Algebra applied to Trigonometry problem

AI Thread Summary
The discussion centers on solving the equation tan²B = 4 for angle B, emphasizing the application of algebra in trigonometry. Participants clarify that the correct interpretation of the equation leads to two primary solutions: tan(B) = ±2. The conversation highlights that while there are four possible angle solutions within the specified domain of 0 ≤ B < 360°, the context of physics may impose further restrictions. The final accepted solutions are approximately 63.43° and 116.57°, with additional angles derived from periodicity. Overall, the thread underscores the importance of understanding trigonometric functions and their applications in physics.
Moroni
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Homework Statement



tansquaredB=4

solving for B

Homework Equations


This is algebra applied to Trigonometry, through simplification, to arrive at several different answers involving the degrees of angles.

I hope this applies to Physics since physics is "applied Mathematics"


The Attempt at a Solution



B=-tansquared+4

I don't know how to arrive at degrees.
 
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Moroni said:

Homework Statement



tansquaredB=4

solving for B

Homework Equations


This is algebra applied to Trigonometry, through simplification, to arrive at several different answers involving the degrees of angles.

I hope this applies to Physics since physics is "applied Mathematics"


The Attempt at a Solution



B=-tansquared+4
With all due respect, your attempt is completely wrong. You are treating "tansquared" is if it had been added on the left side, and have subtracted it from both sides.

Your original equation has nothing to do with addition or subtraction. It is (tan(B))2 = 4. It probably appears in your text as tan2B = 4, which is the same as what I wrote earlier.

Your equation is saying that a certain quantity (tan(B)) squared equals 4. What can you do to get rid of the exponent?
Moroni said:
I don't know how to arrive at degrees.
 


Trigonometry definitely applies to physics. You'll be using it all the time in projectile problems and some other topics.

I recommend you study some trigonometry from a maths textbook.
 


There are 2 answers...
 


zgozvrm said:
There are 2 answers...

You'd need to be more specific. There are two equations to solve, essentially tan(B)=\pm 2 but there are infinite solutions for B.
But the context of this question is for physics, so there will of course be a restriction on the values of B.
 


Mentallic said:
You'd need to be more specific. There are two equations to solve, essentially tan(B)=\pm 2 but there are infinite solutions for B.
But the context of this question is for physics, so there will of course be a restriction on the values of B.

This is a homework help forum ... I don't want to get too specific.

Unless you consider n^\circ and (n+360)^\circ to be different angles (which they are not) then there are only 2 answers.

Angles measured in degrees should be represented by the values: 0 \le n &lt; 360[/tex]<br /> Angles measured in radians should be represented by the values: 0 \le n &amp;lt; 2\pi[/tex]&lt;br /&gt; Angles measured in gradians should be represented by the values 0 \le n &amp;amp;lt; 400[/tex]
 


zgozvrm said:
Unless you consider n^\circ and (n+360)^\circ to be different angles (which they are not) then there are only 2 answers.
Physically, yes, but in most mathematics such angles are considered as different solutions. This is why most trigonometry questions are coupled with restrictions as such: (but it is important to be aware that such questions typically have infinite solutions).

zgozvrm said:
Angles measured in degrees should be represented by the values: 0 \le n &lt; 360[/tex]<br /> Angles measured in radians should be represented by the values: 0 \le n &amp;lt; 2\pi[/tex]&lt;br /&gt; Angles measured in gradians should be represented by the values 0 \le n &amp;amp;lt; 400[/tex]
That&amp;amp;#039;s fine, but this doesn&amp;amp;#039;t support your - still - incorrect statement:&amp;lt;br /&amp;gt; &amp;lt;blockquote data-attributes=&amp;quot;&amp;quot; data-quote=&amp;quot;zgozvrm&amp;quot; data-source=&amp;quot;post: 2546863&amp;quot; class=&amp;quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&amp;quot;&amp;gt; &amp;lt;div class=&amp;quot;bbCodeBlock-title&amp;quot;&amp;gt; zgozvrm said: &amp;lt;/div&amp;gt; &amp;lt;div class=&amp;quot;bbCodeBlock-content&amp;quot;&amp;gt; &amp;lt;div class=&amp;quot;bbCodeBlock-expandContent js-expandContent &amp;quot;&amp;gt; There are 2 answers... &amp;lt;/div&amp;gt; &amp;lt;/div&amp;gt; &amp;lt;/blockquote&amp;gt;With these restrictions, there are 4 answers.
 


Mentallic said:
With these restrictions, there are 4 answers.

Yes, you are right ... there are 4 answers!
 


zgozvrm said:
Yes, you are right ... there are 4 answers!

:-p
 
  • #10


Normally I would factor anything with an exponent, to get rid of it.

The problem calls for degrees.

Are 63.43 and 116.57 degrees acceptable solutions?

Thanks!
 
  • #11


Moroni said:
Normally I would factor anything with an exponent, to get rid of it.

The problem calls for degrees.

Are 63.43 and 116.57 degrees acceptable solutions?

Thanks!

That's 2 out of 4!
 
  • #12


Moroni said:
Normally I would factor anything with an exponent, to get rid of it.

The problem calls for degrees.

Are 63.43 and 116.57 degrees acceptable solutions?

Thanks!

It depends what the question is asking for. You need to solve tanB=2 and tanB=-2 so you'd be correct with those solutions, but there are more answers because tan\theta=tan(\theta+n180^o so you could also have 243.43o and 423.43o and this goes on infinitely. Is the restriction on the domain 0\leq B &lt; 360^o or something similar?
 
  • #13


Yes that is the domain like you said.
 
  • #14


Mentallic said:
Is the restriction on the domain 0\leq B &lt; 360^o or something similar?

Moroni said:
Yes that is the domain like you said.

Like I said, there are 4 answers, and you already got 2 of them:

63.43^\circ and 116.57^\circ


The other 2 would be:

-63.43^\circ = 296.57^\circ and -116.57^\circ = 243.43^\circ
 
  • #15


Thank You both Very Much!
 
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