Trig equation tan(2x+pi/4)=0

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'sin^2(pi/6)+cos^2(pi/3)- tan^2(pi/4)=-1/2'

Evaluate for theta between 0 and 2pi

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Prove sin^2(pi/6) + cos^2(pi/3) - tan^2(pi /4) = -1/2

evaluate tan pi over 4

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Prove that: sin ^2π/6+cos ^2π/3-tan ^2π/4=-1/2 LHS = sin ^2π/6+cos ^2π/3-tan ^2π/4 = (1/2)^2+(1...

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