January 9th, 2007, 12:51 AM  #1 
Newbie Joined: Jan 2007 Posts: 23 Thanks: 0  Exact values
I have 6 exact values questions: 1. Find the exact value of sin(pi/3pi/4)? a) (sqrt2sqrt3)/4 b) (sqrt2sqrt6)/4 c) (sqrt6sqrt2)/4 d) (sqrt2+sqrt6)/4 2. Find exact value of tan(pi/4pi/3)? a) 1+sqrt3 b) 1sqrt3 c) 2sqrt3 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 
January 9th, 2007, 03:43 AM  #2 
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
This is often just an issue of using trig identities, for instance the first problem, which can be solved by this trig identity: sin(AB)=(sin A)(cos B)(cos A)(sin B) 
January 9th, 2007, 02:25 PM  #3 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
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.? 
January 9th, 2007, 07:01 PM  #4 
Newbie Joined: Jan 2007 Posts: 23 Thanks: 0 
I am asking for help big time. I don't know why I am not getting this.

January 10th, 2007, 08:40 AM  #5 
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
Well, for instance: sin(pi/3pi/4) Use the identity sin(AB)=(sin A)(cos B)(cos A)(sin B) sin(pi/3pi/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. 
January 10th, 2007, 02:16 PM  #6 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
The important trig addition identities are as follows: sin(a+b) = sin(a)cos(b) + cos(a)sin(b) sin(ab) = sin(a)cos(b)  cos(a)sin(b) cos(a+b) = cos(a)cos(b)  sin(a)sin(b) cos(ab) = cos(a)cos(b) + sin(a)sin(b) tan(a+b) = [tan(a) + tan(b)] / [1  tan(a)tan(b)] tan(ab) = [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. 
January 10th, 2007, 05:45 PM  #7 
Newbie Joined: Jan 2007 Posts: 23 Thanks: 0 
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? 
January 10th, 2007, 06:36 PM  #8 
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
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))((x3/5)^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. 
January 11th, 2007, 02:48 AM  #9 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
Or, you could just use the identity cos²(x) + sin²(x) = 1, to get the exact value without using a calculator!

January 11th, 2007, 02:53 AM  #10 
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
How would you do that?


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