Question

Which angle(s) have a sine of −1/2? Select all that apply.

Answer 1

Remember the following trigonometric identities:

`\begin{gathered} \sin (x)=\cos (x-(\pi)/(2)) \\ \sin (x+2\pi k)=\sin (x)\text{ for all integer values of k} \\ \cos (x)=a\Rightarrow x=\pm\cos ^(-1)(a) \\ \sin (-x)=-\sin (x) \\ -\cos (x)=\cos (x-\pi) \end{gathered}`

Using these identities, search for an angle x such that:

`\sin (x)=-(1)/(2)`

Write the sine of x in terms of the cosine:

`\Rightarrow\cos (x-(\pi)/(2))=-(1)/(2)`

Multiply both sides by -1:

`\Rightarrow-\cos (x-(\pi)/(2))=(1)/(2)`

Since -cos(x)=cos(x-π), then:

`\begin{gathered} \Rightarrow\cos (x-(\pi)/(2)-\pi)=(1)/(2) \\ \Rightarrow cos(x-(3)/(2)\pi)=(1)/(2) \end{gathered}`

Use the inverse cosine function to isolate x-3π/2:

`\Rightarrow x-(3)/(2)\pi=\pm\cos ^(-1)((1)/(2))`

Since cos(π/3) = 1/2, then cos^-1(1/2) = π/3:

`\begin{gathered} \Rightarrow x-(3)/(2)\pi=\pm(\pi)/(3) \\ \Rightarrow x=(3)/(2)\pi\pm(1)/(3)\pi \\ \Rightarrow x=((3)/(2)\pm(1)/(3))\pi \\ \Rightarrow x_1=(11)/(6)\pi,x_2=(7)/(6)\pi \end{gathered}`

To find other values of x which make sin(x) equal to -1/2, we can add integer multiples of 2π to the values x₁ and x₂. In this case, since all options are less than 2π, that is not necessary.

Therefore, the angles whose sine is -1/2 are:

`\begin{gathered} (7)/(6)\pi \\ (11)/(6)\pi \end{gathered}`