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| 12/2 [Analysis H] 43; Discussion of Analysis Honors Problem 43 for 12/2 | |
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| Tweet Topic Started: Dec 2 2008, 08:13 PM (680 Views) | |
| Anonymous | Dec 2 2008, 08:13 PM Post #1 |
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[u=[color=red]Tycho bloody Brahe[/u][/color]
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Anybody seeking math help is welcome to post here and ask. I for one check this forum regularly and am happy to assist with any math class at LHS. I never took IMP 2 or IMP 3, but I'm still willing to give it a try. Other than that, if it's offered, I've taken a version of it at some point. |
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| Chan | Dec 2 2008, 08:19 PM Post #2 |
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NOOB Poster
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Can you help me find the exact values of the expression: sin15cos15 |
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| Brian | Dec 2 2008, 08:22 PM Post #3 |
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B-Dawg: Probably crazy and Definitely hates everyone
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(sin3x)/sinx-(cos3x)/cosx |
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| Anonymous | Dec 2 2008, 08:25 PM Post #4 |
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[u=[color=red]Tycho bloody Brahe[/u][/color]
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In response to Chan, try using the Double Angle Formula: sin2a=2*sina*cosa . You can rewrite that as (sin2a)/2=sina*cosa . See if that helps. If that doesn't help go ahead and post again. |
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| Saber | Dec 2 2008, 08:29 PM Post #5 |
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Newbie
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ok for 43 I changed sin3x into sinx + sin 2x and the same with cos 3x. I substituted the values for sin2x and cos2x, and it's not simplifying into any simple answer. |
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| Anonymous | Dec 2 2008, 08:36 PM Post #6 |
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[u=[color=red]Tycho bloody Brahe[/u][/color]
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You can't change sin3x into sinx+sin2x. The two are not equivalent. You essentially need a "triple angle formula" which you just get by combing the formula for sin2a and sin(a+b) by rewriting it as sin(x+2x) (not sinx +sin2x). Recall that the double angle formula for sin is sin2a=2*sina*cosa and the formula for sin(a+b)=sina*cosb+cosa*sinb . If you combine these two formulas you can get the formula for sin3a. Use a similar method for the cos3x and then it is much easier to simplify. (Feel free to post requesting clarifcation or additional formulae.) |
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| Anonymous | Dec 2 2008, 08:44 PM Post #7 |
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[u=[color=red]Tycho bloody Brahe[/u][/color]
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I realized I neglected to mention that you will need the cos double angle formula as well for the sin3a derivation, and you also need it for the cos3a derivation. I'll go ahead and post the cos2a formula and cos(a+b) formula for clarifaction: cos2a=[cosa]^2-[sina]^2 and cos(a+b)=cosa*cosb-sina*sinb . I am also happy to clarify any of these formulae or any of the derivations involved in number 43. |
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| Anonymous | Dec 2 2008, 08:48 PM Post #8 |
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[u=[color=red]Tycho bloody Brahe[/u][/color]
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As long as I'm posting, I might as well mention that the identity [sinx]^2+[cosx]^2=1 must be used a few times in this problem. |
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| Chan | Dec 2 2008, 08:55 PM Post #9 |
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NOOB Poster
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It did work, thank you and sorry. |
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| Anonymous | Dec 2 2008, 09:00 PM Post #10 |
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[u=[color=red]Tycho bloody Brahe[/u][/color]
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Sorry for what? It's a good question. When using these trig identities you have to do some weird things sometimes. That was just the tip of the iceberg. |
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