precalc

cos(θ)=-((3)/(7)) where pi<theta <3pi/2 find value sin2 theta

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  1. 1. Get the inverse of each side in cos(θ)=-((3)/(7)).

    Cos^-1 (3/7) = theta.

    2. From trig identities, we know sin2θ = 2sinθcosθ

    Substitute that in for sin2θ and just solve since you figured out θ in part 1 and you were given what cosθ equals, which is 3/7.

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  2. Oh, I forgot to add the negative sign with 3/7.

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