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Exercise: Multiple Angle Identities · Q24

Q.Prove that 1−cos⁡2xsin⁡2x=tan⁡x\dfrac{1-\cos 2x}{\sin 2x} = \tan x.

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Concept understanding — Multiple Angle Identities

Setting y=xy=x in the sum formulas gives the double-angle

identities sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x and cos⁡2x=cos⁡2x−sin⁡2x\cos2x=\cos^2x-\sin^2x, which -- using the fundamental

identity to eliminate one of the two squared terms -- also take the equivalent forms

cos⁡2x=2cos⁡2x−1=1−2sin⁡2x\cos2x=2\cos^2x-1=1-2\sin^2x, each useful in a different situation. Writing 3x=2x+x3x=2x+x and

applying the sum formula again with the double-angle results already known gives the

triple-angle identities sin⁡3x=3sin⁡x−4sin⁡3x\sin3x=3\sin x-4\sin^3x and cos⁡3x=4cos⁡3x−3cos⁡x\cos3x=4\cos^3x-3\cos x, together with …

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