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Verify that sin2x=2cotsin^2x is an identity

2 Answers

1 vote

Explanation:

sin^2x = 2cotx sin^2x

Rewrite right side as fractions:

sin^2x =
(2)/(1) *
(cosx)/(sinx) *
((sinx)(sinx))/(1)

Multiply together
(cosx)/(sinx) and
((sinx)(sinx))/(1) :

sin^2x =
(2)/(1) *
((cosx)(sinx)(sinx))/(sinx)

Cancel out sinx on top and bottom:

sin^2x =
(2)/(1) *
((sinx)(cosx))/(1)

Multiply together 2 and (sinx)(cosx):

sin^2x = 2sinxcosx

Substitute sin^2x in for 2sinxcosx:

sin^2x = sin^2x

User Aarreoskari
by
8.1k points
7 votes

Answer:

Vertify is an identity

Sin2x=2cotx(sin^2x)

starting from the right-hand side

2cotx(sin^2x)

=2(cosx/sinx)(sin^2x)

=2(cosx/sinx)(sin^2x)

=2sinxcosx=sin2x

ans:right-hand side=left-hand side

Explanation:

User Raynold
by
9.1k points

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