3 years ago

Verify (2cos2x)/(sin2x) =cotx - tanx

1 answer:

6 0

Cos2x = (cosx)^2 - (sinx)^2;
sin2x = 2sinxcosx;
Then, (2cos2x)/(sin2x) =2[ (cosx)^2 - (sinx)^2 ] / (2sinxcosx) = [ (cosx)^2 - (sinx)^2 ] / (sinxcosx) = (cosx)^2 / (sinxcosx) - (sinx)^2 / (sinxcosx) = cosx/sinx - sinx/cosx = cotx - tanx;

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m=rise/run OR
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$y-y_1=m(x-x_1)$

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to find "a" we plug in the given y value and solve for x (in this case x=a):

$-4=- \frac{1}{2}x-1$
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so, a=6 and the point becomes (6,-4)

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