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We want sin θ, where θ is the angle whose tangent is ½.θ will be an angle in the first quadrant, so sin θ will be positive.
θ
1
2
And Pythagoras says to me, he says,
5 So…
5
1
2
1tansin 1
We want cos θ, where θ is the angle whose sine is –1/3.θ will be an angle in Quadrant IV, so cos θ will be positive.
θ
1
3
And Pythagoras says to me, he says,
8
So…
3
8
3
1sincos 1
We want tan θ, where θ is the angle whose cosine is –1/3.θ will be an angle in Quadrant II, so tan θ will be negative.
θ
1
3
And Pythagoras says to me, he says,
8So…
1
8
3
1costan 1
We want an angle θ, with
whose cosecant is 2. That is,
an angle θ in Q4 or Q1 whose sine equals 1/2.
62csc 1