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antiseptic1488 [7]
1 year ago
10

PLEASE HELP ME I'M GIVING 20PTS AND MARKING BRAINLIEST!!!!

Mathematics
1 answer:
pishuonlain [190]1 year ago
5 0

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

  • \cos{\theta} = \pm \frac{2\sqrt{2}}{3}
  • \tan{\theta} = \pm \frac{\sqrt{2}}{4}

<h3>What is the trigonometric identity using in this problem?</h3>

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

\sin^{2}{\theta} + \cos^{2}{\theta} = 1

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

\sin{\theta} = \frac{1}{3}

Hence, applying the identity, the cosine is given as follows:

\cos^2{\theta} = 1 - \sin^2{\theta}

\cos^2{\theta} = 1 - \left(\frac{1}{3}\right)^2

\cos^2{\theta} = 1 - \frac{1}{9}

\cos^2{\theta} = \frac{8}{9}

\cos{\theta} = \pm \sqrt{\frac{8}{9}}

\cos{\theta} = \pm \frac{2\sqrt{2}}{3}

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

\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}

\tan{\theta} = \frac{\frac{1}{3}}{\pm \frac{2\sqrt{2}}{3}}

\tan{\theta} = \pm \frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}

\tan{\theta} = \pm \frac{\sqrt{2}}{4}

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

#SPJ1

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