I've been working on this for like half an hour now and don't seem to be getting anywhere. I've tried using the double angle identity to write the LHS with θ/2 and have tried expanding the RHS as a sum. The main issue I'm having is working out how to get a π on the LHS or removing it from the right to equate the two sides.
1+sinθcosθ+cosθ1−sinθ=2tan(θ2+π4)
Answer
hint:
cosθ=cos2(θ/2)−sin2(θ/2)=(cos(θ/2)+sin(θ/2))(cos(θ/2)−sin(θ/2)),1±sinθ=(cos(θ/2)±sin(θ/2))2.
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