1960's basic trig question

Colin67

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Dear All

This question appears in a British textbook aimed at 16 year olds, at the end of a chapter on the "The General Angle".



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I am struggling!. I have tried using sec ^2 =1 +tan^2 identity to no avail, also not sure where beta is small comes into it.

A few clues to start would be appreciated.
 

Colin67

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Just to clarify I do not think beta is another angle, just a numerical constant, in original question I think it used the epsilon symbol, I could not use that when I transcribed the question, could not find it on keyboard!!
 

Dr.Peterson

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Has this book previously taught how to approximate \(\displaystyle \sqrt{1+x}\), when x is small? I would expect the answer to involve applying that to the definition of secant. (It's easiest in the acute case where you can just draw a triangle, but can be extended to the general case.)
 

Colin67

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No mention of that approximation in book, not come across that either
 

Jomo

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Usually when doing such a problem you need to make an approximation so that in the end you get an approximate answer.
Therefore Dr Peterson's suggestion is a very good hint. Possibly it is not the correct approximation but it gets you to think to approximate.

Alternatively you can think limits. As B gets small, then 1+B approaches 1. Therefore tan(A) approaches 1 and A approaches pi/4.
If A is close to pi/4, then since sec(pi/4)= sqrt(2) it should follow that sec(A) approaches sqrt(2).
Then sqrt(2)(1 + pi/8)= 1.969573929642890829678673847967284790896499586548909351062.
My concern is that sqrt(2)(1 + pi/8) - sec(pi/4) = 0.555360367269795780876985123757586712326827711171961277885 which is not very small! I now have doubts about the statement of your problem.
If in fact my work is incorrect can someone please point out my error. Thanks!
 

JeffM

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Jomo

I suspect the problem is misstated.

\(\displaystyle \beta \approx 0 \text { and } tan(\alpha) = 1 + \beta \implies tan(\alpha) \approx 1 \implies\)

\(\displaystyle sin(\alpha) \approx cos(\alpha) \implies \alpha = \dfrac{\pi}{4} \implies cos(\alpha) = \dfrac{1}{\sqrt{2}} \implies\)

\(\displaystyle sec( \alpha) \approx \sqrt{2} = \sqrt{2} * \left ( 1 + \dfrac{0}{2} \right ) \approx \sqrt{2} * \left ( 1 + \dfrac{\beta}{2} \right ).\)
 
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Jomo

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Jomo

I suspect the problem is misstated.

\(\displaystyle \beta \approx 0 \text { and } tan(\alpha) = 1 + \beta \implies tan(\alpha) \approx 1 \implies\)

\(\displaystyle sin(\alpha) \approx cos(\alpha) \implies \alpha = \dfrac{\pi}{4} \implies cos(\alpha) = \dfrac{1}{\sqrt{2}} \implies\)

\(\displaystyle sec( \alpha) \approx \sqrt{2} = \sqrt{2} * \left ( 1 + \dfrac{0}{2} \right ) \approx \sqrt{2} * \left ( 1 + \dfrac{\beta}{2} \right ).\)
Sweet. That is so clear!
So what is wrong in my work? It seems to show that the desired result is not correct
 

Colin67

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Thank you for the replies.

When you say the problem is misstated JeffM do you mean it has been phrased badly or I have transcribed the question incorrectly?

I have checked the book, what I wrote is the question asked.

Also I follow you suggestion until how sec is approx. 2. How then root 2 =root 2(1+0/2)?
 

Jomo

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How then root 2 =root 2(1+0/2)?
You just need to know two things here. The first is that sqrt(2) * 1 = sqrt(2). Then you need to realize that (1+0/2) =1. Putting these two together shows that sqrt(2) = sqrt(2)(1 + 0/2). Where did you get lost?
 

Colin67

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Why it can't be left as sec alpha approx. root2? putting the (1+beta/2) back in
 

Jomo

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Why it can't be left as sec alpha approx. root2? putting the (1+beta/2) back in
You were told what format the equation must be in.
 

Colin67

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Thank you so much, I understand
 

JeffM

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Thank you for the replies.

When you say the problem is misstated JeffM do you mean it has been phrased badly or I have transcribed the question incorrectly?

I have checked the book, what I wrote is the question asked.

Also I follow you suggestion until how sec is approx. 2. How then root 2 =root 2(1+0/2)?
What I meant is that when you intended to substitute beta for epsilon in the final expression, you substituted alpha instead. That results in much weirdness.
 

Subhotosh Khan

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Dear All

This question appears in a British textbook aimed at 16 year olds, at the end of a chapter on the "The General Angle".



View attachment 16629


I am struggling!. I have tried using sec ^2 =1 +tan^2 identity to no avail, also not sure where beta is small comes into it.

A few clues to start would be appreciated.
The problem statement should be:

If

tan(a) = 1 + b, ............where b is small,

prove that

sec(a) = \(\displaystyle \pm \sqrt{2}\) (1 + b/2)

sec2(a) = 1 + tan2(a) = 1 + (1 + b)2 = 2 + 2b + b2 = 2 * (1 + b) .......continue.....
 

Dr.Peterson

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This question appears in a British textbook aimed at 16 year olds, at the end of a chapter on the "The General Angle".

View attachment 16629
I am struggling!. I have tried using sec ^2 =1 +tan^2 identity to no avail, also not sure where beta is small comes into it.
An important thing we need to know in order to help you toward the solution the is expected is, what has the book taught about approximations? In order to even ask the question, they must have said something about the topic. Without a definition of "approximately", or what it takes to "prove" an approximation, it is impossible to do what is asked.

I'm very curious about what was taught!
 

Jomo

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The problem statement should be:

If

tan(a) = 1 + b, ............where b is small,

prove that

sec(a) = \(\displaystyle \pm \sqrt{2}\) (1 + b/2)

sec2(a) = 1 + tan2(a) = 1 + (1 + b)2 = 2 + 2b + b2 = 2 * (1 + b) .......continue.....
Thank you for clarifying that!
 

Jomo

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Jomo

I suspect the problem is misstated.

\(\displaystyle \beta \approx 0 \text { and } tan(\alpha) = 1 + \beta \implies tan(\alpha) \approx 1 \implies\)

\(\displaystyle sin(\alpha) \approx cos(\alpha) \implies \alpha = \dfrac{\pi}{4} \implies cos(\alpha) = \dfrac{1}{\sqrt{2}} \implies\)

\(\displaystyle sec( \alpha) \approx \sqrt{2} = \sqrt{2} * \left ( 1 + \dfrac{0}{2} \right ) \approx \sqrt{2} * \left ( 1 + \dfrac{\beta}{2} \right ).\)
Oh, now I see your point. The OP put alpha for beta.
 
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