Evaluate $\sin, \cos, \tan$ $- 150 °$ without a calculator.

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Solution

Step 1: Calculating $\sin (- 150 °)$
$\sin (- 150 °) = - \sin 150 °$
We can write this because $sinx$ is an odd function.
$- \sin 150 ° = - \sin (180 ° - 30) = - \sin 30 ° = \frac{- 1}{2}$
Step 2: Calculating $\cos (- 150 °)$
We know,
$\begin{array}{rcl} \sin^{2} x + \cos^{2} x & = & 1 \\ {(\frac{- 1}{2})}^{2} + \cos 2 (- 150 °) & = & 1 \\ \cos^{2} (- 150 °) & = & 1 - \frac{1}{4} \\ \cos^{2} (- 150 °) & = & \frac{3}{4} \\ \cos (- 150 °) & = & \frac{- \sqrt{3}}{2} \end{array}$
We choose the negative value because the end arm of the angle lies in the third quadrant, and in this quadrant sin and cos are negative.
Step 3: Calculating $\tan (- 150 °)$ :
$\begin{array}{rcl} \tan x & = & \frac{\sin x}{\cos x} \\ \tan (- 150 °) & = & \frac{\sin (- 150 °)}{\cos (- 150 °)} = \frac{- 1}{2} \div (- \frac{\sqrt{3}}{2}) \\ = & \frac{1}{2} \times \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \end{array}$
Hence, the values are $\sin (- 150 °) = \frac{- 1}{2}, \cos (- 150 °) = \frac{- \sqrt{3}}{2}, \tan (- 150 °) = \frac{\sqrt{3}}{3}$ .

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Evaluate sin,cos,tan -150° without a calculator.

Evaluate $\sin, \cos, \tan$ $- 150 °$ without a calculator.