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tan[1/2 arcsin(-7/25)] |
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| Aug7-11, 10:29 PM | #1 |
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tan[1/2 arcsin(-7/25)]
1. The problem statement, all variables and given/known data
[tex]\tan[\frac{1}{2} \arcsin(\frac{-7}{25})][/tex] 3. The attempt at a solution I'm not sure how to take 1/2 the arcsin, should this use the half-angle formula? Normally I would draw a triangle using the sin value (-7/25), then find the tan value (24/25), but the 1/2 is throwing me off. How do I start this? Is this 1/2 the sin value (-7/25)= -7/50, then solve for the tan(-7/50)? |
| Aug7-11, 10:35 PM | #2 |
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Use the tangent half angle formula. tan(x/2)=??
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| Aug7-11, 10:42 PM | #3 |
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Thanks, so i get
[tex]-\sqrt{26}[/tex] Does that sound right? |
| Aug7-11, 10:55 PM | #4 |
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tan[1/2 arcsin(-7/25)] |
| Aug7-11, 10:59 PM | #5 |
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[tex]-\sqrt{\frac{1+\cos x}{1-\cos x}}[/tex]
[tex]-\sqrt{\frac{1+\frac{24}{25}}{1-\frac{24}{25}}}[/tex] [tex]=-\sqrt{26}[/tex] |
| Aug7-11, 11:07 PM | #6 |
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| Aug7-11, 11:14 PM | #7 |
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Sorry, I now realize I have my signs flipped in my formula.
I think my answer should be:[tex]-\sqrt{\frac{1}{26}}[/tex] |
| Aug7-11, 11:24 PM | #8 |
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| Aug7-11, 11:39 PM | #9 |
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Sorry, I got in a hurry, between typing the tex and working the problem several different ways (wrong of course).
It should equal 1/49, which means my answer should be [itex]-\sqrt{\frac{1}{49}}[/itex] |
| Aug7-11, 11:43 PM | #10 |
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| Aug7-11, 11:50 PM | #11 |
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Thank you for your help.
I realize I still need to rationalize the denominator, and after checking with my calculator both answers come out to -.142857, so that must be correct. My final answer should be -1/7 Once again, thank you for your help. |
| Aug7-11, 11:54 PM | #12 |
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