How do you solve $sin theta = − 1/2$ ?

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Answer 1 · 2015-05-29T21:16:06

First consider a right angled triangle with sides $1$ , $sqrt(3)$ and $2$ . We can tell that it is right angled because is satisfies:

$1^2+sqrt(3)^2 = 1 + 3 = 4 = 2^2$

Notice that this is one half of an equilateral triangle with side $2$ .

The angles of that equilateral triangle are all $pi/3$

So the smallest angle of the right angled triangle is $pi/6$

By definition $sin pi/6 = 1/2$ as that is the length ( $1$ ) of the opposite side divide by the length ( $2$ ) of the hypotenuse.

Now $sin (-alpha) = -sin alpha$ , so $theta = -pi/6$ is a solution of $sin theta = -1/2$ .

Also $sin(pi-alpha) = sin(alpha)$

So $theta = pi - -pi/6 = (7pi)/6$ is another solution.

$sin(alpha+2npi) = sin alpha$ for all $n in ZZ$

So the solutions we have found are:

$theta = -pi/6 + 2npi$ for all $n in ZZ$

$theta = (7pi)/6 + 2npi$ for all $n in ZZ$

How do you solve sin theta = − 1/2sin theta = − 1/2?