Question

write in simplest form 2cosx/sin2x

Answer 1

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Answer 2

Time to haul out our table of trig identities. If you'd find a suitable identity for sin 2x, you'll probably see right away how you could reduce 2cos x / sin 2x.

Answer 3

use the formula for sin2x: \[\sin 2x = 2\sin x \cos x\] in the denominator. Then, cancel the cosines to get 1/(2sinx) = (csc x)/2.

Answer 4

\[\frac{ 2cosx }{ \sin2x }=\frac{ 2cosx }{ 2sinxcosx }=\frac{ 1 }{ sinx }=\sin^{-1}x\]

Answer 5

=cscx

Answer 6

oops. forgot about the 2 in the numerator...:)

Answer 7

thank you

Answer 8

\[\frac{ 2cosx }{ \sin2x }=\frac{ 2cosx }{ 2sinxcosx }=\frac{ 1 }{ sinx }=\sin^{-1}x\] is almost, but not quite, right. Up to 1/sin x, great. But

Answer 9

\[\frac{ 1 }{ \sin x }\neq \sin ^{-1}x.\] Rather, \[\frac{ 1 }{ \sin x }= \csc x\]

Answer 10

and \[\sin ^{-1}x \] is a function in its own right: the INVERSE SINE FUNCTION.

Answer 11

indeed @mathmale