Question

Find two values of Θ, 0< Θ < 2 pi, that satisfy the following equationsin Θ = negative square root 2 over 2

Answer 1

Given:

`\sin\theta=-(√(2))/(2),0\leq\theta<2\pi`

To find:

The two values.

Step-by-step explanation:

It can be written as,

`\begin{gathered} \sin\theta=-(√(2))/(2)*(√(2))/(√(2)) \\ \sin\theta=-(2)/(2√(2)) \\ \sin\theta=-(1)/(√(2)) \\ \theta=\sin^(-1)(-(1)/(√(2))) \end{gathered}`

Since the angle lies between 0 and 2π.

Also, the value of the sine function is negative only when the angle lies in the third and fourth quadrant.

So, the angles are,

`\begin{gathered} \theta=180+45^(\circ)\Rightarrow\text{ }225^(\circ)\text{ }(or)\text{ }(5\pi)/(4) \\ \theta=360-45\Rightarrow\text{ }315^(\circ)\text{ }(or)\text{ }(7\pi)/(4) \end{gathered}`

Final answer:

The values are,

`\theta=(5\pi)/(4),(7\pi)/(4)`