Help with limits of a trig equation, without calculus

[Colin67]

Hi All

Some help please with second part of the attached question. I have proved the fist part using the t-substitution for sin and cos.



The question is at the end of trigonometric functions chapter, no calculus methods used in chapter. I know minimum and maximum values of each individual expression, and happy it is greater than or equal to zero but I don't know how you get the 10/9. I have graphed the expression and it is correct.

Help please!!

 

[Cubist]

Does the book show how to find the max and min using calculus in a previous chapter? Maybe this question is a "crossover" question and you are meant to see that it is easier to find the max/min using calculus when the expression is in the t-substitution form.

 

[Cubist]

A potential way to do this without calculus is to write

[math] \frac{\left(1+t\right)^2}{9+t^2}=\frac{10}{9}+c [/math]
and then attempt to prove that c must be ≤0 for real roots. This would prove 10/9 is the maximum real value of the expression. To do this rearrange into a quadratic

[math] \left(1+9c\right)t^2 - 18t+(81c+81)=0 [/math]
and prove the discriminant is ≥0 only if c≤0. Then do something similar for the lower bound.

I personally don't like this method since it requires the knowledge of the answer, and the question says "deduce" not "prove" (I think this means the maximum value should be found by the chosen method)

 

[Colin67]

Thank you for replies, it has not covered differentiation of trigonometric expressions yet in book, but you could be right that although the question in in the trig section, when the exam question was set, it assumed other knowledge to be applied

 

[Cubist]

A potential way to do this without calculus is to write

[math] \frac{\left(1+t\right)^2}{9+t^2}=\frac{10}{9}+c [/math]

Actually a better method is to simply write...

[math] \frac{\left(1+t\right)^2}{9+t^2}=c [/math]
Expand into a quadratic, similar to above, and work out the range of values of c that give ≥0 discriminant. This does not require knowledge of the limits beforehand

 

[MarkFL]

I would let:

[MATH]y=\frac{(t+1)^2}{t^2+9}[/MATH]
Arrange as:

[MATH](y-1)t^2-2t+9y-1=0[/MATH]
Now, require the discriminant to be non-negative:

[MATH]4-4(y-1)(9y-1)\ge0[/MATH]
[MATH]1-(y-1)(9y-1)\ge0[/MATH]
[MATH]y(10-9y)\ge0[/MATH]
This implies:

[MATH]0\le y\le\frac{10}{9}[/MATH]

 

[MarkFL]

Haha...beaten to the punch!

 

[Colin67]

Dear MarkFL and Cubist

Thank you so much for your help, a very neat method, probably the type of approach they were looking for.

 

[MarkFL]

Dear MarkFL and Cubist

Thank you so much for your help, a very neat method, probably the type of approach they were looking for.

One thing I considered before trying this approach is that:

[MATH]t=\tan\left(\frac{\theta}{2}\right)[/MATH]
has all reals as its range with \(\theta\) unrestricted, and so the domain of \(y\) as I defined it is therefore all reals.

 

[Cubist]

...has all reals as its range with \(\theta\) unrestricted, and so the domain of \(y\) as I defined it is therefore all reals.

I see. Without stating this there is a gap in the proof. Very thorough!

 

[JeffM]

One thing I considered before trying this approach is that:

[MATH]t=\tan\left(\frac{\theta}{2}\right)[/MATH]
has all reals as its range with \(\theta\) unrestricted, and so the domain of \(y\) as I defined it is therefore all reals.

Sorry to be a pain, but

[MATH]\theta = \pi \implies tan \left ( \dfrac{\theta}{2} \right ) \not \in \mathbb R,[/MATH]
or am I missing something again?

 

[JeffM]

By the way, [MATH]y = \dfrac{1 + sin(\theta)}{5 + 4cos(\theta)}[/MATH]
is real for all values of theta. The expression in t is not an exact equivalent.

 

[MarkFL]

Sorry to be a pain, but

[MATH]\theta = \pi \implies tan \left ( \dfrac{\theta}{2} \right ) \not \in \mathbb R,[/MATH]
or am I missing something again?

Yes, the domain of the function \(t(\theta)\) is naturally restricted, but with no other restrictions being applied, the range of \(t\) is all reals.

 

[JeffM]

Ahh. I get you. We forget about how t was derived and just look at (t + 1)^2 / (t^2 + 9).

 

[MarkFL]

Ahh. I get you. We forget about how t was derived and just look at (t + 1)^2 / (t^2 + 9).

To be honest, I was only looking at the range of the expression in \(t\). Let's begin with the identity:

[MATH]t=\tan\left(\frac{\theta}{2}\right)=\frac{\sin(\theta)}{1+\cos(\theta)}[/MATH]
[MATH]t^2=\frac{1-\cos(\theta)}{1+\cos(\theta)}[/MATH]
And we find:

[MATH]\cos(\theta)=\frac{1-t^2}{1+t^2}[/MATH]
[MATH]\sin(\theta)=t\left(1+\frac{1-t^2}{1+t^2}\right)=\frac{2t}{t^2+1}[/MATH]
And so:

[MATH]\frac{1+\sin(\theta)}{5+4\cos(\theta)}=\frac{1+\dfrac{2t}{t^2+1}}{5+4\dfrac{1-t^2}{1+t^2}}=\frac{t^2+2t+1}{t^2+9}=\frac{(t+1)^2}{t^2+9}[/MATH]

 

[JeffM]

I must admit that I am still bothered because that denominator of 1 + cos(theta) can be zero. The unmodified expression of course does not have that issue so I am being persnickety.