Posted by **Joel** on Wednesday, May 8, 2013 at 12:57am.

If tan theta = 9/5 and cot omega = 9/5

Find the exact value of sin (omega-theta)

- PreCalc -
**Joel**, Wednesday, May 8, 2013 at 1:25am
sorry that should be theta-omega not omega-theta

- PreCalc -
**Steve**, Wednesday, May 8, 2013 at 11:07am
Cleverly, you note that since

tanθ = cotω, ω = π/2-θ

Thus, θ-ω = θ-(π/2-θ) = π/2+2θ

sin(θ-ω) = -cos2θ = 2sin^θ-1

tanθ = 9/5, so

sinθ = 9/√106

sin(θ-ω) = 162/106 - 1 = 56/106

or, if you must exercise your sum/difference formulas,

sin(θ-ω) = sinθcosω-cosθsinω

tanθ = 9/5, so

sinθ = 9/√106

cosθ = 5/√106

cotω = 9/5, so

sinω = 5/√106

cosω = 9/√106

sin(θ-ω) = 9/√106 * 9/√106 - 5/√106 * 5/√106 = 56/106

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