Weekly Challenge 25: Trig Trig Trig

7059 /www/nrich/html/content/id/7059/ 1 2011-03-14T09:31:34 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/6481&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/7071&part=">Try this next</a> </li>  <li> <a href="https://nrich.maths.org/discus/messages/27/150785.html">Ask NRICH</a> </li> <li> <a href="http://en.wikipedia.org/wiki/Mean_value_theorem">Read all about it</a> </li> <li> <a href="http://nrich.maths.org/6970&part=solution">Last week's solution</a> </li> </ul> <div> Consider the function $f(x)=\cos(\sin(\cos(x)))$, with $x$ measured in radians. What turning points can you find? What are the maximum and minimum values of the function? </div> <div class="framework"> Did you know ... ? This function is bounded, continuous and differentiable at all points. Mathematicians often use knowledge of conditions such as these to deduce lots of information about the properties of functions without the need for extensive calculation. In first year undergraduate analysis courses theorems are rigorously stated and proved which support intuitive statements such as 'between any two maxima a minimum must be found if the function is finite, continuous and differentiable'. </div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Differentiating gives $$f'(x) = \sin(\sin(\cos x)) \cdot \cos(\cos(x)) \cdot \sin(x)$$ This is zero if and only if $$\sin(\sin(\cos x))=0 \mbox{ or } \cos(\cos(x)) = 0 \mbox{ or } \sin(x)=0$$ Consider the first of these three conditions: $$\sin(\sin(\cos x))=0 \Rightarrow \sin(\cos x) = n\pi, n\in \mathbb{Z}$$ Since $|\sin(X)|\leq 1$ for any real $X$ and $\pi > 1$ we must choose $n=0$ in the previous equation. Thus, $$\sin(\sin(\cos x))=0 \Rightarrow \sin(\cos x) = 0 \Rightarrow \cos x = m\pi, m\in \mathbb{Z}$$ Similarly, we must choose $m=0$ in this expression. We can thus conclude that $$\sin(\sin(\cos x))=0 \Leftrightarrow x = \left(r+\frac{1}{2}\right)\pi, r \in \mathbb{Z}$$ Consider the second of the three conditions: $$\cos(\cos(x)) = 0 \Leftrightarrow \cos(x) = \left(r+\frac{1}{2}\right)\pi, r \in \mathbb{Z}$$ Since $\frac{1}{2}\pi> 1$ there are no real solutions to this condition. Consider the third of the three conditions: $$\sin(x) =0 \Leftrightarrow x = n\pi, n \in \mathbb{Z}$$ Combining all three conditions gives us the locations of the turning points: $$ f'(x)=0 \Leftrightarrow x = \frac{N\pi}{2}, N\in \mathbb{Z} $$ We now need to consider whether they are maxima, minima or something else. We could look at the second derivative, but this will be complicated and the boundedness of $\sin(x)$ and $\cos(x)$ allows us to make shortcuts as follows: Notice that $f(x) = 1$ when $x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \pm \frac{5\pi}{2}, \dots$. Since $f(x)$ is continuous and differentiable and $|f(x)|\leq 1$ these points must be maxima. The even multiples of $\frac{\pi}{2}$ must therefore be minima, at which the function takes the values $f(x) = \cos(\sin 1) \approx 0.666$. A plot of the graph confirms this calculation: <mdo:image height="392" width="929" src="solnGraph.png" alt=""></mdo:image> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Differentiating gives $$f'(x) = \sin(\sin(\cos x)) \cdot \cos(\cos(x)) \cdot \sin(x)$$ This is zero if and only if $$\sin(\sin(\cos x))=0 \mbox{ or } \cos(\cos(x)) = 0 \mbox{ or } \sin(x)=0$$ Consider the first of these three conditions: $$\sin(\sin(\cos x))=0 \Rightarrow \sin(\cos x) = n\pi, n\in \mathbb{Z}$$ Since $|\sin(X)|\leq 1$ for any real $X$ and $\pi > 1$ we must choose $n=0$ in the previous equation. Thus, $$\sin(\sin(\cos x))=0 \Rightarrow \sin(\cos x) = 0 \Rightarrow \cos x = m\pi, m\in \mathbb{Z}$$ Similarly, we must choose $m=0$ in this expression. We can thus conclude that $$\sin(\sin(\cos x))=0 \Leftrightarrow x = \left(r+\frac{1}{2}\right)\pi, r \in \mathbb{Z}$$ Consider the second of the three conditions: $$\cos(\cos(x)) = 0 \Leftrightarrow \cos(x) = \left(r+\frac{1}{2}\right)\pi, r \in \mathbb{Z}$$ Since $\frac{1}{2}\pi> 1$ there are no real solutions to this condition. Consider the third of the three conditions: $$\sin(x) =0 \Leftrightarrow x = n\pi, n \in \mathbb{Z}$$ Combining all three conditions gives us the locations of the turning points: $$ f'(x)=0 \Leftrightarrow x = \frac{N\pi}{2}, N\in \mathbb{Z} $$ We now need to consider whether they are maxima, minima or something else. We could look at the second derivative, but this will be complicated and the boundedness of $\sin(x)$ and $\cos(x)$ allows us to make shortcuts as follows: Notice that $f(x) = 1$ when $x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \pm \frac{5\pi}{2}, \dots$. Since $f(x)$ is continuous and differentiable and $|f(x)|\leq 1$ these points must be maxima. The even multiples of $\frac{\pi}{2}$ must therefore be minima, at which the function takes the values $f(x) = \cos(\sin 1) \approx 0.666$. A plot of the graph confirms this calculation: <mdo:image height="392" width="929" src="solnGraph.png" alt=""></mdo:image> </mdoxml> 2 4 0 0 0 0 1 Weekly Challenge 25: Trig Trig Trig A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students. Collections Weekly Challenge Admin Stage 5 - Core Mapping Stage 5 Core Mapping Document Differentiation and Integration A2 Secondary Mapping Document DisplayCabinet