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HannahSamantha
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Tyro
Asked: 2021-04-08T05:59:21+00:00 In: ,

Mathematics: Trigonometric functions

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Determine the six trigonometric functions of each given angle θ

a. θ = 11π/4

b. θ = 300°

trigonometry #mathematics question

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  2. acedstud
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    This answer was edited.

    a.) The angle \theta=\dfrac{11\pi}{4} is greater than 2\pi so \dfrac{11\pi}{4}\equiv\dfrac{11\pi}{4}-2\pi=\dfrac{3\pi}{4}.

    \dfrac{3\pi}{4} is in the second quadrant so sine and cosecant are positive while other trigonometric functions are negative.

    The reference angle for \dfrac{3\pi}{4}, is \pi-\dfrac{3\pi}{4}=\dfrac{\pi}{4}.

    Thus, \sin\dfrac{3\pi}{4}=\sin\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}, \cos\dfrac{3\pi}{4}=-\cos\dfrac{\pi}{4}=-\dfrac{\sqrt{2}}{2}, \tan\dfrac{3\pi}{4}=-\tan\dfrac{\pi}{4}=-1.

    Similarly, \csc\dfrac{3\pi}{4}=\dfrac{1}{\sin\dfrac{3\pi}{4}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}, \sec\dfrac{3\pi}{4}=\dfrac{1}{\cos\dfrac{3\pi}{4}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}, \cot\dfrac{3\pi}{4}=\dfrac{1}{\tan\dfrac{3\pi}{4}}=\dfrac{1}{-1}=-1.

     

    b.) 300^0 is in the fourth quadrant so cosine and secant are positive while other trigonometric functions are negative.

    The reference angle for 300^o, is 360^o-300^o=60^o.

    Thus, \sin300^o=-\sin60^o=-\dfrac{\sqrt{3}}{2}, \cos300^o=\cos60^o=\dfrac{1}{2}, \tan300^o=-\tan60^o=-\sqrt{3}.

    Similarly, \csc300^o=\dfrac{1}{\sin300^o}=-\dfrac{2}{\sqrt{3}}=-\dfrac{2\sqrt{3}}{3}, \sec300^o=\dfrac{1}{\cos300^o}=\dfrac{2}{1}=2, \cot300^o=\dfrac{1}{\tan300^o}=-\dfrac{1}{\sqrt{3}}=-\dfrac{\sqrt{3}}{3}.

     

    NB. Use the special right triangles (30-60-90 and 45-45-90 right triangles) to get the trigonometric ratios.

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