# How do I find sintheta = -cos^2theta -1 in radians from [0, 2pi]?

## How do I find $\sin \theta$ = $- {\cos}^{2} \theta$ -1 in radians from $\left[0 , 2 \pi\right]$?

Apr 19, 2018

$\frac{3 \pi}{2}$

#### Explanation:

$\sin \theta = - {\cos}^{2} \theta - 1 = - \left(1 - {\sin}^{2} \theta\right) - 1 = {\sin}^{2} \theta - 2 \implies$

${\sin}^{2} \theta - \sin \theta - 2 = 0 \implies \left(\sin \theta - 2\right) \left(\sin \theta + 1\right) = 0$

Since $- 1 \le \sin \theta \le + 1$, the factor $\sin \theta + 2$ can not be zero. Thus

$\sin \theta + 1 = 0 \implies \theta = {\sin}^{-} 1 \left(- 1\right) = \frac{3 \pi}{2}$

(note that $\frac{3 \pi}{2}$ is the only value of $\theta$ in $\left[0 , 2 \pi\right]$ that satisfies $\sin \theta = - 1$)

Apr 19, 2018

$\theta = \frac{3 \pi}{2}$

#### Explanation:

We can get this equation in terms of one trigonometric function using trig identities. In this case, we know the identity

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

$- {\cos}^{2} \theta = {\sin}^{2} \theta - 1$

So, we apply it to the right side:

$\sin \theta = {\sin}^{2} \theta - 1 - 1$

Move everything to the right. This gives us the following:

$0 = {\sin}^{2} \theta - \sin \theta - 2$

Or,

${\sin}^{2} \theta - \sin \theta - 2 = 0$

This strongly resembles a quadratic equation; however, instead of $a {x}^{2} + b x + c$, we observe the form $a {\sin}^{2} \theta + b \sin \theta + c$.

As a result, we can factor this just as we would factor a quadratic:

$\left(\sin \theta - 2\right) \left(\sin \theta + 1\right) = 0$

We then solve the following:
$\sin \theta - 2 = 0$
$\sin \theta + 1 = 0$

For $\sin \theta - 2 = 0 :$

$\sin \theta = 2$

Tells us this one has no solutions, as $- 1 \le \sin \theta \le 1$ for all $\theta .$

$\sin \theta + 1 = 0$

$\sin \theta = - 1$

Holds true for $\theta = \frac{3 \pi}{2}$ in the interval $\left[0 , 2 \pi\right]$

Apr 19, 2018

$\Theta$ = $\frac{3 \pi}{2}$

#### Explanation:

We can use one of the Pythagorean identities to write our equation in terms of a single trigonometric function. In particular, we can use:

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$, rewritten as ${\cos}^{2} \theta = 1 - {\sin}^{2} \theta$.

We get:

$\sin \theta = - {\cos}^{2} \theta - 1$

$\sin \theta = - \left(1 - {\sin}^{2} \theta\right) - 1$

$\sin \theta = - 1 + {\sin}^{2} \theta - 1$

$\sin \theta = {\sin}^{2} \theta - 2$

We'll solve as follows.

$- {\sin}^{2} \theta + \sin \theta + 2 = 0$

${\sin}^{2} \theta - \sin \theta - 2 = 0$ <--Multiplying Both Sides By Negative One

$\left(\sin \theta + 1\right) \left(\sin \theta - 2\right) = 0$ <--Factoring the left-hand side

The product of the factors $\sin \theta + 1 \mathmr{and} \sin \theta - 2$ is zero.
So if the equation has a solution, at least one of the factors must be zero.

$\sin \theta + 1 = 0$
$\sin \theta = - 1$

or

$\sin \theta - 2 = 0$
$\sin \theta = 2$

Sine has the value $- 1$ at $\frac{3 \pi}{2}$.
It NEVER has the value $2$.