Question

question: In the given triangle ABC, prove that..... ​

Answer 1

hi there,

Formula according to double angle

cos2A=1-2sin^2A, cos2B=1-2sin^2B

∴cos2A/a^2-cos2B/b^2

=(1-2sin^2A)/a^2-(1-2sin^2B)/b^1

=1/a^2-1/b^2-2(sin^2A/a^2-sin^B/b^2)

Sine theorem

a/sinA=b/sinB

∴a^2/sin^2A=b^2/sin^2B

∴sin^2A/a^2-sin^B/b^2=0

∴1/a^2-1/b^2-2(sin^2A/a^2-sin^B/b^2)

=1/a^2-1/b^2

That is, cos2A/a^2-cos2B/b^2=1/a^2-1/b^2

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