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The distributive law for real numbers: r(s + t) =? However, if f is a function, and you replace r with f, what happens
 rs + rt
 This isn't true in general for example: cos(π/6 +π/3) ≠ cos π/6 + cos π/3

Addition formulas for Sine and Cosine
sin(s + t) =
sin(s  t) =
cos(s + t) =
cos(s  t) =
 sin(s + t) = (sin s)(cos t) + (cos s)(sin t)
 sin(s  t) = (sin s)(cos t)  (cos s)(sin t)
 cos(s + t) = (cos s)(cos t)  (sin s)(sin t)
 cos(s  t) = (cos s)(cos t) + (sin s)(sin t)

Through the use of addition formulas for sine and cosine we know
cos (π/2  θ) =?
sin (π/2  θ) =?

In the particular case in the previous function we can say that cosine and sine are cofunctions. (explain)
 The angles of interest were "π/2  θ" & "θ" and they both add up to π/2 making them complementary angles.
 So the function value of one function at a number is equal to the cofunctions's value at the complementary number.

How would you find the cos 15°
 Turn it into something in which my addition formulas can work with like:
 cos 15° = cos (45°  30°)

Addition formulas for tangent:
tan (s + t) =
tan (s  t) =
 tan (s + t) = (tan s + tan t) / (1  tan s tan t)
 tan (s  t) = (tan s  tan t) / (1 + tan s tan t)

The DoubleAngle Formulas
sin 2θ =
cos 2θ =
tan 2θ =
 sin 2θ = 2 sin θ cos θ
 cos 2θ = cos^{2}θ  sin^{2}θ
 tan 2θ = (2 tan θ) / (1  tan^{2}θ)

The HalfAngle Formulas *pg 576
sin s/2 =
cos s/2 =
tan s/2 =
 sin s/2 = ± (√1  cos s/2)
 cos s/2 = ± (√1 + cos s/2)
 tan s/2 = (sin s) / (1+ cos s)

In the halfangle formulas the ± symbol is intended to mean either ______ or _____ not _____. The sign before the radical is determined by the ______ in which the angle (or acr) s/2 terminates
 positive or negative
 both
 quadrant

How do you find the identities for sin 2 θ, cos 2 θ, and tan 2 θ
The same way replace 2 θ by (θ + θ) and use the appropriate addition formula

Formulas for cos 3 θ and sin 3 θ
 cos 3 θ = 4 cos^{3}θ  3 cos θ
 sin 3 θ = 3 sin θ  4 sin^{3}θ

Equivalent forms of the formula:
cos 2θ = cos^{2}θ  sin^{2}θ
 cos 2θ = 2 cos^{2}θ  1
 cos 2θ = 1  2 sin^{2}θ
 cos^{2}θ = (1 + cos 2θ)/2
 sin^{2}θ = (1  cos 2θ)/2

cos^{2}2t =
(1 + cos 4t)/2

sin (2x) =
2 cos(x) sin(x)

As indicated in figure 1a (pg 608), the sine function is not ___ ___ ___, therefore there is no _____ _______. The only way around this the _____ _____ function.
 onetoone
 inverse function
 restricted sine function *ex y = sin x (π/2≤ x ≤π/2)

State the domain and range of the restricted sine function has a domain of in the picture. State whether it is one to one, if so or if not, what is the implication?
 Domain: [π/2, π/2]
 Range: [1,1]
 It is one to one, meaning it has an inverse function

What are the two common notations used to denote the inverse sine function:
 y = sin^{1}x
 y = arcsin x

State the range and domain, what do you notice?
 Domain: [1,1]
 Range: [π/2,π/2]
 Domain and range are swapped in inverse functions

*Recurring theme: Inverse functions are symmetrical about the line ______
y = x

sin^{1}x is that number in the interval [π/2,π/2] whose sine is ___. With that in mind, what do you do when asked to "evaluate: sin^{1}(1/2)"?
 x
 Think which number WITHIN the (restricted sine function's) interval [π/2,π/2] has a sine or yvalue of 1/2. The answer is π/6


Are sin^{1}0 and (1/sin 0) the same? (explain)
 No, the quantity sin^{1}0 is that number WITHIN the interval [π/2,π/2] whose sine is 0. Since 0 is in the interval [π/2,π/2] and sin 0 = 0, we conclude sin^{1}0 = 0
 On the other hand, since sin 0 = 0, the expression (1/sin 0) is not even defined

f[f^{1}(x)] = x for every x in the domain of f^{1}
f^{1}[f(x)] = x for every x in the domain of f
How does this apply to sines and arcsines?
 sin(sin^{1}x) = x for every x in the interval [1,1]
 sin^{1}(sinx) = x for every x in the interval [π/2,π/2]

sin^{1}(sin 2)
 Cant be 2 because two is not WITHIN the interval [π/2,π/2]
 However, a) π2 is in that interval & b) sin 2 = sin (π  2). So the original question is equivalent to asking sin^{1}[sin(π  2)] = π  2

sin(sin^{1}2)
2 is not in the in the domain of the inverse sine function so that expression is undefined

Review Example 5 pg 611 if there's time

Restricted cosine function (3)
 y = cos x (0 ≤ x ≤ π)
 must have domain [0,π] and range [1,1]

Two ways to denote inverse cosine function
y = cos^{1}x or arccos x

arccos domain and range
 domain: [1,1]
 range: [0,π]

f[f^{1}(x)] = x for every x in the domain of f^{1}
f^{1}[f(x)] = x for every x in the domain of f
How does this apply to cosines and arccosines?
 cos(cos^{1}x) = x for every x in the interval [1,1]
 cos^{1}(cosx) = x for every x in the interval [0,π]

arccos(cos 4)
Can't be 4, it is not within the restricted interval [0,π]. However, a) 2π  4 is within the interval & b) cos 4 = cos(2π  4)

List the 3 key features that define the restricted tangent function:
 y = tan x (π/2< x <π/2)
 domain: (π/2,π/2)
 range: R

Two common notations for the inverse tangent function
y = tan^{1} x or y = arctan x

f[f^{1}(x)] = x for every x in the domain of f^{1}
f^{1}[f(x)] = x for every x in the domain of f
How does this apply to tangents and arctangents?
 tan(tan^{1}x) = x for every real number
 tan^{1}(tan x) = x for every x in the open interval (π/2,π/2)

