inverse trig functions

inverse trig functions - 6.1: Inverse Trigonometric...

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Unformatted text preview: 6.1: Inverse Trigonometric functions Function Domain −1 y = sin x Range −1 ≤ x ≤ 1 − y = arcsin x If x → −x π π ≤y≤ 2 2 examples sin −1 (0.5) = y → −y quadrant Q1 π 6 sin −1 ( −0.5) = − π 6 ------------------------------------- sin −1 (0.5) = 30° −90° ≤ y ≤ 90° y = cos −1 x −1 ≤ x ≤ 1 0≤ y≤ π y = arccos x y → π−y sin −1 ( −0.5) = −30° π cos −1 (0.5) = 3 π 2π cos −1 ( −0.5) = π − = 3 3 ------------------------------------- 0° ≤ y ≤ 180° y = tan −1 x All Reals − y = arctan x y → 180° − y π π <y< 2 2 y → −y cos −1 (0.5) = 60° cos −1 ( −0.5) = 180° − 60° = 120° π tan −1 (1) = 4 π tan −1 ( −1) = − 4 ------------------------------------- −90° < y < 90° tan −1 (1) = 45° tan −1 ( −1) = −45° Q4 --------Q1 Q4 Q1 Q2 --------Q1 Q2 Q1 Q4 --------Q1 Q4 If an expression contains a trig function evaluated at an inverse trig function, we can think of the inverse trig function as representing an angle with the given trig value. Ex 1: sin(tan −1 3) can be thought of as representing the sine of an angle that has a tangent value of 3. A triangle can be used to compute that the sine of this angle would be 3 10 3 10 −1 , so sin(tan 3) = . 10 10 3 5 Ex 2: sin 2arcsin can be thought of as the sine of a double angle. Using identity 7: 3 3 4 24 . sin 2arcsin = sin(2θ) = 2sin θ cos θ = 2 = 5 5 5 25 Ex 3: tan(arctan 5 + 45° ) can be thought of as the tangent of the sum of two angle, one with a tangent value of 5, and the 5+1 6 other ( 45° ) with a tangent value of 1. Using identity 3, we get = = −1.5 . 1 − (5)(1) −4 Ex 4: cos ( 2sin −1 x ) can be thought of as representing cos(2A) , with A = sin −1 x . Therefore sin A = x , and identity 8c now implies that cos(2A) = 1 − 2sin 2 A , which in this case yields 1 − 2x 2 . Using the calculator and inverse trig functions to find angles in different quadrants 1) Find values for angle θ (nearest tenth of a degree), given that 0° < θ < 90° . Note: If the input is positive, your calculator is programmed to find an angle from Q1. This is why we did not have to talk about this in chapter 2 when we were solving right triangles. All the angles we were looking for were parts of right triangles, hence acute. cos θ = 0.66 θ = sin −1 (0.66) = 41.3° θ = cos −1 (0.66) = 48.7° tan θ = 0.66 θ = tan −1 (0.66) = 33.4° sin θ = 0.66 2) We can use the angles found above as reference angles if we want angles in the other quadrants that have the same trig values, or the negatives of those values. Even if the given trig value is negative, the easiest procedure is to drop the negative sign and use trig −1 x as the reference angle. This is the same use of reference angles that we discussed in section 2.2. a) Find θ in Q2 such that sin θ = 0.66 b) Find θ in Q3 such that sin θ = − 0.66 c) Find θ in Q4 such that sin θ = − 0.66 θ = 180° − 41.3° = 138.7° θ = 180° + 41.3° = 221.3° θ = 360° − 41.3° = 318.7° d) Find θ in Q2 such that cos θ = − 0.66 e) Find θ in Q3 such that cos θ = − 0.66 f) Find θ in Q4 such that cos θ = 0.66 θ = 180° − 48.7° = 131.3° θ = 180° + 48.7° = 228.7° θ = 360° − 48.7° = 311.3° g) Find θ in Q2 such that tan θ = − 0.66 h) Find θ in Q3 such that tan θ = 0.66 i) Find θ in Q4 such that tan θ = − 0.66 θ = 180° − 33.4° = 146.6° θ = 180° + 33.4° = 213.4° θ = 360° − 33.4° = 326.6° ...
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This note was uploaded on 11/16/2011 for the course MAC 1114 taught by Professor Cohen during the Fall '09 term at Santa Fe College.

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