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January 9th, 2007, 12:51 AM   #1
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Exact values

I have 6 exact values questions:

1. Find the exact value of sin(pi/3-pi/4)?
a) (sqrt2-sqrt3)/4
b) (sqrt2-sqrt6)/4
c) (sqrt6-sqrt2)/4
d) (sqrt2+sqrt6)/4

2. Find exact value of tan(pi/4-pi/3)?
a) -1+sqrt3
b) 1-sqrt3
c) -2-sqrt3
d) -2+sqrt3

3. Find the exact value of sin to the -1 power(-sqrt3/2).
a) -pi/3
b) pi/3
c) 2pi/3
d) 4pi/3

4. Find the exact value of cos to the -1 power(cos(4pi/3)).
a) -pi/3
b) pi/3
c) 2pi/3
d) 4pi/3

5. Find the exact value of cos(sin to the -1 power(3/5)).
a) 3/4
b) 4/3
c) 3/5
d) 4/5

6. Find the exact value of tan(sin to the -1 power(-4/5)).
a) -3/4
b) -4/3
c) 3/4
d) 4/3
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January 9th, 2007, 03:43 AM   #2
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This is often just an issue of using trig identities, for instance the first problem, which can be solved by this trig identity:

sin(A-B)=(sin A)(cos B)-(cos A)(sin B)
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January 9th, 2007, 02:25 PM   #3
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I assume that by "<function> to the -1 power", you mean "inverse <function>", which can be stated more clearly and concisely as such. It refers to the number (between 0 and π [for cos, cot, and csc] or -π/2 and π/2 [for sin, tan, and sec]) the <function> of which is the given number.

Are you asking for help or is this for our enjoyment? More specifically, do you already know the identities for sin(a+b), etc.?
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January 9th, 2007, 07:01 PM   #4
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I am asking for help big time. I don't know why I am not getting this.
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January 10th, 2007, 08:40 AM   #5
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Well, for instance:

sin(pi/3-pi/4)

Use the identity sin(A-B)=(sin A)(cos B)-(cos A)(sin B)

sin(pi/3-pi/4)=(sin pi/3)(cos pi/4)-(cos pi/3)(sin pi/4)

=(√3/2)(√2/2)-(1/2)(√2/2)

=(√6/4)-(√2/4)

=(√6- √2)/4

You won't find problems of this sort to be very difficult if you know the right identity.
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January 10th, 2007, 02:16 PM   #6
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The important trig addition identities are as follows:

sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
sin(a-b) = sin(a)cos(b) - cos(a)sin(b)

cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

tan(a+b) = [tan(a) + tan(b)] / [1 - tan(a)tan(b)]
tan(a-b) = [tan(a) - tan(b)] / [1 + tan(a)tan(b)]

For problems 5 and 6, use the facts that sin²(θ) + cos²(θ) = 1 and tan(θ) = sin(θ) / cos(θ), and posibly a little algebra, i.e., if tan(θ) = a/b, then tan(θ) = (a/x) / (b/x) for all values of x. Thus, a²/x² + b²/x² = 1; a² + b² = x². This will give you the value of x. Then sin(θ) = a/x and cos(θ) = b/x.
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January 10th, 2007, 05:45 PM   #7
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You guys are really turning out to be a big help. Thank you very much.

The identities make more sense, I got an answer of -2+sqrt3 on the tan problem.

I will never admit to being very good in math, so please be patient with my question, but how do I apply the identities with the sin and cos to the neg. 1 power?
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January 10th, 2007, 06:36 PM   #8
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Well, you would do it like this:

Find the exact value of cos(sin^-1 (3/5))

=cos(arcsin (3/5))
=4/5 by calculator.

Admittedly, I'm not sure how to prove, longhand, what the exact answer to this problem is. The only way I can think of to do it would be to construct a Taylor Series centered at 3/5, like this:
∑(n=0, n=∞) ((nth derivitive of arcsin(3/5))((x-3/5)^n))/n! and then prove that it converged to about 0.643501108 by evaluating several terms up to a certain accuracy point determined by Taylor's Remainder Theorem, and then constructing another Taylor Series to get the cos of that. In general, quite miserable to do.
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January 11th, 2007, 02:48 AM   #9
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Or, you could just use the identity cos²(x) + sin²(x) = 1, to get the exact value without using a calculator!
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January 11th, 2007, 02:53 AM   #10
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How would you do that?
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