Find the exact value of each trigonometric function for the given angle θ

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Whenever I have to calculate the value of a given trigonometric function for an angle, I always refer to a table similar to this:

But what if I want to find the value for sin$\theta$, where $\theta$ = 32$^{\circ}$ or $\theta$ = 49$^{\circ}$ or for any function, for that matter.

hjpotter92

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asked Aug 18, 2014 at 3:51

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• I think it could be a good idea you switch from degrees to radians.

• You can find more detailed tables than the one you put in your post

• Now, suppose than you need the value of $\sin(\frac{32 \pi}{180})$ (which is your $32$ degrees). One of the ways is to use series centered at a point where you know the values. For this specific case, let use write $$\sin(\frac{32 \pi}{180})=\sin(\frac{30 \pi}{180}+\frac{2 \pi}{180})=\sin(\frac{ \pi}{6}+\frac{\pi}{90})$$ Now, we shall consider the development of $\sin(a+x)$ built at $x=0$ $$\sin(a+x)=\sin (a)+x \cos (a)-\frac{1}{2} x^2 \sin (a)-\frac{1}{6} x^3 \cos (a)+\frac{1}{24} x^4 \sin (a)+O\left(x^5\right)$$ in which you know the values of $\sin(a)$ and $\cos(a)$. For $a=\frac{\pi}{6}$, this just write $$\sin(\frac{ \pi}{6}+x)=\frac{1}{2}+\frac{\sqrt{3} x}{2}-\frac{x^2}{4}-\frac{x^3}{4 \sqrt{3}}+\frac{x^4}{48}+O\left(x^5\right)$$ Now, replace $x=\frac{ \pi}{90}$ and compute; you would obtain a value of $0.529919263860$ while the exact number is $0.529919264233$. For sure, if you need less accuracy, use less terms.

answered Aug 18, 2014 at 5:11

Claude LeiboviciClaude Leibovici

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Question:

Find the exact value of each of the six trigonometric functions of {eq}\theta {/eq}, if {eq}(-3,\ -7) {/eq} is a point on the terminal side of angle {eq}\theta {/eq}.

Trigonometric Functions of an Angle in Standard Position:

Let an angle in standard position whose terminal side passes thorugh the point {eq}P(x_P,y_P) {/eq}. There is a special relationship between the coordinates of this point and the trigonometric ratios of the angle {eq}\theta {/eq}. In fact, if we choose a right triangle with legs {eq}x_P, y_P {/eq} and hypotenuse {eq}r=\sqrt{x_P^2+y_P^2} {/eq} we have that:

$$\begin{aligned} \sin\theta&=\dfrac{y_P}{r}\\[0.2cm] \cos\theta&=\dfrac{x_P}{r}\\[0.2cm] \tan\theta&=\dfrac{y_P}{x_P}\\[0.2cm] \cot\theta&=\dfrac{x_P}{y_P}\\[0.2cm] \sec\theta&=\dfrac{1}{\cos\theta}=\dfrac{r}{x_P}\\[0.2cm] \csc\theta&=\dfrac{1}{\sin\theta}=\dfrac{r}{y_P} \end{aligned} $$

Answer and Explanation: 1

We have to find the trigonometric ratios of an angle whose terminal side passes through the given point.

Since the terminal side of the angle passes...

See full answer below.

Learn more about this topic:

Practice Finding the Trigonometric Ratios

from

Chapter 14 / Lesson 8

Trigonometric ratios can help determine the lengths of the sides of a triangle in different situations. Learn what trigonometric ratios are, learn how to solve for sides and angles, and practice finding these trigonometric ratios in a practice problem.

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