Solving Trigonometric Equations Using Identities

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Homework Help Overview

The discussion revolves around solving trigonometric equations using identities, specifically focusing on equations involving tangent, sine, and cosine functions. Participants are exploring various approaches to manipulate and simplify these equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to simplify two trigonometric equations but expresses uncertainty about their direction. Some participants suggest converting all terms to sine and cosine to facilitate simplification. Others explore the implications of double angle identities and the relationships between sine and cosine.

Discussion Status

Participants are actively engaging with the equations, with some providing detailed algebraic manipulations. There is a sense of exploration, as individuals question the methods used and express curiosity about alternative approaches. However, no consensus on a single method has been reached.

Contextual Notes

Some participants express discomfort with typing mathematical expressions on a computer, indicating a preference for traditional pen and paper methods. This highlights a potential barrier to engagement in the discussion.

Elijah the Wood
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1)
tan(x)-sin(x)/2tan(x) = sin^2(x/2)
1-sin(x)/2 = sin^2(x/2)

(That's as far as I've gotten on this one...but i don't know if I'm in the right direction or what to do next?)

2)
sin2x = 1 / tanx + cot2x
2sin(x)cos(x) = ''
sin(x) * cos(x) * sin(x) = ''

(The same from #1 applies to this problem)
 
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Assuming you mean:
[tex]\frac{\tan(x)-\sin(x)}{2 \tan(x)} = \sin^{2}(\frac{x}{2})[/tex]

Let's change everything to cosinus and sinus

[tex]\frac{\frac{\sin(x)}{\cos(x)} - \sin(x)}{\frac{2 \sin(x)} {\cos(x)}} = \sin^{2}(\frac{x}{2})[/tex]

Simplifying:

[tex]\frac{\frac{\sin(x) - \sin(x) \cos(x)}{\cos(x)}}{\frac{2 \sin(x)} {\cos(x)}} = \sin^{2}(\frac{x}{2})[/tex]

[tex]\frac{\sin(x) - \sin(x) \cos(x)}{2 \sin(x)} = \sin^{2}(\frac{x}{2})[/tex]

[tex]\frac{1 - \cos(x)}{2} = \sin^{2}(\frac{x}{2})[/tex]

Remember [itex]\sin^{2}(\frac{x}{2}) = \frac{1 - \cos(x)}{2}[/itex]

[tex]\frac{1 - \cos(x)}{2} = \sin^{2}(\frac{x}{2})[/tex]

[tex]\sin^{2}(\frac{x}{2}) = \sin^{2}(\frac{x}{2})[/tex]

Assuming you mean:
[tex]\sin (2x) = \frac{1}{\tan(x) + \cot(2x)}[/tex]

Working the right to the left

Changing to sinus and cosinus:

[tex]\sin (2x) = \frac{1}{\frac{\sin(x)}{\cos(x)} + \frac{\cos(2x)}{\sin(2x)}}[/tex]

Applying double angle identities

[tex]\sin (2x) = \frac{1}{\frac{\sin(x)}{\cos(x)} + \frac{\cos^2(x) - \sin^2(x)}{2 \sin(x) \cos(x)}}[/tex]

Simplifying:

[tex]\sin (2x) = \frac{1}{\frac{2 \sin^2(x) \cos(x) + \cos^3(x) - \sin^2(x) \cos(x)}{2 \sin(x) \cos^2(x)}}[/tex]

[tex]\sin (2x) = \frac{1}{\frac{\sin^2(x) \cos(x) + \cos^3(x)}{2 \sin(x) \cos^2(x)}}[/tex]

[tex]\sin (2x) = \frac{1}{\frac{\sin^2(x) + \cos^2(x)}{2 \sin(x) \cos(x)}}[/tex]

Remember [itex]\sin^2(x) + \cos^2(x) = 1[/itex]

[tex]\sin (2x) = \frac{1}{\frac{1}{2 \sin(x) \cos(x)}}[/tex]

Remember [itex]\sin(2x) = 2 \sin(x) \cos(x)[/itex]

[tex]\sin (2x) = 2 \sin(x) \cos(x)[/tex]

[tex]\sin (2x) = \sin (2x)[/tex]
 
Last edited:
damn it... how do u do all these equations on the computer?? i find it hard doing my math on a computer... I am more comfortable with a pen and paper... get tired of using the "^",etc

any words of wisdom ?
 
Jeez...that was incredible. Thanks a lot...but I don't suppose there was an easier way, was there? haha
 
jai6638 said:
damn it... how do u do all these equations on the computer?? i find it hard doing my math on a computer... I am more comfortable with a pen and paper... get tired of using the "^",etc

any words of wisdom ?

You get used to imagining it in your head.
 
lol i imagine it in my head too when possible but i just can't get myself to type it all down...Good ol' pen and paper!
 

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