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I'm having trouble with a real-life application of Trigonometry.

Exam Paper: https://qualifications.pearson.com/content/dam/pdf/A%20Level/Mathematics/2013/Exam%20materials/6665_01_que_20130613.pdf

Question 8b.

enter image description here

If someone could be kind enough to take a screenshot and attach it to this thread, I'd appreciate it.

Right, so we are given the formula for Kate's velocity.

$$ V = \frac{21}{24\sin \theta + 7\cos \theta} $$ It can be shown, $$ 24\sin \theta + 7\cos \theta = 25\cos(\theta -73.74) $$ $$ V = \frac{21}{25\cos(\theta - 73.74)}$$

The question then states: Assuming $ 0<\theta<150 $ find the minimum value of $V$

My first attempt at this question, I did the following: The minimum of $ \cos (f(\theta)) = -1 $ therefore, the minimum of $V = -\frac{21}{25}$ but it says that the minimum of $V = \frac{21}{25} $

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  • $\begingroup$ Quick work, I like it, thanks @mrnovice $\endgroup$ – ddsffsdsdfff Feb 26 '17 at 20:48
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Note that $V=\frac{21}{25\cos(\theta-73.74^{\circ})}$ is never negative on the interval $0< \theta < 150^{\circ}$ as shown on the graph below. Therefore, to minimize $V$, $\cos(\theta-73.74^{\circ})=1$.

enter image description here

Therefore, the minimum value is $V=\frac{21}{25}$.

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  • $\begingroup$ I'm silly, thanks. So, assuming $ V$ could be negative then the minimum would be $-21/25$ right $\endgroup$ – ddsffsdsdfff Feb 26 '17 at 20:58
  • $\begingroup$ Actually that is not true. The minimum value for the full domain would be $-\infty$ since $\cos(f(\theta))$ can get very close to zero (Which means that the denominator will get very close to zero too), and thus the whole expression would tend to $-\infty$ as you can see on the graph. $\endgroup$ – projectilemotion Feb 26 '17 at 21:01
  • $\begingroup$ Was going to post exact same thing, but you beat me to it. Also worth nothing that you can find out that V is positive within the range of $\theta$ provided by seeing when $\cos(\theta-73.74) = 0$ to identify the asmyptotes. If V could be negative then the problem changes. Though in the context of the problem of crossing a road this wouldn't make much sense. So in an exam, you could use the context to conclude your answer could be wrong and then investigate further. $\endgroup$ – mrnovice Feb 26 '17 at 21:02
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    $\begingroup$ Yes velocity can be negative, but why would a person crossing the road move backwards? $\endgroup$ – mrnovice Feb 26 '17 at 21:06
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    $\begingroup$ Thanks both of you, I really need to improve my understanding. But I'm enjoying maths :) $\endgroup$ – ddsffsdsdfff Feb 26 '17 at 21:08

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