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If cos theta=- sqrt 3/ 2 and pi< theta< 3 pi 2 , are the values of sin and tan

Sagot :

Using a trigonometric identity, it is found that the values of the sine and the tangent of the angle are given by:

  • [tex]\sin{\theta} = -\frac{1}{2}[/tex]
  • [tex]\tan{\theta} = \frac{\sqrt{3}}{3}[/tex]

What is the trigonometric identity using in this problem?

The identity that relates the sine squared and the cosine squared of the angle, as follows:

[tex]\sin^{2}{\theta} + \cos^{2}{\theta} = 1[/tex]

In this problem, we have that the cosine is given by:

[tex]\cos{\theta} = -\frac{\sqrt{3}}{2}[/tex]

Hence:

[tex]\sin^{2}{\theta} + \cos^{2}{\theta} = 1[/tex]

[tex]\sin^{2}{\theta} = 1 - \cos^{2}{\theta}[/tex]

[tex]\sin^{2}{\theta} = 1 - \left(-\frac{\sqrt{3}}{2}\right)^2[/tex]

[tex]\sin^{2}{\theta} = 1 - \frac{3}{4}[/tex]

[tex]\sin^{2}{\theta} = \frac{1}{4}[/tex]

[tex]\sin{\theta} = pm \sqrt{\frac{1}{4}}[/tex]

Since the angle is in the third quadrant(between pi and 1.5pi), the sine is negative, hence:

[tex]\sin{\theta} = -\frac{1}{2}[/tex]

The tangent is given by the sine divided by the cosine, hence:

[tex]\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}[/tex]

More can be learned about trigonometric identities at https://brainly.com/question/24496175

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