Trigonometric Half Angle Formula Explained and Applied

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Homework Help Overview

The problem involves evaluating the expression \(\tan\left(\frac{1}{2} \arcsin\left(\frac{-7}{25}\right)\right)\) using trigonometric identities, particularly the tangent half-angle formula. Participants explore the implications of the half-angle in relation to the arcsine function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the tangent half-angle formula and express uncertainty about how to handle the half-angle in conjunction with the arcsine. There are attempts to derive values using triangle properties and the half-angle formula, with some questioning the correctness of their results.

Discussion Status

Multiple interpretations of the problem are being explored, with participants providing different approaches and checking their calculations against known values. Some participants express confusion about sign conventions and the correctness of their answers, while others suggest using calculators for verification.

Contextual Notes

There is an ongoing discussion about the accuracy of calculations and the need to rationalize denominators. Participants reflect on their methods and the importance of self-checking results.

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Homework Statement



[tex]\tan[\frac{1}{2} \arcsin(\frac{-7}{25})][/tex]

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)?
 
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Use the tangent half angle formula. tan(x/2)=??
 
Thanks, so i get

[tex]-\sqrt{26}[/tex]

Does that sound right?
 
jrjack said:
Thanks, so i get

[tex]-\sqrt{26}[/tex]

Does that sound right?

arcsin(-7/25) is about -0.3. If you take half of that and take the tangent, it's nowhere near -sqrt(26) which is about -5. Is it? You can check these solutions using rough estimates or a calculator.
 
[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]
 
jrjack said:
[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]

tan(0/2)=0. If you put x=0 into your supposed half angle formula, what do you get? Does it work?
 
Sorry, I now realize I have my signs flipped in my formula.
I think my answer should be:[tex]-\sqrt{\frac{1}{26}}[/tex]
 
jrjack said:
Sorry, I now realize I have my signs flipped in my formula.
I think my answer should be:[tex]-\sqrt{\frac{1}{26}}[/tex]

That doesn't work either because (1-24/25)/(1+24/25) isn't equal to 1/26. Now what's it really equal to??
 
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]
 
  • #10
jrjack said:
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]

Ok, aside from the fact there is a simpler way to write -1/sqrt(49) could you try and check that using a calculator from your original expression? It's really useful to have a simple way of self-checking whether you are way off or not.
 
  • #11
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.
 
  • #12
jrjack said:
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.

Very welcome and quite right. The main lesson is how easy these answers are to check with a calculator.
 

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