Harmonic form question help

[Theodore Hodson]

Homework Statement

Express $$4sin\theta-3cos\theta$$ in the form $$rsin(\theta-\alpha)$$
Hence find the maximum and minimum values of $$\frac{7}{4sin\theta-3cos\theta+2}$$
State the greatest and least values.

Homework Equations

The Attempt at a Solution

Okay so putting it in the $$rsin(\theta-\alpha)$$ form I get $$5sin(\theta-36.9^{\circ})$$

Which means $$\frac{7}{4sin\theta-3cos\theta+2}=\frac{7}{5sin(\theta-36.9^{\circ})+2}$$Now if I understand correctly it follows that $$\frac{7}{5sin(\theta-36.9^{\circ})+2}=\frac{7}{5}cosec(\theta-36.9^{\circ})+2$$

I understand the greatest and least values part of the question- they would be +infinity and -infinity as it's a cosec graph. It's just I can't seem to arrive at the answer that's in my textbook for the maximum and minimum values bit.
The textbook says the max value is $$\frac{-7}{3}$$ and that the min value is $$\frac{7}{8}$$

By my own calculations the max value would be 17/5 and the min value would be 3/5. I got this by evaluating $$\frac{7}{5}cosec(\theta-36.9^{\circ})+2$$ when $$cosec(\theta-36.9^{\circ})$$ is equal to 1 and -1

However I know this is wrong because I drew the graph on some graph drawing software which showed a maximum value of -7/3 (what my textbook got) and a minimum value of 1 (curiously my textbook said the minimum value was 7/8).

So basically I'm a bit confused, got conflicting information from my textbook and from the graph drawing software. I actually think that there might be something that I don't understand quite fundamentally here so I would appreciate any help to clear up this misunderstanding and solve the problem.

Thanking you

 

[Dr. Courtney]

This stuff is easy to double check by graphing.

 

[PeroK]

Homework Statement

Express $$4sin\theta-3cos\theta$$ in the form $$rsin(\theta-\alpha)$$
Hence find the maximum and minimum values of $$\frac{7}{4sin\theta-3cos\theta+2}$$
State the greatest and least values.

Homework Equations

The Attempt at a Solution

Okay so putting it in the $$rsin(\theta-\alpha)$$ form I get $$5sin(\theta-36.9^{\circ})$$

Which means $$\frac{7}{4sin\theta-3cos\theta+2}=\frac{7}{5sin(\theta-36.9^{\circ})+2}$$Now if I understand correctly it follows that $$\frac{7}{5sin(\theta-36.9^{\circ})+2}=\frac{7}{5}cosec(\theta-36.9^{\circ})+2$$

This is not correct. You cannot split the fraction like that.

I understand the greatest and least values part of the question- they would be +infinity and -infinity as it's a cosec graph. It's just I can't seem to arrive at the answer that's in my textbook for the maximum and minimum values bit.
The textbook says the max value is $$\frac{-7}{3}$$ and that the min value is $$\frac{7}{8}$$

By my own calculations the max value would be 17/5 and the min value would be 3/5. I got this by evaluating $$\frac{7}{5}cosec(\theta-36.9^{\circ})+2$$ when $$cosec(\theta-36.9^{\circ})$$ is equal to 1 and -1

However I know this is wrong because I drew the graph on some graph drawing software which showed a maximum value of -7/3 (what my textbook got) and a minimum value of 1 (curiously my textbook said the minimum value was 7/8).

So basically I'm a bit confused, got conflicting information from my textbook and from the graph drawing software. I actually think that there might be something that I don't understand quite fundamentally here so I would appreciate any help to clear up this misunderstanding and solve the problem.

Thanking you

The question makes no sense unless you are restricting $\theta$ to a certain range, as the function is unbounded.

Also how can $-7/3$ be a maximum and $7/8$ be a minimum? That makes no sense either.

 

[SammyS]

... The

The Attempt at a Solution

Okay so putting it in the $$rsin(\theta-\alpha)$$ form I get $$5sin(\theta-36.9^{\circ})$$
Which means $$\frac{7}{4sin\theta-3cos\theta+2}=\frac{7}{5sin(\theta-36.9^{\circ})+2}$$It's fine up to this point. The next step is incorrect.

Now if I understand correctly it follows that $$\frac{7}{5sin(\theta-36.9^{\circ})+2}=\frac{7}{5}cosec(\theta-36.9^{\circ})+2$$

As the denominator → 0, $\displaystyle \ \left|\frac{7}{5\sin(\theta-36.9^{\circ})+2}\right|\$ gets very large, so there is no global minimum or maximum.

This expression does have local minima & maxima .

 

[Theodore Hodson]

As the denominator → 0, $\displaystyle \ \left|\frac{7}{5\sin(\theta-36.9^{\circ})+2}\right|\$ gets very large, so there is no global minimum or maximum.

This expression does have local minima & maxima .

Right okay I see my mistake now. It's the $+2$ in the denominator that complicates things right? Say if it were just $\frac{7}{5sin(\theta-36.9^{\circ})}$ then you could simplify that to $\frac{7}{5}cosec(\theta-36.9^{\circ})$.

Right so going back to the correct form of the function $\frac{7}{5\sin(\theta-36.9^{\circ})+2}$.

As you say there is no global maximum or minimum because the function tails off to plus and minus infinity at certain points. The value of the function at its local minima will be when $5sin(\theta-36.9^{\circ})$ is equal to $5$ and the value at its local maxima will be when $5sin(\theta-36.9^{\circ})$ is equal to $-5$.

So for local minima, f(x) is $1$ and for local maxima f(x) is $-\frac{7}{3}$. And this is consistent with what I see when I graph this on my computer.

Thank-you very much for helping me