Solve tan(30)=(sin(theta))/(1+cos(theta)) Without a Graph

  • Thread starter trajan22
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In summary, the conversation explores different ways to solve the equation tan(30)=(sin(theta))/(1+cos(theta)) and suggests using a graphing calculator or trigonometric identities. The half-angle formula for tangent is mentioned, but it is pointed out that the equation is not in the correct form. The conversation also touches on the importance of taking into account both positive and negative solutions. Eventually, a solution of theta=60 is found.
  • #1
trajan22
134
1
tan(30)=(sin(theta))/(1+cos(theta))

The only way I can solve this is by using the graph on the calculator. There must be a way to solve it by hand though but I can't find it. Maybe I am just not thinking straight but its really getting to me.
 
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  • #2
What about

[tex]\cot (30) = \frac {1 + \cos (\theta)}{\sqrt {1 - \cos^{2} \theta}} [/tex]

?
 
  • #3
Isn't that original equation exactly in the form of one of the trig identities?
 
  • #4
I tried to find a trigonometric identity that reduces the expression to one unknown, but couldn't.
 
  • #5
I think I found it. It came to me when I went to get the mail. :)
basically I have this
[tex] tan(30)=\frac {\sqrt {1 - \cos^{2} \theta}}{1+\cos(\theta)}[/tex]
From here I just put it in a quadratic form and solved.
 
  • #6
The original expression is essentially one of the half-angle formulas for tangent.
 
  • #7
Almost, but not quite. It looks pretty tough to put it in that particular form. The way I mentioned works so I guess Ill just stick with that, especially since remembering all those identities is a pain.

Thanks for the suggestions though.
 
  • #8
but not knowing them is obviously more painful.
 
  • #9
trajan22 said:
Almost, but not quite. It looks pretty tough to put it in that particular form.

It's tough to put
[tex]\tan 30^\circ = \frac{\sin \theta}{1 + \cos \theta}[/tex]

into the form

[tex]\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}[/tex]

?
 
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  • #10
trajan22 said:
I think I found it. It came to me when I went to get the mail. :)
basically I have this
[tex] tan(30)=\frac {\sqrt {1 - \cos^{2} \theta}}{1+\cos(\theta)}[/tex]
From here I just put it in a quadratic form and solved.

By the way, that expression isn't right -- you don't know that the numberator is the positive root. You also have to consider

[tex]tan(30)=-\frac {\sqrt {1 - \cos^{2} \theta}}{1+\cos(\theta)}[/tex]
 
  • #11
Hurkyl said:
It's tough to put
[tex]\tan 30^\circ = \frac{\sin \theta}{1 + \cos \theta}[/tex]

into the form

[tex]\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}[/tex]

?

Unless I am missing something I thought the half angle formula was [tex]

\tan \frac{\theta}{2} = \frac{\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}}[/tex]

Thats not the same as [tex]\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}[/tex] right? Or am I making a mistake?
 
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  • #12
And just what multiplying [tex] \frac{\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}} [/tex] by

[tex]\frac {\sqrt{ 1 + \cos \theta}}{\sqrt{ 1 + \cos \theta}} [/tex]

gives?
 
  • #13
Werg22 said:
And just what multiplying [tex] \frac{\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}} [/tex] by

[tex]\frac {\sqrt{ 1 + \cos \theta}}{\sqrt{ 1 + \cos \theta}} [/tex]

gives?

Well, it gives:
[tex]... = \frac{\sqrt{1 - \cos ^ 2 \theta}}{1 + \cos \theta} = \frac{\textcolor{red} {|} \sin \theta \textcolor{red} {|}}{1 + \cos \theta}[/tex] (Notice that it's never negative, since there's an absolute value in it)

What you should use is Half-Angle and Power Reduction Identities:

[tex]\sin (2 \theta) = 2 \sin ( \theta ) \cos ( \theta )[/tex]

and: [tex]\cos ^ 2 \theta = \frac{1 + \cos (2 \theta)}{2}[/tex]
 
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  • #14
?

One solution is found by taking the positive root and the other by taking the negative, as such

[tex] \tan \frac{\theta}{2} = \frac{-\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}}[/tex]
 
  • #15
Werg22 said:
?

One solution is found by taking the positive root and the other by taking the negative, as such

[tex] \tan \frac{\theta}{2} = \frac{-\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}}[/tex]

What Hurkyl, and I are (?, is it is, am, or are should be used here?) trying to say is that, the expression:
[tex]\tan \left( \frac{\theta}{2} \right) = \frac{\sqrt{1 - \cos \theta}}{\sqrt{1 + \cos \theta}}[/tex] is, indeed, incorrect.

It'll be correct if you take into account its negative part as well, i.e using the (+/-) sign as you did.
 
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  • #16
VietDao29 said:
What Hurkyl, and I are (?, is it is, am, or are should be used here?) trying to say is that, the expression:
[tex]\tan \left( \frac{\theta}{2} \right) = \frac{\sqrt{1 - \cos \theta}}{\sqrt{1 + \cos \theta}}[/tex] is, indeed, incorrect.

It'll be correct if you take into account its negative part as well, i.e using the (+/-) sign as you did.

Technicalities, you say tomato, I say tomato.
 
  • #17
trajan22 said:
tan(30)=(sin(theta))/(1+cos(theta))

The only way I can solve this is by using the graph on the calculator. There must be a way to solve it by hand though but I can't find it. Maybe I am just not thinking straight but its really getting to me.

tan30=sin(@)/(1+cos@)
=2sin(@\2)cos(@\2)/2cossquare(@/2)
=tan(@/2)
implies 30=@/2
@=60

here @=theta
 
  • #18
pradeep reddy said:
@=60
Of course, that's merely one solution.

(p.s. use [ code ] ... [ /code ] tags for preformatted text)
 

1. How do I solve tan(30)=(sin(theta))/(1+cos(theta)) without a graph?

To solve this equation without a graph, you can use the trigonometric identity tan(x)=(sin(x))/(cos(x)). This means that tan(30)=(sin(30))/(cos(30)). Since sin(30)=1/2 and cos(30)=√3/2, we can substitute these values into the equation to get tan(30)=(1/2)/((√3/2)+1)=1/√3=√3/3. Therefore, the solution for theta is 30 degrees.

2. What is the trigonometric identity used to solve this equation?

The trigonometric identity used to solve this equation is tan(x)=(sin(x))/(cos(x)). This identity relates the tangent function to the sine and cosine functions.

3. Is a calculator necessary to solve this equation?

No, a calculator is not necessary to solve this equation. You can use the trigonometric identity tan(x)=(sin(x))/(cos(x)) and substitute in the values for sin(30) and cos(30) to solve the equation without a calculator.

4. Can this equation be solved without any prior knowledge of trigonometry?

No, this equation requires prior knowledge of trigonometry in order to be solved. You need to know the trigonometric identity tan(x)=(sin(x))/(cos(x)) and the values for sin(30) and cos(30) in order to solve this equation.

5. Are there any alternate methods for solving this equation without a graph?

Yes, there are alternate methods for solving this equation without a graph. One method is to use the double angle formula for tangent, which states that tan(2x)=(2tan(x))/(1-tan^2(x)). By setting 2x=30 degrees, we can solve for x and get the same solution of 30 degrees. Another method is to use the unit circle to find the values of sin(30) and cos(30) and then substitute them into the equation.

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