Trigonometric Identities: Proving \frac{sin3x}{sinx}-\frac{cos3x}{cosx} = 2

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Homework Help Overview

The discussion revolves around proving the trigonometric identity \(\frac{\sin 3x}{\sin x} - \frac{\cos 3x}{\cos x} = 2\). Participants are exploring various approaches to manipulate the left-hand side of the equation using trigonometric identities and properties.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to simplify the left-hand side by using angle addition formulas and exploring common denominators. Some are questioning whether to expand \(\sin(3x)\) and \(\cos(3x)\) using double angle formulas. Others suggest rewriting the expression in terms of sine and cosine products.

Discussion Status

The discussion is ongoing, with participants providing insights and suggestions for manipulation of the expression. Some guidance has been offered regarding the use of trigonometric identities, but no consensus or resolution has been reached yet.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding the manipulation of trigonometric identities rather than arriving at a final answer.

GrandMaster87
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Homework Statement


Prove that
[tex]\frac{sin3x}{sinx}[/tex]-[tex]\frac{cos3x}{cosx}[/tex] = 2

Homework Equations


The Attempt at a Solution


LHS:[tex]\frac{sin(2x+x)}{sin}-\frac{cos(2x+x)}{cosx}[/tex]

=[tex]\frac{sin2x.cosx + cos2x.sinx}{sin}-\frac{cos2xcosx - sin2x.sinx}{cosx}[/tex]
 
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thats as far i got...im really new with trig ..caught a wake up call at school so i started working with it...
 
By multiplying [itex]\frac{sin3x}{sinx}[/itex] with [itex]\frac{cos(x)}{cos(x)}[/itex] and [itex]\frac{cos(3x)}{cos(x)}[/itex] with [itex]\frac{sin(x)}{sin(x)}[/itex] you got:

[tex]\frac{cos(x)sin(3x)-sin(x)cos(3x)}{sin(x)cos(x)}[/tex]

What can you spot now? :smile:
 
GrandMaster87 said:

Homework Statement


Prove that
[tex]\frac{sin3x}{sinx}[/tex]-[tex]\frac{cos3x}{cosx}[/tex] = 2


Homework Equations





The Attempt at a Solution


LHS:[tex]\frac{sin(2x+x)}{sin}-\frac{cos(2x+x)}{cosx}[/tex]

=[tex]\frac{sin2x.cosx + cos2x.sinx}{sin}-\frac{cos2xcosx - sin2x.sinx}{cosx}[/tex]

Did you try getting a common denominator at this point?
 
njama said:
By multiplying [itex]\frac{sin3x}{sinx}[/itex] with [itex]\frac{cos(x)}{cos(x)}[/itex] and [itex]\frac{cos(3x)}{cos(x)}[/itex] with [itex]\frac{sin(x)}{sin(x)}[/itex] you got:

[tex]\frac{cos(x)sin(3x)-sin(x)cos(3x)}{sin(x)cos(x)}[/tex]

What can you spot now? :smile:

can we expand sin(3x) and cos(3x) using double angle formula?
 
GrandMaster87 said:

Homework Statement


Prove that
[tex]\frac{sin3x}{sinx}[/tex]-[tex]\frac{cos3x}{cosx}[/tex] = 2


Homework Equations





The Attempt at a Solution


LHS:[tex]\frac{sin(2x+x)}{sin}-\frac{cos(2x+x)}{cosx}[/tex]

=[tex]\frac{sin2x.cosx + cos2x.sinx}{sin}-\frac{cos2xcosx - sin2x.sinx}{cosx}[/tex]

It can also be done onwards from here. Do you know how to expand cos(2x) and sin(2x)?
 
GrandMaster87 said:
can we expand sin(3x) and cos(3x) using double angle formula?

No.

You can write the nominator:

[tex]cos(x)sin(3x)-sin(x)cos(3x)[/tex]

as

[tex]sin(3x-x)=sin(2x)[/tex]

using the sum difference formula.

Also you can write the denominator

[tex]cos(x)sin(x)[/tex]

as

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

:smile:
 

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