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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
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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)
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Through the use of addition formulas for sine and cosine we know
cos (π/2 - θ) =?
sin (π/2 - θ) =?
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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.
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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°)
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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)
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The Double-Angle Formulas
sin 2θ =
cos 2θ =
tan 2θ =
- sin 2θ = 2 sin θ cos θ
- cos 2θ = cos2θ - sin2θ
- tan 2θ = (2 tan θ) / (1 - tan2θ)
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The Half-Angle 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)
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In the half-angle 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
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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
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Formulas for cos 3 θ and sin 3 θ
- cos 3 θ = 4 cos3θ - 3 cos θ
- sin 3 θ = 3 sin θ - 4 sin3θ
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Equivalent forms of the formula:
cos 2θ = cos2θ - sin2θ
- cos 2θ = 2 cos2θ - 1
- cos 2θ = 1 - 2 sin2θ
- cos2θ = (1 + cos 2θ)/2
- sin2θ = (1 - cos 2θ)/2
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sin (2x) =
2 cos(x) sin(x)
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As indicated in figure 1a (pg 608), the sine function is not ___ ___ ___, therefore there is no _____ _______. The only way around this the _____ _____ function.
- one-to-one
- inverse function
- restricted sine function *ex y = sin x (-π/2≤ x ≤π/2)
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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
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What are the two common notations used to denote the inverse sine function:
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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
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*Recurring theme: Inverse functions are symmetrical about the line ______
y = x
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sin-1x 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 y-value of 1/2. The answer is π/6
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Are sin-10 and (1/sin 0) the same? (explain)
- No, the quantity sin-10 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-10 = 0
- On the other hand, since sin 0 = 0, the expression (1/sin 0) is not even defined
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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-1x) = x for every x in the interval [-1,1]
- sin-1(sinx) = x for every x in the interval [-π/2,π/2]
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sin-1(sin 2)
- Cant be 2 because two is not WITHIN the interval [-π/2,π/2]
- However, a) π-2 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
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sin(sin-12)
2 is not in the in the domain of the inverse sine function so that expression is undefined
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Review Example 5 pg 611 if there's time
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Restricted cosine function (3)
- y = cos x (0 ≤ x ≤ π)
- must have domain [0,π] and range [-1,1]
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Two ways to denote inverse cosine function
y = cos-1x or arccos x
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arccos domain and range
- domain: [-1,1]
- range: [0,π]
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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 arc-cosines?
- cos(cos-1x) = x for every x in the interval [-1,1]
- cos-1(cosx) = x for every x in the interval [0,π]
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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)
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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
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Two common notations for the inverse tangent function
y = tan-1 x or y = arctan x
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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 arc-tangents?
- tan(tan-1x) = x for every real number
- tan-1(tan x) = x for every x in the open interval (-π/2,π/2)
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