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My first line of reasoning was to try directly evaluating $$\lim\limits_{h \to 0}\frac{\cos^2 (x+h) - \cos^2 (x)}{h}$$ and showing such a limit existed for any x, but when $\cos^2(x)$ evaluates to zero (e.g. when $x = \frac{\pi}{2}$), then directly evaluating the limit yields the indeterminate form $\frac{0}{0}$. The same thing happens even after transforming $\cos^2 x$ to $\frac{1}{2} \cos (2x)$, so I don't think this is the correct strategy. I considered using L'Hopitals rule, but don't think that is the route that the question intended, as it is suggested to use the identity $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$ along with standard trigonometric identities. I am not sure where to begin. What would be a step in the right direction?

James Hiew
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  • Well, the chain rule tells you it is everywhere differentiable... – Amateur May 14 '14 at 23:23
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    What do you know at this point about the unsquared cosine? And what is known about binomial and trigonometric identities? Also note that this limit is for any $x$ of the form $\frac 00$, which is the nature of differential quotients as limit of difference quotients. – Lutz Lehmann May 14 '14 at 23:27
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    Factor the top. Then for the $\frac{\cos(x+h)-\cos x}{n}$ part, use $\cos(x+h)=\cos x\cos h-\sin x\sin h$. – André Nicolas May 14 '14 at 23:34

3 Answers3

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Use the trigonometric identity $$ \cos A - \cos B = -2 \sin\frac{A+B}2\sin\frac{A-B}2, $$ either directly or for the $\cos 2x$ form.

Lutz Lehmann
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  • This identity was very helpful. As soon as I saw $\frac{sin(h)}{h}$ appear I realised it could be substituted for 1, removing the $h$ in the denominator that was causing so much trouble. Thank you. – James Hiew May 14 '14 at 23:55
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$$\begin{align}L &= \lim\limits_{h \to 0}\frac{\cos^2 (x+h) - \cos^2 (x)}{h} \\ &= \lim\limits_{h \to 0}\frac{(\cos(x)\cos(h)-\sin(x)\sin(h))^2 - \cos^2 (x)}{h}\\ &= \lim\limits_{h \to 0}\frac{\cos^2(x)\cos^2(h) + \sin^2(x)\sin^2(h) - 2\cos(x)\cos(h)\sin(x)\sin(h) - \cos^2 (x)}{h} \\ &= \lim\limits_{h \to 0} \dfrac{\cos^2(x)(\cos^2(h)-1)}{h} + \lim\limits_{h \to 0} \sin^2(x)\sin(h)\dfrac{\sin(h)}{h}- \lim\limits_{h \to 0} \dfrac{\sin(2x)\sin(2h)}{2h} \\ &= \cos^2(x)\lim\limits_{h \to 0} \dfrac{\cos^2(h)-1}{h} + \sin^2(x)\lim\limits_{h \to 0}\sin(h)- \sin(2x)\lim\limits_{h \to 0} \dfrac{\sin(2h)}{2h}\\ &= \cos^2(x)\lim\limits_{h \to 0} \dfrac{(\cos(h)-1)}{h}\lim\limits_{h \to 0}(\cos(h)+1) - \sin(2x)\\ &= 2\cos^2(x)\lim\limits_{h \to 0} \dfrac{(\cos(h)-1)}{h} - \sin(2x) \\ &= -\sin(2x)\end{align}$$

Brad
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HINT:

$$\cos^2B-\cos^2A=1-\sin^2B-(1-\sin^2A)$$

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $, this becomes $$\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$$

So, $$\cos^2(x+h)-\cos^2x=\sin(2x+h)\sin(-h)=-\sin(2x+h)\sin(h)$$

Community
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lab bhattacharjee
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