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Verify the identity:

$$\frac{(\cos(x+h) - \cos x)}{h} = \cos x \left(\frac{\cos h - 1}{h}\right)- \sin x \left(\frac{\sin h }{h}\right)$$

=(Cosxcosh - sin x sin h -cos x)/h.

I can't think of where to go from here.

Thanks

KKendall
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  • Note on the close vote as duplicate: the cosine of sum formula is the only nontrivial identity that needs to be used. Anyone who struggles to derive equality of these terms is probably helped best by being pointed to that formula. Ref'd question was the best one I could find for the addition formula on MSE. As an external reference, see ProofWiki. – Lord_Farin Oct 09 '13 at 16:59
  • Related: http://math.stackexchange.com/questions/519096/limits-problem-with-trig-factoring-cos-ab – lab bhattacharjee Oct 09 '13 at 17:57

2 Answers2

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If I understood correctly what you have written: $$\frac{(\cos(x+h) - \cos(x))}{h}\stackrel{?}{=} \cos x \left(\frac{\cos h - 1}{h}\right)- \sin x \left(\frac{\sin h }{h}\right)$$ Or $$\frac{\cos(x+h) - \cos(x)}{h}\stackrel{?}{=}\frac{\cos(x)(\cos(h)-1)-\sin(x)\sin(h)}{h} $$ Or $$ \begin{eqnarray} \cos(x+h) - \cos(x)&=&\cos(x)(\cos(h)-1)-\sin(x)\sin(h)=\\ &=&\cos(x)\cos(h)-\cos(x)-\sin(x)\sin(h) \end{eqnarray} $$ $$\cos(x)\cos(h)-\sin(x)\sin(h)=\cos(x+h)$$ and so you have $$\cos(x+h) - \cos(x)=\cos(x+h) - \cos(x)$$

Caran-d'Ache
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Hint: You're nearly there. Which terms of the numerator have a common factor?

Cameron Buie
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