SOLUTION: Prove that the given equations are identities: a)cos (30°+x)+ cos (150°-x)=0 b)cos (60°+x)- cos (300°-x)=0

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Question 203979: Prove that the given equations are identities:
a)cos (30°+x)+ cos (150°-x)=0
b)cos (60°+x)- cos (300°-x)=0
Answer by jsmallt9(2063) About Me  (Show Source):
You can put this solution on YOUR website!
Both of these problems require the sum and difference formulas for cos:
cos(a+b) = cos(a)*cos(b)-sin(a)*sin*(b)
cos(a-b) = cos(a)*cos(b)+sin(a)*sin*(b)

a) cos (30°+x)+ cos (150°-x)=0

(cos(30)*cos(x) - sin(30)*sin(x)) + (cos(150)*cos(x) + sin(150)*sin(x)) = 0
%28%28sqrt%283%29%2F2%29cos%28x%29+-+%281%2F2%29sin%28x%29%29+%2B+%28%28-sqrt%283%29%2F2%29cos%28x%29+%2B+%281%2F2%29sin%28x%29%29+=+0
And since everything cancels out...
0+=+0

b)cos (60°+x)- cos (300°-x)=0
(cos(60)cos(x) - sin(60)*sin(x)) - (cos(300)*cos(x) + sin(300)sin(x)) = 0
%28%281%2F2%29cos%28x%29+-+%28%28sqrt%283%29%29%2F2%29sin%28x%29%29+-+%28%281%2F2%29cos%28x%29+%2B+%28%28-sqrt%283%29%29%2F2%29sin%28x%29%29+=+0
Note the "-" in front of the second "half" of the left side of the equation. As a result, everything cancels out (again) leaving...
0+=+0