Trig Functions of Angles Beyond Right Triangles (5.2)(3)

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Presentation transcript:

Trig Functions of Angles Beyond Right Triangles (5.2)(3)

SAT Prep Quick poll! is what percent of 20? (Free response)

SAT Prep Quick poll! (34)(66) = (Free response)

SAT Prep Quick poll! 3.

POD– if we have time Express csc θ in terms of cos θ.

POD Express csc θ in terms of cos θ.

Another look at trig ratios Consider an angle θ in standard position. We know that sin θ is the ratio y/r. In a unit circle, what is r? So, what is sin θ? Using these letters, what are the other five trig ratios? θ x y (x,y) r

Another look at trig ratios These ratios hold, even beyond the acute angles in the first quadrant. functionratio unit circle sin θ y/ry cos θx/rx tan θy/xy/x csc θ r/y1/y sec θ r/x1/x cot θ x/yx /y θ x y (x,y) r r r

Another look at trig ratios These ratios hold, even beyond the acute angles in the first quadrant. functionratio unit circle sin θ y/ry cos θx/rx Can you see why the coordinates of the point traveling around the unit circle are (cos θ, sinθ)? θ x y (x,y) r r r

Another look at trig ratios Consider an angle θ in standard position. We’ve considered the idea that tan θ = sin θ/cos θ. Given this ratio, how else could we think of tangent? θ x y (x,y) r

Another look at trig ratios Consider an angle θ in standard position. We’ve considered the idea that tan θ = sin θ/cos θ. Given this ratio, how else could we think of tangent? Tangent represents the slope of the terminal side, or the “slope” of the angle, if you will. θ x y (x,y) r

Another look at trig ratios Although the trig functions are all positive in a right triangle (or in the first quadrant), they change signs depending on where the terminal side lands. In which quadrant(s) are sine positive and negative? What about cosine and tangent? θ x y (x,y) r r r

Trig ratios as functions We can consider these trig ratios as functions of the rotation of an angle. Suppose we trace the values of the y-coordinate as the angle rotates. Or trace the values of the x-coordinate? What are we graphing? What are the independent and dependent variables? How do we think of the domain and range of the functions?

An animated look at sine We’re working toward the graphs of these trig functions, where the angle is x, and the trig ratio is y. Here’s a quick look at y = sin x.y = sin x

Trig ratios as functions What are the domains of each of the six functions? sin θ csc θ cos θ sec θ tan θ cot θ θ x y (x,y) r r r

Trig ratios as functions What are the domains of each of the six functions? sin θ R csc θ R ≠ 0°± 180n°, R ≠ ± πn cos θ R sec θ R ≠ 90°± 180n°, R ≠ π/2 ± πn tan θ R ≠ 90°± 180n°, R ≠ π/2 ± πn cot θ R ≠ 0°± 180n°, R ≠ ± πn θ x y (x,y) r r r

Trig ratios as functions What are the ranges of each of the six functions? sin θ csc θ cos θ sec θ tan θ cot θ Graph the sine and tangent functions on calculators. θ x y (x,y) r r r

Trig ratios as functions What are the ranges of each of the six functions? sin θ |R| ≤ 1 csc θ |R| ≥ 1 cos θ |R| ≤ 1 sec θ |R| ≥ 1 tan θR cot θR θ x y (x,y) r r r

Use it Find the values of the six trig functions for 3π/2.

Use it Find the values of the six trig functions for 3π/2. sin θ = -1 cos θ = 0 tan θ = undefined csc θ = -1 sec θ = undefined cot θ = 0

Use it Find the values of the six trig functions for θ. θ (-15,8)

Use it Find the values of the six trig functions for θ. sin θ = 8/17 cos θ = -15/17 tan θ = -8/15 csc θ = 17/8 sec θ = -17/15 cot θ = -15/8 θ (-15,8)

Use it An angle θ is in standard position, with its terminal side in quadrant III, along the line y = 3x. Find the values of the six trig functions. (Hint, draw a picture, find a point.)

Use it An angle θ is in standard position, with its terminal side in quadrant III, along the line y = 3x. Find the values of the six trig functions. (Hint, draw a picture, find a point.) θ (-1, -3) r r = what?

Use it An angle θ is in standard position, with its terminal side in quadrant III, along the line y = 3x. Find the values of the six trig functions. (Hint, draw a picture, find a point.) sin θ = -3/√10 = -3√10/ 10 cos θ = -1/√10 = -√10/ 10 tan θ = 3 csc θ = -√10/3 sec θ = -√10 cot θ = 1/3