Mathematics

Mathematics

Problem

Which of the following identities rewrites 1−cos⁡x1-\cos x in a form involving sin⁡\sin that is particularly handy when evaluating limits as x→0x\to 0? a) 1−cos⁡x=sin⁡2x1-\cos x = \sin^{2}x b) 1−cos⁡x=sec⁡2x−11-\cos x = \sec^{2}x-1 c) 1−cos⁡x=2sin⁡2(x2)1-\cos x = 2\sin^{2}\left(\dfrac{x}{2}\right) d) 1−cos⁡x=2sin⁡(x2)1-\cos x = 2\sin\left(\dfrac{x}{2}\right)

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