sin2θ1+cosθ=1−cosθ
Right Side: 1−cosθ either stays the same, or can be 1−1secθ
Left Side: =sin2θ1+cosθ=1−cos2θ1+cosθ=(1−cosθ)(1+cosθ)1+cosθ=1−cosθ
Is this correct?
Answer
Perhaps slightly simpler and shorter (FYI, what you did is correct): sin2x1+cosx=1−cosx⟺sin2x=(1−cosx)(1+cosx)⟺sin2x=1−cos2x And since the last equality is just the trigonometric Pytahgoras Theorem we're done.
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