engomavaw10b

2023-02-25

How to calculate $\mathrm{sin}(-150\xb0)$?

objesilo5lr1

Beginner2023-02-26Added 4 answers

Given solution: $\mathrm{sin}(-{150}^{\circ})$

$=-\mathrm{sin}\left({150}^{\circ}\right)$

$=-\mathrm{sin}({180}^{\circ}-{30}^{\circ})$

$=-\mathrm{sin}\left({30}^{\circ}\right)(\because \mathrm{sin}({180}^{\circ}-\theta )=\mathrm{sin}\theta )$

$=-\frac{1}{2}$

$=-\mathrm{sin}\left({150}^{\circ}\right)$

$=-\mathrm{sin}({180}^{\circ}-{30}^{\circ})$

$=-\mathrm{sin}\left({30}^{\circ}\right)(\because \mathrm{sin}({180}^{\circ}-\theta )=\mathrm{sin}\theta )$

$=-\frac{1}{2}$

daneetuhxxtj

Beginner2023-02-27Added 8 answers

Recall the identification of the negative angle.

$\mathrm{sin}(-\theta )=-\mathrm{sin}\left(\theta \right)$

With this in mind, we can rewrite #sin(-150)# as #-sin(150)#

$150}^{\circ$ has a reference angle of $30}^{\circ$, which means it will have the same trig values as $30}^{\circ$

On the Unit Circle, we know the coordinates for $30}^{\circ$ are $(\frac{\sqrt{3}}{2},\frac{1}{2})$, where the #y#-coordinate is the $\mathrm{sin}$ value.

This implies $\mathrm{sin}(-150)=-\frac{1}{2}$

$\mathrm{sin}(-\theta )=-\mathrm{sin}\left(\theta \right)$

With this in mind, we can rewrite #sin(-150)# as #-sin(150)#

$150}^{\circ$ has a reference angle of $30}^{\circ$, which means it will have the same trig values as $30}^{\circ$

On the Unit Circle, we know the coordinates for $30}^{\circ$ are $(\frac{\sqrt{3}}{2},\frac{1}{2})$, where the #y#-coordinate is the $\mathrm{sin}$ value.

This implies $\mathrm{sin}(-150)=-\frac{1}{2}$