Can \cos(\theta-\lambda) Be Used to Show Sin C in Addition Formula/Sine Rule?

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Homework Help Overview

The discussion revolves around the use of trigonometric identities, specifically the sine and cosine functions, in the context of a triangle's angles and their relationships. Participants are exploring how to express \(\sin C\) in terms of \(\cos(\theta - \lambda)\) within the framework of the sine rule and addition formulas.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to derive expressions for angle C and its relationship to other angles in the triangle. Questions about the interpretation of angle C and the attempts to apply the sine addition formula are raised.

Discussion Status

There is ongoing exploration of the relationships between the angles in the triangle, with some participants providing their reasoning and attempts. The discussion includes attempts to apply trigonometric identities, but there is no explicit consensus on the approach or final expression.

Contextual Notes

Some participants express confusion regarding the diagram and the specific angles involved, indicating a need for clarification on the visual representation of the problem. The angles in a triangle are noted to sum to \(\pi\) radians, which is a fundamental assumption in the discussion.

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http://img117.imageshack.us/img117/1482/graph015tt6.png
I know an alternative way of showing Sin C is cos 'theta'.
Thanks
 
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What is angle C? I can't make it out in the picture... Also could you show your attempt of the problems?
 
Angle C is pi/2 - theta (radians) "pi over two minus theta" ie. 90 deegrees - theta.
I did show my attempt at it, i couldn't get further.
 
Certainly, since the angles in a triangle add to [itex]2\pi[/itex] radians, [itex]B= 2\pi- \lambda- (\pi/2- \theta)= 3\pi/2- (\theta- \lambda)[/itex].
Okay, using sin(a+ b)= sin(a)cos(b)- cos(a)sin(b)[/itex], what is
[itex]sin(3\pi/2+ (-(\theta- \lambda)))[/itex]?
 
HallsofIvy said:
Certainly, since the angles in a triangle add to [itex]2\pi[/itex] radians,

The angles of a triangle add upto [itex]\pi[/itex] radians
[itex]B= 2\pi- \lambda- (\pi/2- \theta)= 3\pi/2- (\theta- \lambda)[/itex].
thus [itex]B=\pi-\lambda-(\pi/2-\theta)=\pi/2+(\theta-\lambda)[/itex]
Okay, using sin(a+ b)= sin(a)cos(b)- cos(a)sin(b)[/itex], what is
[itex]sin(3\pi/2+ (-(\theta- \lambda)))[/itex]?

The expression you require is therefore [itex]\sin(\pi/2+(\theta-\lambda))[/itex] = ?
 
[itex]\cos(\theta-\lambda)[/itex]

Thanks
 

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