若1+tan a/1-tan a=3+2根号2 则sin2a=?
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若1+tan a/1-tan a=3+2根号2 则sin2a=?
怎么算呢?
怎么算呢?
(1+tan a)/(1-tan a)=3+2√2→
(tan π/4 + tan a)/(1-tan π/4 · tan a)= tan (π/4 + a)
即tan (π/4 + a)=3+2√2
根据万能公式,cos(π/2 + 2a)
[1-tan^2 (π/2 + 2a)]/[1+tan^2 (π/2 + 2a)]
=[1-(3+2√2)^2]/[1+(3+2√2)^2]
=[1-(17+12√2)]/[1+(17+12√2)]
=-[16+12√2]/[18+12√2]
=-[8+6√2]/[9+6√2]
=-(2√2)/3
则cos(π/2 + 2a)
=-cos(2a-π/2)
=-cos(π/2-2a)
=-sin2a
即-sin2a=-(2√2)/3
则sin2a=(2√2)/3
(tan π/4 + tan a)/(1-tan π/4 · tan a)= tan (π/4 + a)
即tan (π/4 + a)=3+2√2
根据万能公式,cos(π/2 + 2a)
[1-tan^2 (π/2 + 2a)]/[1+tan^2 (π/2 + 2a)]
=[1-(3+2√2)^2]/[1+(3+2√2)^2]
=[1-(17+12√2)]/[1+(17+12√2)]
=-[16+12√2]/[18+12√2]
=-[8+6√2]/[9+6√2]
=-(2√2)/3
则cos(π/2 + 2a)
=-cos(2a-π/2)
=-cos(π/2-2a)
=-sin2a
即-sin2a=-(2√2)/3
则sin2a=(2√2)/3
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