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Trigonometric Functions and the Unit Circle. Six Trig Functions. We will now define the 6 trig functions for ANY angle. (Not just positive acute angles.) Let θ be any angle in standard position and ( x,y ) a point on its terminal ray. Let the distance from the point to the origin be r. Then
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Six Trig Functions • We will now define the 6 trig functions for ANY angle. (Not just positive acute angles.) Let θ be any angle in standard position and (x,y) a point on its terminal ray. Let the distance from the point to the origin be r. Then NameAbbreviationDefinition cosine coscosθ = x/r sine sin sin θ = y/r tangent tan tanθ = y/x x≠0 secant sec sec θ = r/x x≠0 cosecant csccscθ = r/y y≠0 cotangent cot cot θ = x/y y≠0
Find the value of the six trig functions for the angle whose terminal ray passes through: a) (-5,12) b) (-3,-3)
The Unit Circle • The unit circle is given by x2+ y2 = 1. • Graph this. • Now choose an angleθ which intersects the unit circle at (x,y). • Draw it.
We will now define the 6 trig functions for any NUMBER t on a number line. Let (x,y) be the point on t when the number line is wrapped around the unit circle. (What happened to r?) NameAbbreviationDefinition cosine coscosθ = x sine sin sin θ = y tangent tan tanθ = y/x x≠0 secant sec sec θ = 1/x x≠0 cosecant csccscθ = 1/y y≠0 cotangent cot cot θ = x/y y≠0
Compute the six trigonometric functions for θ = π. cosπ = -1 sin π = 0 tan π = 0 sec π = -1 cscπ is undefined cot π is undefined Check your answers on your calculator. Make sure you are in the correct mode.
A few more… • Find the tan 450° • Find the cos of 7π/2
Can you make a chart showing in which quadrants sin, cos, and tan have positive values and in which quadrants they have negative values?
Reference Angles • For an angle θ in standard position, the reference angle is the acute, positive angle formed by the x-axis and the terminal side of θ. • Give the reference angles for 135° 5π/3 210° –π/4.
An angle will share the same x and y coordinates with its reference angle, but the signs may be different. (Can you see why?) • Find cos 150° tan 135° cot (-120°) cos (11π/6) csc (-7π/4)
Given that tanӨ = -3/4 and cosӨ > 0, find sinӨ and secӨ. • Given that π/2 < Ө < π and that sin Ө = 1/3, find cosӨ and tanӨ.
To think about… • Can you explain why sine and cosine must be between -1 and 1? Must the other functions also lie in that interval?
Which is greater sin 2 or sin 2°? • Which is greater cos 2 or cos 2°?
Let θ = 30°, find sec θ using your calculator. Answer: 1.15 Make sure you are in degree mode. Then either 1/ (cos 30) or (cos 30)-1. NOT cos (30-1) or cos-1 (30). Why don’t these work?
TRUE or FALSE? cos (-θ) = cos(θ) sin (-θ) = -sin(θ) Both statements are true. Justify with a picture.