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5.1 Using Fundamental Identities Fundamental Trigonometric Identities (on page 376)
Reciprocal Identities
sinu = 1/cscu
cosu = 1/secu
tanu = 1/cotu
Quotient Identities
tanu = sinu/cosu
cotu = cosu/sinu
Pythagorean Identities
sin2u + cos2u = 1 1 + tan2u = sec2u
Cofunction Identities
sin(π/2 ­ u) = cosu
tan(π/2 ­ u) = cotu
sec(π/2 ­ u) = cscu
1 + cot2u = csc2u
cos(π/2 ­ u) = sinu
cot(π/2 ­ u) = tanu
csc(π/2 ­ u) = secu
Even/Odd Identities
sin(­u) = ­sinu
cos(­u) = cosu
csc(­u) = ­cscu
sec(­u) = secu
tan(­u) = ­tanu
cot(­u) = ­cotu
One use of trig identities is to use give values of trig functions to evaluate other trig functions. We have already done a little of this.
Remember: If I know that sinu = 1/2, I can find cosu using the Pythagorean Identity sin2u + cos2u = 1.
(1/2)2 + cos2u = 1
Example: Use the values of secu = ­3/2 and tanu > 0 to find the values of the 6 trig functions without drawing a triangle.
Reciprocal Identity: cosu = 1/secu cosu = ­2/3
Pythagorean Identity: sin2u + cos2u = 1
sin2u = 1 ­ (­2/3)2
sin2u = 5/9
sinu = √5/3
Because cosu < 0 and tanu > 0, we know that u is in Quadrant III. Therefore sinu must be negative.
sinu = ­√5/3
Reciprocal Identity: cscu = 1/sinu
cscu = ­3/√5 = ­3√5/5
Quotient Identity: tanu = sinu/cosu
tanu = ­√5/3/­2/3
tanu = √5/2
Reciprocal Identity: cotu = 1/tanu
cotu = 2/√5 = 2√5/5
We may also want to simplify a trigonometric function. This will be helpful to practice before we get to proofs and verifying trig identities.
Example: Simplify sinxcos2x ­ sinx
sinxcos2x ­ sinx
sinx(cos2x ­ 1)
­sinx(1­cos2x)
­sinx(sin2x)
­sin3x
Factor out what is alike
Convert to use an identity
Pythagorean Identity
Multiply
Independent Practice
Use the values and trig identities to find all 6 trig functions.
1. secx = ­3 and tanx < 0
Simplify the expression using the fundamental identities.
2. secθ/cscθ
3. cot(π/2 ­ x)cosx
Now let's verify an identity. When you verify you can only change on side, just like a proof. Start with the side with more "stuff" on it!
Example: Verify sinx/(1 + cosx) + cosx/sinx = cscx
Start by making it one fraction with a common denominator.
Homework: pg381 #5­60 by 5's use trig identities NOT triangles
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5.1 Using Fundamental Identities Continued
When factoring trigonometric expressions, it can be helpful to find a polynomial form that fits the expression.
Examples: Factor
1. sec2x ­ 1
Difference of perfect squares
(secx ­ 1)(secx + 1)
2. 4tan2x + tanx ­ 3 Factor just like it was 4x2 + x ­ 3
(4tanx ­ 3)(tanx + 1) Sometimes, it may help to rewrite the expression in terms of one trigonometric function, or in terms of sine or cosine alone so you can factor.
Examples:
1. Factor csc2x ­ cotx ­ 3
Use the Pythagorean Identity csc2x = 1 + cot2x
1 + cot2x ­ cotx ­ 3
cot2x ­ cotx ­ 2
(cotx ­ 2)(cotx + 1)
2. Simplify sinx + cotxcosx
sinx + (cosx/sinx)cosx
Quotient Identity
(sin2x + cos2x)sinx
Add fractions
1/sinx
Pythagorean Identity
cscx
Reciprocal Identity
There are also times we just may want to rewrite expressions into forms that we can use.
Example: Rewrite 1/(1 + sinx) so it is not in fraction form.
1 * (1 ­ sinx)
Multiply numerator and (1 + sinx) (1 ­ sinx)
denominator by (1 ­ sinx) (1 ­ sinx) / (1 ­ sin2x)
FOIL
2
(1 ­ sinx) / cos x
Pythagorean Identity
1 ­ sinx * 1 Write as separate cos2x cosx cosx
fractions
sec2x ­ tanxsecx
Reciprocal and Quotient Identities
Independent Practice:
1. Perform the subtraction and simplify.
1 ­ 1 secx + 1 secx ­ 1
2. Factor and simplify
sin4x ­ cos4x
3. Rewrite so it is not in fractional form
3 secx ­ tanx
Homework pg 382 #65­85 by 5's and #110­125 by 5's
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How can we check identities by graphing?
Graph both sides and see if they are the same.
Example: cos3x = 4cos3x ­ 3cosx
You can also look at the table
You can use trig substitution and triangles
Example: If x = 2tanθ and 0 < θ < π/2 and the opposite of θ = x, adjacent = 2 and the hypotenuse = √(4 + x2), express √(4 + x2) as a trigonometric function of θ.
Using the Pythagorean Theorem
2 2
2
2
√(4 + x ) = 2 + x
Substitute in x = 2tanθ
2 2
2
2
√(4 + x ) = 2 + (2tanθ)
Solve for √(4 + x2)
2
2
2
√(4 + x ) = √(2 + (2tanθ) )
2
2
√(4 + x ) = √(4 + 4tan θ)
2
2
√(4 + x ) = √(4(1 +tan θ))
2
2
√(4 + x ) = √(4sec θ)
2
√(4 + x ) = 2secθ
Independent Practice:
The rate of change of the function f(x) = ­cscx ­ sinx is given by the expression cscxcotx ­ cosx. Show that this expression can also be written as cosxcot2x.
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More Techniques for Verifying Trig Identities
Guidelines (on page 384)
1. Work with one side of the equation at a time. it is often better to work with the more complicated side first.
2. Look for opportunities to factor an expression, add fractions, square a binomial, or create a monomial denominator.
3. Look for opportunities to use the fundamental identities. Note which functions are in the final expression you want. Sines and cosines pair up well, as do secants and tangents, and cosecants, and cotangents.
4. If the preceding guidelines do not help, try converting all terms to sines and cosines.
5. Try something! Even making an attempt that leads to a dead end gives insight!
Examples: Verify the Identities
1. sec2x ­ 1 = sin2x
Hint: The sec2x do not 2
sec x
cancel!
Of course there is more than one way to do proofs, so let's look at both ways.
2. 1 + 1 = 2sec2x Hint: Combine 1­sinx 1+sinx
fractions
3. (tan2x + 1)(cos2x ­ 1) = ­tan2x
Note: Although graphing utilities can be useful in helping to verify an identity, you must use algebraic techniques to produce a valid proof. For example the graphs of the two functions y1 = sin50x and y2 = sin2x in a trigonometric viewing window look identical. However, sin50x ≠ sin2x.
4. tanx + cotx = secxcscx Hint: convert to sines and cosines.
Note: Remember that rationalizing the denominator using conjugates is, on occasion, a powerful simplification technique. A related form of this technique works for simplifying trig expressions as well.
Independent Practice
1. (tanxcotx)/cosx = secx
2. (1/tanx) + (1/cotx) = tanx +cotx
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Even more examples of verifying!
Remember there are numerous tricks and the more practice you get the better you will be at identifying which tricks will work with which problems. Examples: Verify the trig identities.
1. secx + tanx = cosx Hint: Always start
1­ sinx
with the fraction.
The "illegal method"
So we can only touch one side! Right?! Well not necessarily. You can not jump the fence but on occasion it is practical to work with each side separately to obtain one common form equivalent to both sides. This method works best if there is a fraction on both sides.
2. cot2x = 1 ­ sinx
1 + cscx sinx
In the next example, the powers of trig functions are rewritten as more complicated sums of products of trig functions. This is a common procedure used in calculus!
3. Verify each identity starting with the less complex side!
a. tan4x = tan2xsec2x ­ tan2x
b. sin3xcos4x = (cos4x ­ cos6x)sinx
c. sec4xtan2x = (tan2x + tan4x)sec2x
Writing About Math Error Analysis
Suppose you are tutoring a friend in trig. One of the homework problems asks whether the following statement is an identity.
tan2xsin2x = (5/6)tan2x
Your friend does not attempt to verify the equivalence algebraically, but mistakenly uses only a graphical approach. Using window settings of ­3π < x < 3π with an x­scale of π/2, and ­20 < y < 20 with a y­scale of 1. They graph both sides of the expression on the graphing utility and conclude that the statement is an identity. Write a short paragraph explaining what is wrong with you friend's reasoning. Homework pg 389 #5­80 by 5's skip #55 & 60
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5.3 Solving Trigonometric Expressions
To solve a trig equation, used methods such as collecting like terms and factoring. Your goal is to isolate the trig function involved in the equation.
Solving a Trig equation
Example: Solve 2sinx ­ 1 = 0
2sinx = 1
sinx = 1/2
where is the sinx = 1/2? at x = π/6 and x = 5π/6
so there is an infinite number of solutions at
x = π/6 + 2nπ and x = 5π/6 + 2nπ also note that any angles that are co­terminal to these two are also solutions
Collecting Like Terms
Example: Find all solutions of sinx + √2 = ­sinx in the interval [0, 2π).
sinx + sinx = ­√2
Isolate sinx on one side
2sinx = ­√2
Combine like terms
sinx = ­√2/2
The solutions on the interval are x = 5π/4 and x = 7π/4.
Extracting a Square Root
Example: Solve 3tan2x ­ 1 = 0
3tan2x = 1
tan2x = 1/3
tanx = ±√(1/3) = ±√3/3
The solutions are x = π/6 + nπ and x = 5π/6 + nπ because tanx has a period of [0, π). Factoring with Two Different Trig Functions
Try to separate the functions by factoring or using appropriate identities.
Example: Solve cotxcos2x = 2cotx
cotxcos2x ­ 2cotx = 0
cotx(cos2x ­ 2) = 0
Just like factoring put each part equal to zero
cotx = 0 cos2x ­ 2 = 0
x = π/2
cosx = ±√2
This is outside the range of cosine
So x = π/2 + nπ
so there is no solution for this part.
Note: Do not divide out cotx or you lose solutions.
Equations of the Quadratic Type Factoring an Equation
Example: Find all the solutions of 2sin2x ­ sinx ­1 from [0, 2π).
2sin2x ­ sinx ­1 = (2sinx + 1)(sinx ­ 1)
2sinx + 1 = 0
sinx ­ 1 = 0
sinx = ­1/2
sinx = 1
Solutions are x = π/2 + 2nπ, x = 7π/6 + 2nπ, x = 11π/6 + 2nπ
Rewriting with a Single Trig Function
Example: 2sin2x + 3cosx ­ 3 = 0
2(1 ­ cos2x) + 3cosx ­ 3 = 0
2 ­ 2cos2x + 3cosx ­ 3 = 0
2cos2x ­ 3cosx +1 = 0
(2cosx ­ 1)(cosx ­ 1) = 0
2cosx ­ 1 = 0
cosx ­ 1 = 0
cosx = 1/2
cosx = 1
Solutions are x = 2nπ, x = π/3 + 2nπ, x = 5π/3 + 2nπ
Squaring and Converting to a Quadratic Type
Example: cosx + 1 = sinx
(cosx + 1)2 = sin2x
Square Both Sides
cos2x + 2cosx +1 = 1 ­ cos2x Convert to cos
2cos2x + 2cosx = 0
Factor
2cosx(cosx + 1) = 0
2cosx = 0
cosx +1 = 0
cosx = 0
cosx = ­1
x = π/2, 3π/2
x = π Because we square the original we check for extraneous solutions and find the 3π/2 is extraneous.
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Functions Involving Multiple Angles And Inverse Functions
Functions of Multiple Angles
Example: Find all the solutions of 2cos3t ­ 1 = 0
Write the original equation
2cos3t ­ 1 = 0
Add 1 to each side
2cos3t = 1
Divide each side by 2
cos3t = 1/2
In the interval [0, 2π), you know that 3t = π/3 and 3t = 5π/3 are the only solutions. So in general you have
3t = π/3 + 2nπ and 3t = 5π/3 + 2nπ Divide by 3 (or multiply by 1/3) and you get
t = π/9 + (2nπ)/3 and t = 5π/9 + (2nπ)/3
Example: Find all solutions of 3tan(x/2) + 3 = 0
Write the original equation
3tan(x/2) + 3 = 0
Subtract 3 from each side
3tan(x/2) = ­3
Divide each side by 3
tan(x/2) = ­3/3 = ­1
In the interval [0, 2π), you know that x/2 = 3π/4 is the only solution. So in general you have
(x/2) = 3π/4 + nπ
Multiply by 2 and you get
x = 3π/2 + 2nπ Using Inverse Functions
Example: Find all solutions of sec2x ­ 2tanx = 4
Write the original equation
sec2x ­ 2tanx = 4
Pythagorean Identity
(1 + tan2x) ­ 2tanx ­ 4 = 0
Combine like terms
tan2x ­ 2tanx ­ 3 = 0
Factor
(tanx ­ 3)(tanx + 1) = 0
Setting each factor equal to zero, you obtain two solutions in the range of tangent (­π/2, π/2).
tanx ­ 3
tanx + 1
x = arctan3
x = arctan(­1) = ­π/4
Add multiples of π, to obtain the general solution
x = arctan3 + nπ x = ­π/4 + nπ There are some trig equations that have no reasonable way to solve algebraically. In these cases, you have to use a graph to approximate the solutions. You will not be assessed on this type.
Application of Solving Trig Equations
Example: The surface area of a honeycomb is given by the equation S = 6hs +(3/2)s2 [(√3­ cosθ)/(sinθ)], 0 < θ < 90o
where h = 2.4 inches, s = 0.75 inch and θ is the angle needed.
a. What value of θ gives a surface area of 12 square inches?
b. What value of θ) gives the minimum surface area?
When we trace the graph we find the x­intercepts to be θ ≈ 59.9o and θ ≈ 49.9o
Here is the minimum point found through the trace feature of the calculator.
HOMEWORK: pg 400 #5­40 by 5's & #60, 72, 77, 80
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Verify that the x­values are solutions of the equation
5. 2cos2x + 3 cosx + 1 = 0
(a) x = 4π/3 (b) x = π
Do #10 at home to look at book picture
Solve the equation for 0 < x < 2π
15. 3csc2x ­ 4 = 0
20. cosx(cosx ­ 1) = 0
Find all Solutions of the equation in the interval [0, 2π) algebraically.
25. cos3x = cosx
30. (secx)(cscx) = 2cscx
35. cos(x/2) = √(2)/2
40. cosx + sinxtanx = 2
Approximate the solutions(to 3 decimal places) of the equation in the given interval.
60. 2sec2x + tanx ­ 6 = 0 [­π/2, π/2]
#72, 77, & 80 do at home with book and pictures
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Sum and Difference Formulas
(Proofs are in Appendix A)
sin(u + v) = sinucosv + cosusinv
sin(u ­ v) = sinucosv ­ cosusinv
cos(u + v) = cosucosv ­ sinusinv
cos(u ­ v) = cosucosv + sinusinv
tan(u + v) = (tanu + tanv) / (1­tanutanv)
tan(u ­ v) = (tanu ­ tanv) / (1 + tanutanv)
THINGS TO NOTE
1. For sin the sign stays the same and sin and cos are together.
2. For cos the sign switches and sines and cosines are together.
3. For tan the top sign stays the same and tangents are added and for the bottom the sign switches and tangents are multiplied and added or subtracted from 1.
Evaluating Trig Functions with Sum and Difference Formulas
Example: Find the exact value of cos75o.
To find the value we are going to use the fact that 75o = 30o + 45o.
So consequently, cos(u + v) = cos(30o + 45o)
cos(30o + 45o) = cos30ocos45o ­ sin30osin45o
= (√3/2)(√2/2) ­ (1/2)(√2/2)
= (√6 ­ √2) / 4
You can check this by plugging the answer in the calculator and finding the decimal value and then plugging in cos75o making sure that it is in degrees on the calculator.
Example: Find the exact value of sin(π/12).
To find the value we are going to use the fact that
π/12 = π/3 ­ π/4.
So consequently, sin(u ­ v) = sin(π/3 ­ π/4).
sin(π/3 ­ π/4) = sin(π/3)cos(π/4) ­ cos(π/3)sin(π/4)
= (√3/2)(√2/2) ­ (1/2)(√2/2)
= (√6 ­ √2) / 4
Notice the correction!
This is because π/12 is actually 15o. Lke 30o and 60o switch there sin and cos values, so do any two angles that add up to 90o. This is because of there relationship in the triangle.
Independent Practice
1. Find the sine, cosine, and tangent for 195o = 225o ­ 30o.
2. Use the sum and difference formulas to write the expression cos(π/7)cos(π/5) ­ sin(π/7)sin(π/5).
Example: Find the exact value of sin(u + v) given that sinu = 4/5, where 0 < u < π/2 and cosv = ­12/13, where π/2 < v < π.
Because sinu = 4/5 and u is in Quad I, cos u = 3/5 if we draw a triangle. Because cosv = ­12/13 and v is in Quad II, sin v = 5/13 if we draw a triangle.
Now you can find sin(u + v) = sinucosv + cosusinv
= (4/5)(­12/13) + (3/5)(5/13)
= (­48/65) + (15/65) = ­33/65
Applications of the Sum Formula
Example: Evaluate cos(arctan1 + arccosx)
This expression fins the formula for cos(u + v). Angles u = arctan1 & v = arccosx can be drawn as triangles.
Then use the values of the triangles to solve.
cos(u + v) = cos(arctan1)cos(arccosx) ­ sin(arctan1)sin(arccosx)
= (1/√2)(x) ­ (1/√2)(√(1­x2))
= (x ­ √(1­x2)) / √2
Independent Practice
1. If sinu=5/13, where 0< u<π/2 & cosv=­3/5 where π/2<v<π. Find cos(v ­ u)
HOMEWORK: pg 408 #5­40 by 5's
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More with the Sum and Difference Formulas
Proving Identities
Example: Prove the cofunction identity cos(π/2 ­ x) = sinx.
Use the formula for cos(u ­ v) cos(π/2 ­ x) = cos(π/2)cosx + sin(π/2)sinx
= 0(cosx) + 1sinx
= sinx
Note: Sum and difference formulas can be used to derive reduction formulas involving expressions such as sin(θ + nπ/2) and cos(θ + nπ/2) where n is an integer!
Example: Simplify each expression
1. cos(θ ­ 3π/2) Use the formula for cos(u ­ v) = cosucosv + sinusinv
cos(θ ­ 3π/2) = cosθcos(3π/2) + sinθsin(3π/2)
= cosθ(0) + sinθ(­1)
= ­sinθ 2. tan(θ + 3π)
Use the formula for tan(u + v) = (tanu + tanv) / (1 ­ tanutanv)
tan(θ + 3π) = (tanθ + tan3π) / (1 ­ tanθtan3π)
= (tanθ + 0) / (1 ­ tanθ(0))
= tanθ / 1
Note that the period of tanθ is π, so the period of tan(θ­3π) is the same.
Applications from Calculus
Example:Verify sin(x+h)­sinx = cosx[(sinh)/h] ­ sinx[(1­cosh)/h]
h
Use the formula for sin(u + v) = sinucosv + cosusinv
sin(x+h) ­ sinx = sinxcosh + cosxsinh ­ sinx
h
h
= cosxsinh ­ sinx(­cosh + 1)
h
= cosx[(sinh)/h] ­ sinx[(1­cosh) / h]
Solving a Trigonometric Equation
Example: Find all solutions of sin(x + π/4) + sin(x ­ π/4) Using both the sum and difference formulas of sin we can say
[sinxcos(π/4)+cosxsin(π/4)]+[sinxcos(π/4)­cosxsin(π/4)] = ­1
2sinxcos(π/4) = ­1
2sinx(√2/2) = ­1
√2sinx = ­1
sinx = ­1/√2
sinx = ­√2/2
Therefore, the solutions are x = 5π/4 + 2nπ and x = 7π/4 + 2nπ
Independent Practice
1. Verify the identity
sin(x + y) + sin(x ­ y) = 2sinxcosy
2. Find all the solutions in the interval [0, 2π)
tan(x + π) + 2sin(x + π) = 0
HOMEWORK: pg 409 #45­60 by 5's & #67, 70, 88, 92
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Multiple­Angle and Power Reducing Formulas
Double Angle Formulas (Proofs in Appendix A)
1. sin2u = 2sinucosu
2. cos2u = cos2u ­ sin2u = 2cos2u ­ 1 = 1 ­ 2sin2u
3. tan2u = 2tanu 1 ­ tan2u
NOTE: sin2u ≠ 2sinu, cos2u ≠ 2cosu, tan2u ≠ 2tanu
Solving a Multiple­Angle Equation
Example: Find all solutions of 2cosx + sin2x = 0
2cosx + sin2x = 0 2cosx + 2sinxcosx = 0
2cosx(1 + sinx) = 0
cosx = 0
1 + sinx = 0
x = π/2, 3π/2
sinx = ­1
x = 3π/2
Therefore: x = π/2 + 2nπ and x = 3π/2 +2nπ Using Double­Angle Formulas in Sketching Graphs
Example: Analyze the graph of y = 4cos2x ­ 2 from [0, 2π]
y = 4cos2x ­ 2
= 2(2cos2x ­1)
= 2(cos2x)
We can graph this knowing that the new amplitude is 2 and the new period is 2π/2 = π. Now graph!
Evaluating Functions Involving Double Angles
Example: Use the following to find the sin2x, cos2x, and tan2x.
cosx = 5/13, (3π)/2 < x < 2π * Using the Pythagorean Theorem, we find that the other side is 12, so the sinx = ­12/13, since it is in the 4th quadrant!
Therefore, sin2x = 2sinxcosx = 2(­12/13)(5/13) = ­120/169
cos2x = 2cos2 ­ 1 = 2(25/169) ­ 1 = ­119/169
tan2x = (sin2x)/(cos2x) = 120/119
NOTE: These are not restricted to angles 2θ and θ. They work for combinations such as 4θ and 2θ, or 6θ and 3θ.
Examples: sin4x = 2sin2xcos2x and cos6x = cos23x­sin23x
Deriving Triple Angle Formulas
Example: Express sin3x in terms of sinx.
sin3x = sin(2x + x)
= sin2xcosx + cos2xsinx
= 2sinxcosxcosx + (1­2sin2x)sinx
= 2sinxcos2x + sinx ­ 2sin3x
= 2sinx(1 ­ sin2x ) + sinx ­ 2sin3x
= 2sinx ­ 2sin3x + sinx ­ 2sin3x
= 3sinx ­ 4sin3x
Power­Reducing Formulas
1. sin2u = 1­cos2u 2. cos2u = 1+cos2u 3. tan2u = 1­cos2u
2
2
1+cos2u
Reducing a Power
Example: Rewrite sin4x as a sum of first powers of the cosines of multiple angles.
sin4x = (sin2x)2 = [(1 ­ cos2x)/ 2]2 = (1/4)[1 ­ 2cos2x + cos22x] = (1/4){1 ­ 2cos2x + [(1 + cos4x)/2]}
= (1/4) ­ (1/2)cos2x + (1/8) + (1/8)cos4x
= (3/8) ­ (1/2)cos2x + (1/8)cos4x
= (1/8)(3 ­ 4cos2x + cos4x)
HOMEWORK: pg418 #3­30 by 5's
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Half­Angle, Product­to­Sum, and Sum­to­Product Formulas
Half­Angle Formulas
1. sin(u/2) = ±√[(1 ­ cosu)/2]
2. cos(u/2) = ±√[(1 + cosu)/2]
3. tan(u/2) = (1 ­ cosu) / sinu = sinu / (1 + cosu)
NOTE: the signs of sin(u/2) and cos(u/2) depend on quadrant!
Using a Half­Angle Formula
Example: Find the exact values of sin105o.
Begin by noting that 105o is half of 210o. Then using the half­angle formula for sin(u/2) and the fact that 105o lies in Quadrant II, you have the following:
sin105o = + √[(1 ­ cos210o) /2]
= + √{[1 ­ (­cos30o)] /2}
= + √{[1 + (√3/2)] /2}
= + √(2 + √3) / 2
You can then use a calculator to verify!
Solving a Trigonometric Equation
Example: Find all solutions of 2 ­ sin2x = 2cos2(x/2) in [0, 2π)
2 ­ sin2x = 2cos2(x/2)
2 ­ sin2x = 2{±√[(1 + cosx) / 2]}2
2 ­ sin2x = 2[(1 + cosx) / 2]
2 ­ sin2x = 1 + cosx
2 ­ (1 ­ cos2x) = 1 + cosx
cos2x ­ cosx = 0
cosx(cosx ­ 1) = 0
cosx = 0
cosx ­ 1 = 0
x = π/2, 3π/2, and 0
Product­to­Sum Formulas
1. sinusinv = (1/2)[cos(u ­ v) ­ cos(u + v)]
2. cosucosv = (1/2)[cos(u ­ v) + cos(u + v)]
3. sinucosv = (1/2)[sin(u + v) + sin(u ­ v)]
4. cosusinv = (1/2)[sin(u + v) ­ sin(u ­ v)]
Writing Products as Sums
Example: Rewrite the product cos5xsin4x as a sum or difference. cos5xsin4x = (1/2)[sin(5x + 4x) ­ sin(5x ­ 4x)]
= (1/2)[sin9x ­ sinx]
Sum­to­Product Formulas
1. sinx + siny = 2 sin[(x + y) / 2] cos[(x ­ y) / 2]
2. sinx ­ siny = 2 cos[(x + y) / 2] sin[(x ­ y) / 2]
3. cosx + cosy = 2 cos[(x + y) / 2] cos[(x ­ y) / 2]
4. cosx ­ cosy = 2 sin[(x + y) / 2] sin[(x ­ y) / 2]
Using a Sum­to­Product Formula
Example: Find the exact value of cos195o + cos105o.
cos195o + cos105o = 2cos[(195o+105o)/2]cos[(195o­105o)/2]
= 2cos150ocos45o = 2(­√3/2)(√2/2) = ­√6/2
Solving a Trigonometric Equation
Example: Find all solutions of sin5x+sin3x = 0 from [0, 2π).
sin5x+sin3x = 2 sin[(5x + 3x) / 2] cos[(5x ­ 3x) / 2] = 0
2sin4xcosx = 0
sin4x = 0 cosx = 0 x = π/2, 3π/2
sin4x = 2sin2xcos2x = 0
sin2x = 2sinxcosx = 0 and cos2x = 2cos2x­ 1 = 0
sinx = 0 and cosx = 0
cosx = ±√2/2
x = 0, π and x = π/2, 3π/2 and x = π/4, 3π/4, 5π/4, 7π/4
Verifying a Trigonometric Identity
Example: Verify the identity (sint + sin3t)/(cost + cos3t) = tan2t.
sint + sin3t = 2sin[(t+3t)/2]cos[(t­3t)/2] = 2sin2tcos(­t) cost + cos3t 2cos[(t+3t)/2]cos[(t­3t)/2] 2cos2tcos(­t)
= sin2t / cos2t = tan2t
HOMEWORK: pg 418 #35­90 by 5's &
REVIEW: pg 422 #1­125 odd due Monday
Extra Credit: Evens due Wednesday!
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