If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Precalculus II
Asst # 5
09-23-2001
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08:05:46
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**** Query problem 6.2.8
exact value of tan(195 deg)
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08:46:26
Your solution:
Confidence Assessment:
Given Solution:
** tan(alpha+beta)=
[tan(alpha)+tan(beta)]/[1-tan(alpha)tan(beta].
So tan(195 deg) = tan(150 deg + 45 deg) =
[ tan(150 deg) + tan(45 deg) ] / [ 1 - tan(150 deg) * tan(45 deg) ] =
[ -sqrt(3) / 3 + 1 ] / [ 1 - (-sqrt(3)) * 1 ] =
[1 - sqrt(3)/3] / (1 + sqrt(3)/3 )=
(1 - sqrt(3)/3)(1 - sqrt(3)/3) / [ (1+sqrt(3)/3)(1-sqrt(3)/3) ] =
(1 - 2 sqrt(3)/3 + 1/3) / (1 - 1/3) =
(4/3 - 2 sqrt(3)/3) / (2/3) =
2 - sqrt(3). **
STUDENT SOLUTION:
we
know that the sum and difference formula will allow us to say that
tan(alpha+beta)=tan(alpha)+tan(beta)/1-tan(alpha)tan(beta).
so for this
problem we can say
tan(180+15)=tan(180)+tan(15)/1-tan(180)tan(15).
but since we do not know the exact value of tan 15 deg we have use the sum and
difference formula to find it. we can say that tan(alpha-beta)= tan(alpha)-
tan(beta)/1+tan(alpha)tan(beta). i chose to use alpha=60 and beta=45 there
is another way this can be done. but i said
tan(60-45)=tan60-tan45/1+tan60tan45
and we know that the tan60=sq.rt.3 from
the chart in chapter 5. we also know that tan 45=1 . so we can say
(sq.rt.3-1)/1+(sq.rt.3)(1).
we can multiply through by the conjugate to get the radical out of the
denominator and we have (2sq.rt.3-4)/(-2) .
we could go ahead and say that this is the answer given the fact that tan 180=0
and
we know that 0+tan 15 divided by 1+0 is going to give the same answer but it
will be more profitable to go through all of the steps. so we can say
tan(alpha+beta)=tan(alpha)+tan(beta)/1-tan(alpha)tan(beta). alpha=180 and
beta=15 so we can say that tan(180+15)= 0+(2sq.rt.3-4)/(-2)/
1-(0)(2sq.rt.3-4)/(-2)= ((2sq.rt.3)-4)/(-2)
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08:46:27
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Self-critique (if necessary):
Self-critique Rating:
**** what is the exact
value of tan(195 deg)?
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Your solution:
Confidence Assessment:
Given Solution:
**** to get this result you
used the exact values of two angles to get
tan(15 deg). Which two angles were these, what were the exact values, and
what formula did you use to get your result?
to find the value of tan(15 deg) as i stated in the first problem i used the
two angles 60 deg and 45 deg. i used the sum and difference formula for
tan(alpha-beta) and i let alpha=60 and beta=45. the exact value of tan
60=(sq.rt.3) or approximately 1.73. the value of tan 45=1.
Self-critique (if necessary):
Self-critique Rating:
**** For which angles
between 0 and 90 degrees do you know the exact
value of the trigonometric functions? How many ways are there to combine
angles you have listed to obtain 15 degrees?
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08:55:24
Your solution:
Confidence Assessment:
Given Solution:
for all six trigonometric
functions we know the exact values of 5 angles
including 90 deg. we know the exact value for 0 deg. and 30 deg,45 deg, 60
deg,90 deg. there are two ways to obtain an angle of 15 deg. the first way
is the one i used in solving the previous problem which is 60deg-45deg and
the other is 45deg-30deg.
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08:55:25
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Self-critique (if necessary):
Self-critique Rating:
**** Query problem 6.2.20
cos(5 `pi / 12) cos(7`pi/12) -
sin(5`pi/12)sin(7`pi/12)
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09:16:29
Your solution:
Confidence Assessment:
Given Solution:
STUDENT SOLUTION: to begin
solving this equation it is obvious that we need to simplify.
because each part to this equation can be simplified using the sum and
difference formulas. first we take cos(5pi/12) this can be written as
cos(75deg) we then know that
cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta).
so we can say
cos(45+30)=cos45cos30-sin45sin30.
(sq.rt.2)/2(sq.rt.3)/2- (sq.rt.2)/2(1/2)
which gives us
(sq.rt.6-sq.rt.2)/4.
we can then simplify the second term in the equation. cos7pi/12. which can
be written as cos(105deg). we can then say
cos(60+45)=cos60cos45-sin60sin45.
which is
(1/2)(sq.rt.2/2)-(sq.rt.3/2)(sq.rt.2/2)=(sq.rt.2-sq.rt.6)/4.
we can now solve for the third term in the equation which is sin(5pi/12) which
can
be written as sin75deg. we can then say that
sin(45+30)=sin45cos30+cos45sin30. which is equal to
(sq.rt.2/2)(sq.rt.3/2)+(sq.rt.2/2)(1/2)=(sq.rt.6+sq.rt.2)/4.
we can now solve the fourth term in the equation which is sin7pi/12. which can
be
written as sin 105 deg. we can then say
sin(60+45)=sin60cos45+cos60sin45=(sq.rt.3/2)(sq.rt.2/2)+(1/2)(sq.rt.2/2)
=(sq.rt.6+sq.rt.2/4).
we can now solve the equation. (sq.rt.6-sq.rt.2)/4 (sq.rt.2-sq.rt.6)/4
-(sq.rt.6+sq.rt.2)/4(sq.rt.6+sq.rt.2)/4=
(2sq.rt.12)-8/16- (2sq.rt.12)+8/16= 0 .
** great until the last step, which should read
(2sq.rt.12)-8/16- [(2sq.rt.12)+8/16]= -1. **
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09:16:29
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Self-critique (if necessary):
Self-critique Rating:
**** Which sum or
difference formula does this expression fit?
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09:22:06
Your solution:
Confidence Assessment:
Given Solution:
STUDENT SOLUTION if we look
at the entire expression before it has been simplified at all, we
see that it says cos(cos)-sin(sin) which fits the form of the formula for
cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta).
Self-critique (if necessary):
Self-critique Rating:
**** If this expression is
the right-hand side of a sum or difference
formula, what then is the left-hand side and what is its exact value?
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09:22:54
Your solution:
Confidence Assessment:
Given Solution:
STUDENT SOLUTION if this
expression is the right hand side of the formula then the left hand
side of the formula has to be cos(alpha-beta) and it has to be equal to
zero. because in the sum and distance formulas the two equations must equal
each other.
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09:22:55
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Self-critique (if necessary):
Self-critique Rating:
**** Query problem 6.2.38
tan(2 `pi - `theta) = - tan(`theta)
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09:25:15
Your solution:
Confidence Assessment:
Given Solution:
to prove this identity we
can say that tan(2pi-theta)=-tan theta. to figure
out this problem we use the sum and difference formula for
tan(alpha-beta)=tan 2pi-tan theta/1+tan2pi tan theta
we know that the value of tan of 2pi =0 so we can say
0-tan theta/1+(0)(tan theta=(- tan theta)/1
which is -tan theta so the identity is true.
** Good solution. An alternative would be to use the periodicity and even/odd
properties of sine and cosine:
sin(-`theta) = - sin(`theta)
cos(-`theta) = - cos(`theta)
so
tan (-`theta) = - tan(`theta)
all functions have the
same values at 2 `pi + `theta as at `theta, so the result follows. **
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09:25:16
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Self-critique (if necessary):
Self-critique Rating:
**** how do you establish
the given identity?
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09:25:38
Your solution:
Confidence Assessment:
Given Solution:
STUDENT SOLUTION we
establish the identity by using the sum and difference formula for
tan(alpha-beta) where alpha is equal to 2pi and beta is theta. we then say
that (0-tan theta)/1+(0)(tan theta is equal to -tan theta.
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09:25:39
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Self-critique (if necessary):
Self-critique Rating:
**** Query problem 6.3.8
find exact values of sin & cos of 2`theta,
and of `theta/2 if csc(`theta) = -`sqrt(5)
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09:27:35
Your solution:
Confidence Assessment:
Given Solution:
csc ' theta = 1/sin 'theta = - sqrt(5) = 1/(-sqrt(5)) = -sqrt(5)/5 = sin
'theta = y/r so y = -sqrt(5) and r = 5.
(-sqrt(5))^2 + x^2 = 5^2 => 5+x^2=25 => c = 2sqrt(5)
cos 'theta = x/r = 2sqrt(5)/5.
sin 2'theta = 2(-sqrt(5)/5)(2sqrt(5)/5) = -20/25 = - 4/5
cos 2'theta = (2sqrt(5)/5)^2 - (-sqrt(5)/5)^2 = 20/25 - 5/25 = 15/25 = 3/5.
sin 'theta/2 = -sqrt[(1-2sqrt(5)/5)/2] = - sqrt[1/2 - sqrt(5)/5]
cos 'theta/2 =sqrt[(1+2sqrt(5)/5)/2] = sqrt[1/2 + sqrt(5)/5]
STUDENT SOLUTIONS
I used the following formulas:
sin(2'theta) = 2 sin'theta * cos'theta
cos(2'theta) = cos^2'theta - sin^2'theta
sin (theta/2) = + - sqrt[(1-cos'theta)/2]
and cos ('theta/2) = + - sqrt[(1+cos'theta)/2]
we know that the csc theta is 1/sin theta or r/b so we can say that
*******
ADDITIONAL STUDENT SOLUTIONS INTERSPERSED WITH INSTRUCTOR COMMENTARY
sin theta=(-sq.rt.5/5)
and if the sin is (-sq.rt.5/5) we can find the cosine by saying that
(-sq.rt.5+x)^2=5^2 so 5+x=25
** Watch the algebra. (a+b)^2 = a^2 + 2 ab + b^2, so
(-sq.rt.5+x)^2 = 5 - 2 x `sqrt(5) + x^2. **
** However you should use sin^2(`theta) + cos^2(`theta) = 1 so
cos^2(`theta) = 1 - sin^2(`theta) and
cos(`theta) = `sqrt[ 1 - sin^2(`theta) ] = `sqrt(1 - (`sqrt(5) / 5)^2 ) =
`sqrt(1 - 1/5)^2 = `sqrt(4/5) = 2 `sqrt(5) / 5. **
and the cosine is (sq.rt.20/5) which is 2(sq.rt.5)/5.
so
sin 2 theta=2(-sq.rt.5)(2(sq.rt.5)=2(-2sq.rt25)=-20
next we need to find cos2theta. so we say cos^2 theta-sin^2 theta. so we
have
(2sq.rt.5)^2-(-sq.rt.5)^2=10-5=5
** right strategy but wrong information from previous steps. What should have
tipped you off here is that you got 5. The cosine and sine functions can't have
absoluted values greater than 1. **
the third part of the problem asks us to solve for sin(theta/2). so we can
use the formula for half angles. which is
sin(alpha/2)=+or- (sq.rt. 1-cos alpha/2).
so we can say
sin(alpha/2)=+or-(sq.rt. 1-2(sq.rt.5)/2) which equals approximately 1.74.
**this strategy won’t work; the correct result can't be > 1 **
the fourth part of the equation asks us to find the value of cos (theta/2).
we can also use the half angle formula to find the value of this expression.
we can say
cos(alpha/2)=+or -(sq.rt. 1+cos alpha/2) which is equal to cos
(alpha/2)=+or- (sq.rt.1+2sq.rt.5)/2= approximately 1.65
** same comment as previous **
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09:27:35
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Self-critique (if necessary):
Self-critique Rating:
**** Query problem 6.3.20
exact value of csc(7`pi/8)
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Your solution:
Confidence Assessment:
Given Solution:
STUDENT SOLUTION
we can say that csc(7pi/8)
can be written as csc(4pi/8+3pi/8) we know that
the csc of 4pi/8 is equal to 1.
so we can say csc(1+sq.rt.2) because 2pi/8 is equal to sq.rt. 2 .
we can then say that pi/8 is equal to sq.rt.2/2 so we
can say that the csc7(pi/8) equals 1/1+sq.rt.2/1+sq.rt.2/2=5sq.rt.2/2
** sin(7`pi/8) = sin( 1/2 * 7`pi/4). You know the exact value of sin(7`pi/4).
Use the half-angle formula. You get `sqrt(2`sqrt(2)+4) ). **
Self-critique (if necessary):
Self-critique Rating:
**** query problems 6.3.42
tan(`theta/2) = csc(`theta) - cot(`theta)
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09:29:37
Your solution:
Confidence Assessment:
Given Solution:
** The formula in the text is
tan(alpha/2)= (1-cos (alpha)) / sin (alpha) which is the same as
1 / sin(alpha) - cos(alpha) / sin(alpha) =
csc(alpha) - cot(alpha). **
now we know that the
csc theta=1/sin theta
and we also know that
cot theta=cos theta/sin theta.
so we can say
tan(theta/2)=1/sin theta-cos theta/sin theta=1-cos theta/sin theta=sin
theta/1+cos theta.
therefore the identiity is true.
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09:29:37
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Questions about some odd-numbered problems:
Section 2
63. cos(csc^-1(u) Theta = csc^-1 u so that csc theta = u
csc^-1(u) is the base angle of a right triangle whose hypotenuse is u
(corresponding to a circle of radius u about the origin) and whose opposite side
(the 'y' leg) is 1 (thus csc(theta) = r / y = u / 1 =u).
The adjacent side of this triangle (the 'x' leg) is thus sqrt(u^2 - 1), by the
Pythagorean Theorem.
The cosine of the angle is adjacent side / hypotenuse = sqr(1 - u^2) / 1 =
sqrt(1 - u^2) (this corresponds to x / r in the circular model).
A similar scheme is used for a variety of problems involving a trig function of
an inverse trig function.
69. g^-1(f(7pi/4)) cos^-1(sin(7pi/4)) 0 is less than or equal to x is less than
or equal to pi cos^-1(sin7pi/4) = x = 2.356
By the unit circle picture for the multiples of pi / 4, we have sin(7 pi / 4)
= -sqrt(2) / 2.
cos^-1(sqrt(2) / 2) is the angle between 0 and pi whose cosine in -sqrt(2) / 2.
The unit circle picture tells us that this angle is 3 pi / 4. That is, cos(3 pi
/ 4) = -sqrt(2) / 2.
Section 5
41. Show that sin^4(theta) = 3/8 – 1/2cos(2theta) +1/8cos(4theta)
sin^2(theta) = 1 - cos^2(theta), so
sin^4(theta) = (1 - cos^2(theta) ) ^ 2 = 1 - 2 cos^2(theta) + cos^4(theta).
cos(2theta) = cos^2(theta) - sin^2(theta)
4 theta = 2 ( 2 theta) so
cos(4 theta) = cos^2(2 theta) - sin^2(2 theta)
= (cos(2 theta))^2 - sin(2 theta)^2
= ( cos^2(theta) - sin^2(theta))^2 - (2 sin(theta) cos(theta))^2
= cos^4(theta) - 2 cos^2(theta) sin^2(theta) + sin^4(theta) - 4 sin^2(theta) cos^2(theta)
= cos^4(theta) - 2 cos^2(theta) ( 1 - cos^2(theta)) + (1 - cos^2(theta))^2 - 4 (
1 - cos^2(theta)) cos^2(theta)
Thus you can write sin^4(theta) in terms of powers of cos(theta). You can do the
same with cos(2 theta) and cos(4 theta).
This allows you to write the identity
sin^4(theta) = 3/8 – 1/2cos(2theta) +1/8cos(4theta)
in terms of cos(theta). Having done so you can fairly easily verify that the
equation holds.
45. Find an expression for sin(5theta) as a 5th-degree polynomial in the
variable sin(theta)
Sin(5theta) = sin^2(theta)+sin^2(theta) +sin(theta) ?
sin(5 theta) = sin(4 theta + theta), which is simplified using the formula
for sin(a+b).
sin(4 theta) and cos(4 theta) are simplified as the sine and cosine of (2 * 2
theta), using double-angle formulas.
then sin(2 theta) and cos(2 theta) are simplified using the same formula.
any term involving cos^2 can be written as 1 - sin^2
you end up with a 5th-degree polynomial
Establish the identity:
51. sec(2theta) = sec^2(theta)/2-sec^2(theta) = 1/cos^2(theta)
*(1/1+cos(2theta)/2) =
sec(2 theta) = 1 / cos(2 theta) = 1 / (cos^2(theta) - sin^2(theta) )
sec^2(theta) / 2 = 1 / (2 cos^2(theta) )
sec^2(theta) = 1 / cos^2(theta)
It would be easy enough to write out the two sides, multiply by a common
denominator, etc.
However I don't think you have the signs of grouping right in your first
expression. As you're written it the right-hand side would just simplify to
-sec^2(theta) / 2.
Can you resubmit this question and clarify the signs of grouping?
For example, could you mean one of the following?
sec^2(theta/2)-sec^2(theta)
sec^2(theta)/(2-sec^2(theta))
63. sin(3theta)/sin(theta)-cos(3theta)/cos(theta) = 2
sin(3 theta) = sin(2 theta + theta) =
sin(2 theta) cos(theta) + cos(2 theta) sin(theta) =
2 sin(theta) cos(theta) * cos(theta) + (cos^2(theta) - sin^2(theta) ) sin(theta)
so the first term (dividing this by sin(theta) ) is 2 cos^2 (theta) + cos^2(theta)
- sin^2(theta) = 3 cos^2(theta) - sin^2(theta).
cos(3 theta) = cos(2 theta + theta) =
cos(2 theta) cos(theta) - sin(2 theta) sin(theta) =
(cos^2(theta) - sin^2(theta) ) cos(theta) - 2 sin(theta) cos(theta) sin(theta) .
Dividing by cos(theta) and simplifying we find that the second term is
- (cos^2(theta) - 3 sin^2(theta) ) .
Combining the two terms we get
2 cos^2(theta) + 2 sin^2(theta) = 2 ( cos^2(theta) + sin^2(theta) ) = 2 * 1 = 2.
Self-critique (if necessary):
Self-critique Rating: