已知tan(π/4+Q)=3,求sin2Q-2cos^2Q

Question

已知tan(π/4+Q)=3,求sin2Q-2cos^2Q

Answer 1

tan(π/4+Q)=(tanπ/4+tanQ)/(1-tanπ/4tanQ)=3
1+tanQ=3-3tanQ
tanQ=1/2
sinQ/cosQ=tanQ=1/2
cosQ=2sinQ
代入(sinQ)^2+(cosQ)^2=1
(sinQ)^2=1/5
(cosQ)^2=4/5
sinQ/cosQ=1/2>0
所以sinQcosQ>0
所以sinQcosQ=根号(sinQ)^2(cosQ)^2=2/5
所以sin2Q-2(cosQ)^2
=2sinQcosQ-2(cosQ)^2
=4/5-2*4/5
=-4/5