzzt! Wrong, but thanks for playing! :-)
No, the same transformation as for sin and cos â€” 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

oh ok

- whpalmer4

now we still have to find those pesky asymptotes, or "retriceytotes" as my old calc teacher used to say

- whpalmer4

ah, the openstudy censor has made that much less funny! a s s m y t o t e s

- 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

So, we can find the asymptotes by finding the places where \(\cos(2x+\pi/4) = 0\)

- whpalmer4

Normally, \(\cos x = 0\) for \(x = \pm k\pi+ (\frac{\pi}{2}), k = 0,1,2,3...\)

- anonymous

it kicked me out sorry lol

- 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

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

Did the problem specify the domain of the tan function?

- anonymous

so it would be period pi/3 and asymptote at 0?

- 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

Asymptotes occur wherever \(\cos(3x-\pi/2) = 0\)

- 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 \(y = \tan(3x-\pi/2)\)

##### 1 Attachment

- whpalmer4

You can see wherever cos crosses the x-axis, the tan function has a vertical asymptote

- whpalmer4

Have your eyes glazed over permanently yet? :-)

- anonymous

lol probably

- whpalmer4

Were there any more you wanted to go through?

- anonymous

which are identities
\[tanxcosxcscx=1\]
\[\frac{ secx-cosx }{ secx } = \sin^2x\]
\[1-tanxtany=\frac{ \cos(x+4) }{ cosxcosy }\]
\[4cosxsinx=2cosx+1-2sinx\]

- anonymous

\[\cos^22x + \sin^22x = 1 \]
\[\sec^2x + \csc^2x = 1 \]
\[\sec^2x - \tan2x = 1 \]
\[\tan^2x - \sec^2x = 1 \]

- 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

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

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

Also handy in this stuff is a decent table of trig identities, such as http://www.sosmath.com/trig/Trig5/trig5/trig5.html

- anonymous

thanks!

- whpalmer4

Try to do those yourself, but if you get stuck, let me know and I can make suggestions

- anonymous

okie dokie!

- anonymous

as far as the question about the asymptotes and the period question, i forgot to write down the answer

- 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

##### 1 Attachment

- anonymous

ok thx. do you know if\[Sin \frac{ a }{ 2 } = \pm \frac{ \sqrt{1-cosa} }{ 2 }\] for all values of a?

- 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

okay thanks!

- anonymous

we can finish the original paper we were working on

- whpalmer4

Pick the most interesting/important question, I reserve the right to call it a night after each question :-)

- anonymous

okay how bout number 7?

- 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

They should give you expansions in just terms of sin and cos x

- 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

so what would go in the first line?

- 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

the left hand side simplifies to \[\frac{1}{\cos(x)}\] which is the definition of \(\sec(x)\)...

- whpalmer4

See how that worked?

- anonymous

yes!

- 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

lol thank you!

- whpalmer4

I think the result is worth it, mind you...

- anonymous

lol very true!

- whpalmer4

but I could scribble it out on paper faster, for sure!

- anonymous

\[\frac{2\cos^2(x)}{\cos(x)}-\frac{(2\cos^2(x)-1)}{\cos(x)}=\sec(x)\]

- 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

wouldnt the denominator be 0?

- 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

so the denominator would be cosx - cosx?

- whpalmer4

no, that's two separate fractions...

- anonymous

uh! im tired lol! cosx?

- 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

thank you! i don't think i'm normally this stupid but i'm not sure lol.

- 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

lol thanks. and ya its 2am ill be up till like 9 ha FUN!

- whpalmer4

next question?

- anonymous

how about cos^2(x)csc(x)-csc(x)

- anonymous

=___ x

- 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

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

and what if it is ----^2 x = (csc x-1)(csc x+1)

- 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

and how bout pi/4cospi/6=1/2(___pi/12+cos5pi/12)

- 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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.