Solve Trigonometry Proofs: Tan(x) – ½sin(2x) = tan(x)sin2(x)

[Mathhelp77]

Homework Statement

I need help with trigonometry proofs. the question asks me to prove the following and show all my steps.

Prove that: Tan(x) – ½sin(2x) = tan(x)sin2(x)

Homework Equations

I don't know :(

The Attempt at a Solution

No attempt as I don't get it.Any help at all would be greatly appreciated! Thanks so much!

 

[Mark44]

Homework Statement

I need help with trigonometry proofs. the question asks me to prove the following and show all my steps.

Prove that: Tan(x) – ½sin(2x) = tan(x)sin2(x)

Homework Equations

I don't know :(

The Attempt at a Solution

No attempt as I don't get it.


Any help at all would be greatly appreciated! Thanks so much!

What identities do you know? The double angle formula for sine is one that you need in this problem.

 

[Mark44]

Homework Statement

I need help with trigonometry proofs. the question asks me to prove the following and show all my steps.

Prove that: Tan(x) – ½sin(2x) = tan(x)sin2(x)

I'm pretty sure this should be tan(x) - (1/2)sin(2x) = tan(x)sin²(x)

Homework Equations

I don't know :(

The Attempt at a Solution

No attempt as I don't get it.


Any help at all would be greatly appreciated! Thanks so much!

 

[Mathhelp77]

Yes your right... sorry that should be

tan(x) - (1/2)sin(2x) = tan(x)sin^2(x)

and as for identities i know I have a formula sheet with quite a few.

such as:
csc(x) = 1/sin(x)
cot(x) = 1/tan(x)
sec(x) = 1/cos(x)
tan(x) = sin(x)/cos(x)
cot(x) = cos(x)/sin(x)
sin^2(x) +cos^2(x) = 1
1 +tan^2(x) = sec^2(x)
1 +cot^2(x) = csc^2(x)
sin(A+B) = (sinA)(cosB) + (cosA)(sinB)
sin(A-B) = (sinA)(cosB) - (cosA)(sinB)
cos(A+B) = (cosA)(cosB) - (sinA)(sinB)
cos(A-B) = (cosA)(cosB) + (sinA)(sinB)
sin(2A) = 2(sinA)(cosA)
cos(2A) = cos^2A - sin^2A

 

[AlexChandler]

Try writing tangent as sine over cosine. Also use the double angle formula for sin2x. Then get a common denominator for the two terms. You will need to use a form of the pythagorean identity. If you don't know what these things are, look them up on wikipedia.

 

[Mathhelp77]

Okay I got...

2sin(x) - 2sin(x)cos2(x) = tan(x)sin2(x)
2cos(x) 2cos(x)Am I on the right track? and where would I go from there?

 

[AlexChandler]

Okay I got...

2sin(x) - 2sin(x)cos2(x) = tan(x)sin2(x)
2cos(x) 2cos(x)


Am I on the right track? and where would I go from there?

Well yes but it is hard to understand your math. It is not quite clear that cos2(x) is cosine squared, and it is not clear that the LHS of the equation is being divided. Try to use LaTeX. It is much easier to understand. For example, you equation in LaTeX is

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

Now you should be able to combine the two terms on the LHS and factor out sin(x). From here you should get to the answer after using the pythagorean identity.

 

[Mathhelp77]

alright then...

but would the 2sin(x)'s just cancel out to leave me with

cos^2(x)
2cos(x)

and this would just turn into

cos(x)
2

but that can't be right because I have to get it to be tan(x)sin^2(x) ?

*and the LaTex thing wasn't working for me sorry

 

[AlexChandler]

alright then...

but would the 2sin(x)'s just cancel out to leave me with

cos^2(x)
2cos(x)

and this would just turn into

cos(x)
2

I don't know why the 2sin(x)'s would cancel... try this...

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

That was combining the two terms with a common denominator. now... get rid of the 2's... you don't need them

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

now factor the numerator

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

From here you should be able to use a pythagorean identity, and some canceling to finish the proof.

From your comment about the 2 sin(x)'s canceling... I think you may need to go back and work on algebra. Is this what you were thinking would happen?

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

This is wrong. Do you see why?

 

[Mathhelp77]

okay yes I see where I went wrong...that was silly! But how did you get the numerator to equal

six(x)(1 - cos^2(x)

I got the numerator as sin(x) - ((sin(x))(sin(x)-1))

this would then simplify to

2sin(x) +sin^2(x) (still all over cos(x)) but this isn't right?

I'm sorry, I am actually decent at trig, just really not showing it here!
You are helping lots though!

 

[eumyang]

okay yes I see where I went wrong...that was silly! But how did you get the numerator to equal

six(x)(1 - cos^2(x)

I got the numerator as sin(x) - ((sin(x))(sin(x)-1))

this would then simplify to

2sin(x) +sin^2(x) (still all over cos(x)) but this isn't right?

I'm sorry, I am actually decent at trig, just really not showing it here!
You are helping lots though!

No, you don't want to do that.

We go from this:
[tex]\sin x - \sin x \cos^2 x[/tex]

to this:
[tex]\sin x(1 - \cos^2 x)[/tex]

by factoring out the GCF, which is sin x. It's sort of like factoring this algebraic expression:
[tex]y - yz^2 = y(1 - z^2)[/tex]

Do you see it now?