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

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In summary, the homework statement is to prove that \frac{sin3x}{sinx}-\frac{cos3x}{cosx} = 2. The Attempt at a Solution shows how to calculate \frac{sin(2x+x)}{sin}-\frac{cos(2x+x)}{cosx} using the sum difference formula.
  • #1
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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  • #2
thats as far i got...im really new with trig ..caught a wake up call at school so i started working with it...
 
  • #3
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:
 
  • #4
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?
 
  • #5
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?
 
  • #6
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)?
 
  • #7
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:
 

1. What are trignometric identities?

Trignometric identities are mathematical equations that involve trigonometric functions, such as sine, cosine, and tangent, and are true for all values of the variables involved. They are used to simplify and manipulate trigonometric expressions.

2. How many types of trignometric identities are there?

There are three main types of trignometric identities: Pythagorean identities, reciprocal identities, and quotient identities. Each type has multiple variations and can be used to solve different types of problems.

3. What is the Pythagorean identity?

The Pythagorean identity is a trignometric identity that relates the three basic trigonometric functions: sine, cosine, and tangent. It states that sin²θ + cos²θ = 1, and can be used to solve problems involving right triangles.

4. How can trignometric identities be used in real life?

Trignometric identities have many real-life applications in fields such as engineering, physics, and astronomy. They can be used to solve problems involving angles, distances, and other geometric concepts.

5. Are there any common mistakes when using trignometric identities?

One common mistake when using trignometric identities is forgetting to check the domain of the function. Trignometric identities are only valid for certain values of the variables, so it is important to make sure that the values being used fall within the domain of the function.

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