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TheLindsy · Feb 18, 2009, 02:24 PM

I'm stuck on a problem, and I want to figure it out so that I will know it later if it might appear on a test or something.

cot2A = 1/2(cotA-tanA)

Thanks!

galactus · Feb 18, 2009, 02:36 PM

$cot(2A)=\frac{1}{2}(cot(A)-tan(A))$

Note the identities: $cot(2A)=\frac{cos(2A)}{sin(2A)}$

$cos(2A)=cos^{2}(A)-sin^{2}(A)$

$sin(2A)=2sin(A)cos(A)$

$\frac{cos^{2}(A)-sin^{2}(A)}{2sin(A)cos(A)}$

$\frac{1}{2}\frac{cos(A)}{sin(A)}-\frac{1}{2}\frac{sin(A)}{cos(A)}$

$=\frac{1}{2}cot(A)-\frac{1}{2}tan(A)$

$\frac{1}{2}(cot(A)-tan(A))$