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The six trigonometric function of [tex]\theta[/tex] are [tex]\sin \theta \approx 0.781[/tex], [tex]\cos \theta = - \frac{5}{8}[/tex], [tex]\tan \theta \approx -1.250[/tex], [tex]\cot \theta \approx -0.800[/tex], [tex]\sec \theta = - \frac{8}{5}[/tex], [tex]\csc \theta \approx 1.280[/tex], respectively.
In this question, we assume that x-component of the terminal point is part of a unit circle. Then, we can find the value of y by means of the Pythagorean theorem:
[tex]y = \sqrt{1-x^{2}}[/tex] (1)
If we know that [tex]x = -\frac{5}{8}[/tex] and P is in the second quadrant, then the value of y is:
[tex]y = + \sqrt{1-\left(-\frac{5}{8} \right)^{2}}[/tex]
[tex]y \approx 0.781[/tex]
By trigonometry, we remember the following definitions for the six basic trigonometric functions:
[tex]\sin \theta = \frac{y}{1}[/tex] (1)
[tex]\cos \theta = \frac{x}{1}[/tex] (2)
[tex]\tan \theta = \frac{y}{x}[/tex] (3)
[tex]\cot \theta = \frac{1}{\tan\theta}[/tex] (4)
[tex]\sec \theta = \frac{1}{\cos \theta }[/tex] (5)
[tex]\csc \theta = \frac{1}{\sin \theta}[/tex] (6)
If we know that [tex]x = -\frac{5}{8}[/tex] and [tex]y \approx 0.781[/tex], then the six basic trigonometric functions are, respectively:
[tex]\sin \theta \approx 0.781[/tex], [tex]\cos \theta = - \frac{5}{8}[/tex], [tex]\tan \theta \approx -1.250[/tex], [tex]\cot \theta \approx -0.800[/tex], [tex]\sec \theta = - \frac{8}{5}[/tex], [tex]\csc \theta \approx 1.280[/tex]
The six trigonometric function of [tex]\theta[/tex] are [tex]\sin \theta \approx 0.781[/tex], [tex]\cos \theta = - \frac{5}{8}[/tex], [tex]\tan \theta \approx -1.250[/tex], [tex]\cot \theta \approx -0.800[/tex], [tex]\sec \theta = - \frac{8}{5}[/tex], [tex]\csc \theta \approx 1.280[/tex], respectively.
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