Finding sine and cosine formulas

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Homework Help Overview

The discussion revolves around proving a relationship involving sine and cosine functions given specific ratios of sine and cosine values. The subject area includes trigonometric identities and relationships.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss expressing cosine and sine in terms of each other based on given ratios, and substituting these into the left-hand side of the equation to prove the identity.

Discussion Status

Some participants have provided hints and suggestions for manipulating the equations, while others have expressed uncertainty about how to proceed with substitutions and maintaining the integrity of the proof. There is an ongoing exploration of how to incorporate sine and cosine values effectively.

Contextual Notes

Participants are navigating the constraints of the problem, including the need to prove an identity without losing track of the original sine and cosine relationships established in the problem statement.

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[SOLVED] Finding sine and cosine formulas

Homework Statement


If sinx/siny = 1/2 and cosx/cosy = 3 prove:

sin (x + y) = 7/3 sinx cosx


Homework Equations


sin (x + y) = sinx cosy = cosx siny


The Attempt at a Solution



Can someone please give me a hint so that I can start? Thanks.
 
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Well fine cosy in terms of cosx from the 2nd formula and find siny in terms of sinx from the 1st formula and sub into the LHS of the proof
 
Ok, here it goes:

* sinx/siny = 1/2

siny = 2sinx

* cosx/cosy = 3

cosy = cosx/3sin(x + y) = sinx cosy + cosx siny
= cosx/3 + 3 X 2sinx
= cosx/3 + 6sinx
= 7/3 sinx cosxIs this possibly right?
Please help. Thanks
 
well that is correct...but you should put in the sinx and cosx in the 2nd and 3rd lines...or else it may seem weird that you simply got back the sinx and cosx at the end
 
Well, If I put back sinx and cosx, then how will I get rid of them in the end?
 
sin(x + y) = sinx cosy + cosx siny
= (sinxcosx)/3 + 2sinxcosx
=[itex]\frac{7sinxcosx}{3}[/itex]
 
Thanks very much rock.freak667. I finally get this.
 

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