If tan theta = 9/5 and cot omega = 9/5
Find the exact value of sin (omega-theta)
2 answers
sorry that should be theta-omega not omega-theta
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
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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