Trig help!!!! (1 Viewer)

blackops23

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Show that (i) cos 46 (degrees) ~ [(root 2)/360)(180 - pi)]

(ii) sin 29 (degrees) ~ (1/360)(180(root 3) - 1))

obv it has something to do with 45 + 1 = 46, and 30 - 1 = 29...

i then tried using like...limx--> 0 cosx - 1/x = 1 and lim x-->0 sinx/x = 1....because u no - 1 degree is very close to 0.

so for part (i) = (1/(root 2))(cos 1 - sin 1)....

now cos 1 = cos (pi/180), therefore [cos(pi/180) -1 ]divided by (pi/180) is almost equal to 1.....so i replaced cos 1 with = pi/180 + 1....

and similarly replaced sin 1 with pi/180....and for part (i) I got 1/(root 2)....=( and for part 2 I got (1/360)(180 - (pi)(1- root(3)).???

any help please?
 
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Shadowdude

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Pretty sure you have to use the results:

cos(x+y) = cos(x)cos(y) - sin(x)sin(y)

and

sin(x-y) = cos(x)sin(y) - sin(x)cos(y)

Where x and y are in radians.
 

blackops23

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Pretty sure you have to use the results:

cos(x+y) = cos(x)cos(y) - sin(x)sin(y)

and

sin(x-y) = cos(x)sin(y) - sin(x)cos(y)

Where x and y are in radians.
yeah thats what i did - my answer wasn't close enough
 

Shadowdude

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Are you sure it's cos(46 degrees) = [(root 2)/360)(180 - pi)]

WolframAlpha says they aren't equal.
 

Shadowdude

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Whoever wrote this question should be shot. Approximations use approximation signs, not equal signs...!
 

blackops23

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Whoever wrote this question should be shot. Approximations use approximation signs, not equal signs...!
my bad....i swear when i type this on BOS I used ~ instead of =, yeah its approximations....
 

Shadowdude

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Well, here's how I did part (i):

Need to show

Now

Convince that of yourself by doing basic algebra.

We have:



Now, do some algebra so you get:



Now for small theta and

Result follows.
 

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