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#Post#: 7--------------------------------------------------
Unit 5 forum
DIR By: dluvbell
Date: September 25, 2014, 12:38 pm
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Unit 5
#Post#: 39--------------------------------------------------
HW
DIR By: dluvbell
Date: October 8, 2014, 8:18 am
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P393 #3,5,7
Compound angle
P400 #3-6,8-13
Double angle
P407#2-8,11,12,13
Solving trig equation
P426 #3,7-10,11, 13
P435 #1-10 (a,c,e,...), 14
#Post#: 103--------------------------------------------------
Re: Unit 5 forum
DIR By: dadougiedude
Date: November 27, 2014, 1:43 pm
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find the exact value of:
tan (23pi/12)
do you do tan(2pi - pi/12)?
#Post#: 104--------------------------------------------------
Re: Unit 5 forum
DIR By: dluvbell
Date: November 28, 2014, 12:16 pm
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Yes. Then deal with tan(pi/12) using compoumd angle. Once you
fine the value, apply CAST inorder to determine the sign of the
value.
#Post#: 105--------------------------------------------------
Re: Unit 5 forum
DIR By: dadougiedude
Date: November 28, 2014, 8:06 pm
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why is cos(11pi/12)'s answer [-sqrt6-sqrt2]/4 instead of
[sqrt6+sqrt2]/4?
since cos(11pi/12) is in quadrant 2 don't we apply CAST rule by
adding negative sign in front?
#Post#: 106--------------------------------------------------
Re: Unit 5 forum
DIR By: dluvbell
Date: November 29, 2014, 4:28 pm
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--- Quote from: DaDougieDude link ---
>
> why is cos(11pi/12)'s answer [-sqrt6-sqrt2]/4 instead of
[sqrt6+sqrt2]/4?
>
> since cos(11pi/12) is in quadrant 2 don't we apply CAST rule
by adding negative sign in front?
>
--- End Quote ---
As you said, you will put (-) in front of (sqrt6+sqrt2)/4 since
cos is (-) in the second quadrant. This becomes
(-sqrt6-sqrt2)/4.
Cos(pi/12) is (sqrt6+sqrt2)/4. But -cos(pi/12) is
-(sqrt6+sqrt2)/4 = (-sqrt6-sqrt2)/4.
#Post#: 107--------------------------------------------------
Re: Unit 5 forum
DIR By: dadougiedude
Date: November 29, 2014, 7:17 pm
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but i got[sqrt6+sqrt2]/4 first by using cos(210-45) and after
that i apply the CAST rule which i put a negative sign infront i
got [-sqrt6-sqrt2]/4!
so i still dont understand why final answer is [sqrt6+sqrt2]/4
#Post#: 108--------------------------------------------------
Re: Unit 5 forum
DIR By: dluvbell
Date: November 30, 2014, 10:25 am
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Show me in class. You are saying the answer in the back of the
book is positive, not negative right?
It should be negative. So show me the qiestion in class.
#Post#: 109--------------------------------------------------
Re: Unit 5 forum
DIR By: dadougiedude
Date: November 30, 2014, 12:58 pm
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actually the book's answer is negative, i made a mistake in the
last post. i actually got a negative answer before applying the
CAST rule, anyway i'll show you in class!
i can't wait to find out where i made a mistake
#Post#: 110--------------------------------------------------
Re: Unit 5 forum
DIR By: dadougiedude
Date: December 2, 2014, 1:14 pm
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solve x for 0 <= x <= 2pi
3sinx = sinx +1
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