Math 1113: Pre-Calculus
Graphs Of Inverse Trig Functions
y = sin−1 (x) = arcsin(x)
y = cos−1 (x) = arccos(x)
• Domain: [−1, 1]
• Domain: [−1, 1]
• Range: [−π/2, π/2] (i.e. Quads I and IV)
• Range: [0, π] (i.e. Quads I and II)
• Increasing
• Decreasing
• Odd: arcsin(−x) = − arcsin(x)
• No symmetry!
• Domain: (−∞, ∞)
y = tan−1 (x) = arctan(x)
• Range: (−π/2, π/2) (i.e. Quads I and IV)
• Increasing
• Odd: arctan(−x) = − arctan(x)
• Asymptotes: y = ±π/2
QUADRANT I NOTE: When x > 0, arctrig(x) is always the reference angle solution θR to trig(θ) = x.
QUADRANT IV NOTE: With arcsin and arctan, Quad IV gets negative values! For that quadrant, θ = −θR .
(Draw a picture with θR and −θR to see the symmetry.)
Tips for Simplifying Expressions with Inverse Trigs
1. Make sure the inverse trig makes sense!
Sample: sin(sin−1 (2)) is undefined since 2 is not in domain of sin−1 . However, sin(sin−1 (−3/5)) = −3/5.
2. Would cancelling trig and arctrig give you an angle in the correct range? Think about reference angles.
Sample: sin−1 (sin(7π/6)) is not 7π/6, since that’s not in [−π/2, π/2].
• Say θ = 7π/6. Then θR = π/6.
• Since sin(θ) < 0 (θ is in Quad III), we want our arctrig in Quad IV.
Answer: −θR = −π/6 .
3. When finding an inverse trig first, make a reference triangle and mark the correct quadrant.
Sample: cos(tan−1 (−3/4)). We first work with θ = tan−1 (−3/4), so tan(θ) = −3/4.
• Since −3/4 is negative, tan−1 must use Quad IV for θ.
• Reference equation: tan(θR ) = 3/4. By SOHCAHTOA, opp = 3 and adj = 4... find hyp = 5.
• Find cos(θR ) first: cos(θR ) = adj/hyp = 3/5
Lastly, in Quad IV, cos is positive, so our answer is cos(θ) = 3/5 .