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Trig/Calculus Help

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RevealingSkirt
Posts: 528 - Karma: 62
So we had a huge test in class, and everyone bombed. No one passed, at all. So we're doing a make up test right now, and its the exact same test, with the answers already on it, and we need to show the work of how to achieve the answer.
I'm stuck on one question, and supposedly it's "really easy" but I can't seem to figure it out. So I was hoping someone on here can show me the work on how to achieve the answer.
It seems like a quick and easy one, according to everyone, so it shouldn't take more than a moment of your time.

Here's the problem

2cos(squared)22.5(deg)-1

The cosine is squared and the 22.5 is in degrees. The answer is supposed to be sqrt2/2 so square root of 2 over 2.

If someone can show me the work to do this, I would be forever grateful, and may just give you +K or help you out with something you need!
-Thanks!
Mar 22 2010 5:16PM
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Conker
Posts: 1379 - Karma: 6
whats the problem, just plug that shit in your calculator.
Mar 22 2010 5:23PM
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Microcosm
Posts: 4765 - Karma: 153
http://www.wolframalpha.com/

your new best friend
Mar 22 2010 5:27PM
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RevealingSkirt
Posts: 528 - Karma: 62
I guess neither of you read the thread. I have the answer. I need the work that gives the answer. Wolframalpha is great. I've known about it forever. When you need an answer, it's sweet. When you need the work, not so much. I can plug that into my calculator, and it will return the answer, not the work. Thanks for the effort but its no good.
Mar 22 2010 5:30PM
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HIRO
Posts: 548 - Karma: 34
so just to clarify, you need to justify that:

Two times the cosine squared of 22.5 degrees, minus one equals the square root of 2 over 2?


Mar 22 2010 5:46PM
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blackoutbill
Posts: 2918 - Karma: 282
I think it uses a double angle formula which is:

cos(2theta)=2cos^2theta-1

so cos(2(22.5))=2cos^2(22.5)-1

I think that should work
Mar 22 2010 5:47PM
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An10a_Eng
Posts: 2505 - Karma: 26
this looks right
Mar 22 2010 5:49PM
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RevealingSkirt
Posts: 528 - Karma: 62
You my friend get +K for realizing something that I had written down on my test and said no that doesn't work. Thank you very much.
Mar 22 2010 5:50PM
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blackoutbill
Posts: 2918 - Karma: 282
I just tried it out and it does work. Simplified it would be:

cos(45)=2cos^2(22.5)-1
cos(45)=sqrt2/2
Mar 22 2010 5:50PM
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Microcosm
Posts: 4765 - Karma: 153
Sorry, realized after.

In any event, how does this look..
A shift of 1 on the x-axis [as suggested by the cos^2(22.5-1) ? ] should make the equation [cos^2(45)] I think? And cos(45)=1/√(2)
Mar 22 2010 5:53PM
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An10a_Eng
Posts: 2505 - Karma: 26
i think its cos(22.5) - 1 not cos(22.5 -1), and cos(22.5-1) = cos(21.5) which does not equal cos(45)
Mar 22 2010 5:54PM
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blackoutbill
Posts: 2918 - Karma: 282
cos(45)=1/√2=√2/2
Mar 22 2010 5:59PM
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Chickendodo
Posts: 1426 - Karma: 33
Another way is the half angle identity,


2cos^2(22.5)-1
= 2* 1/2(1 + cos(2*22.5)) - 1
= -cos(45)
= sqrt(2)/2

Yay math.
Mar 22 2010 6:05PM
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TheButterHashira
Posts: 9691 - Karma: 1,702
Take a look at this. It has all the trig identities and stuff.
http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf
Mar 22 2010 6:06PM
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chapz
Posts: 10182 - Karma: 66
getting my cal 3 test back tomorrow, totally bombed too. maybe my teacher will give us a retake also!

just curious,
can anybody get the arclength of (sin(theta/2))^3 from 0 to 2pi?
Mar 22 2010 6:54PM
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Chickendodo
Posts: 1426 - Karma: 33
hahaha, shit dude, I had to google that, and I'm a math major.

you take
sin^3(x/2) = sin(x/2)*sin^2(x/2)
then do the integral,

∫(sin(x/2) * sin^2(x/2)dx)

Then use integration by parts to do

sin^2(x/2) = u
2(sin(x/2)cos(x/2)) = du

and
sin(x/2) = dv
-cos(x/2) = v

then you do that uv - integral(vdu) shit
= -sin^2(x/2)cos(x/2) + 2*∫cos^2(x/2)sin(x/2)

= -sin^2(x/2)cos(x/2) + 2∫ [1-sin^2(x/2)] * sin(x/2)

= -sin^2(x/2)cos(x/2) + 2∫sin(x/2) - 2∫sin^3(x/2)

and here's where it gets stupid and nobody in their right mind would make this connection, but here goes.

  ∫sin^3(x/2) = -sin^2(x/2)cos(x/2) + 2∫sin(x/2) - 2∫sin^3(x/2)
+2∫sin^3(x/2)                                                     + 2∫sin^3(x/2)

3∫sin^3(x/2) = -sin^2(x/2)cos(x/2) - cos(x/2)

so...

∫sin^3(x/2) = -sin^2(x/2)cos(x/2) - cos(x/2)

and from 0 to 2pi

= -1 - (1) = -2

so, the answer to all that fucking work is -2.  Should be right unless I fucked up some algebra
Mar 22 2010 9:49PM
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Chickendodo
Posts: 1426 - Karma: 33
Cool, the end was wrong

∫sin^3(x/2) = -(1/3)sin^2(x/2)cos(x/2) - (1/3)cos(x/2)

so the answer is actually, -2/3.  but still, the rest of the work should be right.
Mar 22 2010 9:50PM
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chapz
Posts: 10182 - Karma: 66
hahaha can you imagine that being on a cal3 test, the first problem? my teacher is retarded. thanks for the work though! now i can tell all my friends how to do it! haha
Mar 22 2010 10:19PM
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rach
Posts: 917 - Karma: 507
no joke took the same test this morning.
pretty positive i got 20/50 and im pretty good at math usually.
i don't think this way...problem solving is hard.
Mar 22 2010 10:27PM
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Chickendodo
Posts: 1426 - Karma: 33
haha yeah man, I'm in calc 3 right now, and I haven't seen anything even close to that hard, that's seriously some grade a like crazy shit right there haha, glad to help, math is my major O_o
Mar 23 2010 12:47AM
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squibble
Posts: 4569 - Karma: 104
thats not calculus? thats shitty trig
Mar 23 2010 1:51AM
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Gnarshred
Posts: 5670 - Karma: 18
I would not recommend taking Calculus. Without a strong trig/ algebra/ problem solving skill set, you are fucked
Mar 23 2010 1:56AM
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C0LE
Posts: 316 - Karma: 475
mmhmmm. this is senior math for me. identities and shit
Mar 23 2010 1:56AM
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