anonymous
  • anonymous
Find the exact value of each expression: a. sin20degcos80de-cos20degsin80deg b. cos 5pi/12cos7pi/12-sin5pi/12sin7pi/12
Mathematics
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anonymous
  • anonymous
Find the exact value of each expression: a. sin20degcos80de-cos20degsin80deg b. cos 5pi/12cos7pi/12-sin5pi/12sin7pi/12
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
@jim_thompson5910 does that make sense or should i write out again?
jim_thompson5910
  • jim_thompson5910
you don't need to write "deg"
jim_thompson5910
  • jim_thompson5910
hint: see page 2 of http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf specifically the "Sum and Difference Formulas" section

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anonymous
  • anonymous
so it would be the formula stating that sin(a +or-b)=sinacosb-cosasinb?
jim_thompson5910
  • jim_thompson5910
correct
jim_thompson5910
  • jim_thompson5910
you'll use the cosine version for part b
anonymous
  • anonymous
oh okay so for the test tomorrow I should just memorize those formulas? but so now to solve it, it would be (20+80) or (20-80)? so the answer is 100 or -60?
jim_thompson5910
  • jim_thompson5910
sin(20)cos(80)-cos(20)sin(80) = sin(20-80) = sin(-60) = ???
anonymous
  • anonymous
oh wait oops
anonymous
  • anonymous
would it be sin(60)?
anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
sin(-60) is not equal to sin(60)
anonymous
  • anonymous
oh .. i thought you put the question marks as in that it was wrong
jim_thompson5910
  • jim_thompson5910
no it's my way of putting a blank
anonymous
  • anonymous
it's pi/3
jim_thompson5910
  • jim_thompson5910
so you'd fill in the blank
anonymous
  • anonymous
ohh ok sorry
jim_thompson5910
  • jim_thompson5910
sin(-60) = -sin(60) = _______________
anonymous
  • anonymous
-pi/3
anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
no not -pi/3
jim_thompson5910
  • jim_thompson5910
you're thinking of -60 degrees = -pi/3 radians
jim_thompson5910
  • jim_thompson5910
use the unit circle and determine -sin(60)
anonymous
  • anonymous
okay i've been looking at the unit circle for 5 minutes and because you're determining -sin60 you would be looking in the 3 and fourth quadrants right?
anonymous
  • anonymous
and looking for a distance of 60 degrees the opposite way? I'm just trying to remember what my teacher said.. He said something like that @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
|dw:1433398564125:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1433398585790:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1433398602906:dw|
jim_thompson5910
  • jim_thompson5910
determine the y coordinate of that point, then make it negative that will be the value of -sin(60)
anonymous
  • anonymous
root 3/2?
jim_thompson5910
  • jim_thompson5910
\[\Large \sin(60) = \frac{\sqrt{3}}{2}\] \[\Large -\sin(60) = -\frac{\sqrt{3}}{2}\]
anonymous
  • anonymous
okay so I just look for it on the unit circle and determine it through the coordinate?
jim_thompson5910
  • jim_thompson5910
yes, x = cos(theta) and y = sin(theta)
anonymous
  • anonymous
ok! got it!
anonymous
  • anonymous
now for part b do you find pi/6 on the unit circle and look at x coordinate to determine the value?
anonymous
  • anonymous

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