SOLUTION: What is {{{x}}} in the equation {{{sin(pi/3)=cos(pi/2 -x)}}}?

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Question 1120265: What is x in the equation sin%28pi%2F3%29=cos%28pi%2F2+-x%29?
Found 2 solutions by josmiceli, greenestamps:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+sin%28+pi%2F3+%29+=+cos%28+pi%2F2+-+pi%2F3+%29+
+sin%28+pi%2F3+%29+=+cos%28+3%2Api%2F6+-+2%2Api%2F6+%29+
+sin%28+pi%2F3+%29+=+cos%28+pi%2F6+%29+
Now I can say:
+cos%28+pi%2F6+%29+=+cos%28+pi%2F2+-+x+%29+
+pi%2F6+=+3%2Api%2F6+-+x+
+x+=+3%2Api%2F6+-+pi%2F6+
+x+=+2%2Api%2F6+
+x+=+pi%2F3+
---------------------
check:
The other way to do it is:
+cos%28+pi%2F2+-+x+%29+=+sin%28+x+%29+
And
+sin%28+pi%2F3+%29+=+sin%28+x+%29+
+x+=+pi%2F3+

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


(1) If x is positive and less than pi/2, then x and (pi/2-x) are complementary acute angles; the sine of one is equal to the cosine of the other.

Specifically, cos(pi/2-x) = sin(x); the equation then says sin(pi/3) = sin(x); that makes the answer x = pi/3.

(2) Using the formula for the cosine of the difference of two angles gives the same result:

cos(pi/2-x) = cos(pi/2)*cos(x)+sin(pi/2)*sin(x) = 0+sin(x) = sin(x)

And again the equation says sin(pi/3) = sin(x), so the answer is = pi/3.

(3) However, x=pi/3 is only what might be called the "principal" answer; there are infinitely many more.

sin(pi/3) = sqrt(3)/2) = cos(pi/6) --> pi/2-x = pi/6 --> x = pi/3
sin(pi/3) = sqrt(3)/2) = cos(11pi/6) --> pi/2-x = 11pi/6 --> x = -4pi/3
sin(pi/3) = sqrt(3)/2) = cos(13pi/6) --> pi/2-x = 13pi/6 --> x = -5pi/3
sin(pi/3) = sqrt(3)/2) = cos(-pi/6) --> pi/2-x = -pi/6 --> x = 2pi/3
sin(pi/3) = sqrt(3)/2) = cos(-11pi/6) --> pi/2-x = pi/6 --> x = 7pi/3
sin(pi/3) = sqrt(3)/2) = ..... etc., etc., ....