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Okay so, started going back to school again (yay).
And atm in Math were going back over some stuff we have seen in the past, but since that has been 3 years ago I have some trouble with a certain thing:
Note: This isn't really homework help, these are some exercices we get, so we can see if we can solve them, and we get some points for the effort even if the answer is wrong. Atm I have put in the work to get these points, so basicly this is just for myself, because else I'll be thinking about this the rest of the day.
Cos 2 a = -sqrt(2) / 2
1) what is alpha (a, since I don't know how to write the greek letter alpha on the forum) 2) What are all the possible corners/angles (don't know the exact english word) if Cos 2 a = -sqrt(2) / 2
And, the solution is not important, I basicly want to understand how you get there.
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konadora
Singapore66063 Posts
cosine gets a value after dividing the adjacent side with the hypotenuse. so you are finding: cos(2a) = -(1/sqrt2)
since the absolute value is 1/sqrt2, the angle is 45degrees. however, it is a negative value, so it has to be in the 90 degrees < 2a < 270 degrees range, which means: 2a = 135 degrees or 225 degrees
divide that by 2 to get 67.5 degrees or 112.5 degrees
if the value of cos 2a is positive, the value of 2a must be between 0 and 90 degrees or between 270 degrees and 360 degrees.
edit: if you want to find in radian mode, just convert using 1 degree = 2pi/360
edit edit: here are some values you should remember, they are helpful in a lot of school maths questions:
for 60 degree/30 degree right-angle triangles (draw out a equilateral triangle and split it down the middle to get a 60 degree angle on one side, a right angle and a 30 degree on the last side, then give the hypotenuse a value of 2, the longer of the other two side a sqrt(3) and the last side a 1. (gotta visualise/draw) sin60 = (sqrt3)/2 cos60 = 1/2 tan60 = (sqrt3)
sin30 = 1/2 sin30 = (sqrt3)/2 tan30 = 1/(sqrt3)
for 45 degrees, draw an isosceles triangle with a right angle. give the hypotenuse a value of sqrt2, other 2 sides a 1
sin45 = 1/sqrt2 sin45 = 1/sqrt2 tan45 = 1
hope this helps lol
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well you go to wikipedia, check out the "significant values" table on the cosinus/sinus pages and then you have one possible alpha. Then you just draw the cosinus curve and make a vertical line at -1/sqrt(2) and see where it intersects the cosinus curve. one of these is the value you looked up. You should be able to derive the alphas of all the other points where this equation holds true simply by looking at that diagram.
/edit: konadora is boring and posts solutions.
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-sqrt(2)/2 is actually one of the "known" values of cos that you can look up directly from a unit circle!
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Okay thanks for the help, seems like I was looking way to far into it and I was using things that I shouldn't have :p
Especially the 2a = 135 degrees, for some reason (because of how you add up 2 angles) I was looking way to far for the solution ^^
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konadora
Singapore66063 Posts
i added some extra edit at the end of my post regarding certain values that you should remember. help it helps!
edit: On October 17 2011 00:01 Icx wrote: Okay thanks for the help, seems like I was looking way to far into it and I was using things that I shouldn't have :p
Especially the 2a = 135 degrees, for some reason (because of how you add up 2 angles) I was looking way to far for the solution ^^
were you looking at double angles? lol
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yep, I was it when I refreshed.
Altough these are in my math book, I really appreciate the effort you put into helping me (same with the others), and in the timeframe (you guys posted solutions in 5mins or so) So a big <3 to you all
edit: And ye in the exercise before this one we had to "prove" the formula for sin(2a) by using sin(a+a)=sin a*cos a + cos a * sin a, ( and cos(2a), tan(2a)) and that caused a "brainfart"
Don't know if I am explaining it well, my english in term of "expressions" in math isn't to good :p
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konadora
Singapore66063 Posts
ah alright haha, no problem!
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On October 17 2011 00:01 Louuster wrote:-sqrt(2)/2 is actually one of the "known" values of cos that you can look up directly from a unit circle!
The unit circle is incredibly helpful in so many situations. Not only in trigonometry, but also when you run into imaginary exponentials, which you will do all the time with differential equations, not to mention physics. Time well spent to really understand what is going on there. Check it up on wiki for example http://en.wikipedia.org/wiki/Unit_circle. Notice that the x-coordinate is cosine(a), and the y-coordinate is sinus(a).
glgl
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