Verify the following: a) sin π/3 = cos π/6 b) sin π/6 = cos π/3

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You need to remember that `pi/3` and `pi/6` are complementary angles such that `pi/3 + pi/6 = (2pi+pi)/6 = 3pi/6 = pi/2`

a) You need to check if `sin (pi/3) = cos (pi/6)`

You may write `pi/3 = pi/2 - pi/6, hence sin (pi/3) = sin (pi/2 - pi/6)`

You need to use the formula `sin (alpha-beta) = sin alpha*cos beta- cos alpha*sin beta`

Plugging `alpha= pi/2 and beta = pi/6` in the formula yields:

`sin (pi/2-pi/6) = sin (pi/2)*cos(pi/6) + cos (pi/2)*sin (pi/6)`

Since `sin (pi/2) = 1 andcos (pi/2) = 0=gtsin (pi/2-pi/6) =cos(pi/6)`

Hence, `sin (pi/3) =cos(pi/6)`

b) You need to use the fact that `pi/6 = pi/2 - pi/3` such that:

`sin (pi/6) = sin (pi/2 - pi/3) = sin (pi/2)*cos(pi/3) + cos (pi/2)*sin (pi/3)`

`sin (pi/6) = cos(pi/3)`

**Hence, evaluating the values of sine and cosine of complementary angles yields: a) sin `(pi/3) =cos(pi/6);` b) `sin (pi/6) = cos(pi/3).` **

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