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cosine and sine
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Cosine and Sinus: Courses and exercises

This article aims to make a course with examples on sines and cosines. If you are looking for properties, go and see this article.

Definitions

By the trigonometric circle (high school level)

Consider a point on the trigonometric circle, ie the circle whose center is the origin and radius 1. Let us take an angle x with respect to the abscissa axis. The cosine is then the abscissa of this point and the sine is the ordinate. Here is a diagram to better understand how to define sine and cosine via the trigonometric circle.

With a right-angled triangle (college level)

Triangle
Triangle rectangle

We then have as formulas for the sine and the cosine:

\begin{array}{l}\cos(x) = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \\ \sin(x) = \frac{\text{opposite}}{ \text{hypotenuse}}\end{array}

From an entire series (prep)

We can define cosine and sine as an integer series:

\begin{array}{l}\cos\left(x\right)=\displaystyle \sum_{n=0}^{+\ \infty}\left(-1\right)^n\ \frac{x^ {2n}}{\left(2n\right)!}\\ \\ \sin\left(x\right)=\displaystyle \sum_{n=0}^{+\infty}\left(-1\right )^n\ \frac{x^{2n+1}}{\left(2n+1\right)!}\end{array}

Graphs

Here is the cosine graph

Cosine
Cosine graph

And that of the sine

Graph of the sine function

Manufacturing

The sine and cosine functions are 2π-periodic, which means that

\begin{array}{l}\forall x \in\mathbb{R}, \cos(x+2\pi) = \cos\left(x\right)\\ \forall x \in\mathbb{R} , \sin\left(x+2\pi\right)=\sin(x)\end{array}

The cosine function is even, the sine function is odd:

\begin{array}{l}\forall x \in\mathbb{R},\cos\left(-x\right) = \cos(x)\\ \forall x\in\mathbb{R},\sin \left(-x\right) = -\sin(x)\end{array}

The following formula is also to be remembered:

\cos^2(x) +\sin^2(x)=1

The sine and cosine values ​​to know

\begin{array}{ |c|c|c| c| c| c| c| } \hline x & 0 & \frac{\pi}{6}& \frac{\pi}{4}& \frac{\pi}{3}& \frac{\pi}{2}&\pi \ \ \hline \cos(x) & 1 & \frac{\sqrt 3}{2} & \frac{\sqrt 2}{2} & \frac{1}{2} & 0 & -1 \\ \hline \sin(x) & 0 & \frac{1}{2}& \frac{\sqrt 2}{2} & \frac{\sqrt 3}{2} &1&0 \\ \hline \end{array}

It is of course essential to very quickly find the values ​​of sine of 0 and cosine of 0:

  • cos(0) = 1
  • sin(0) = 1

On the other hand, sine of 1 and cosine of 1 do not have a precise value:

  • cos(1) ≈ 0.54030230586
  • sin(1) ≈ 0.8414709848

Derivatives

The sine and cosine functions are differentiable on their definition set and have for derivative:

\begin{array}{l}\cos^{\prime}(x)=-\sin(x)\\ \sin^{\prime}(x) = \cos\left(x\right)\end{ array}

limitations

\begin{array}{l}
\displaystyle\lim_{x\to0}\ \frac{\sin\left(x\right)}{x}=1\\
\displaystyle \lim_{x\to0}\ \frac{\cos\left(x\right)-1}{x^2}=\frac{1}{2}\end{array}

For the rest, sine and cosine have a large number of properties that you will find here in this article.

Examples

Example 1
Simplify the expression

\cos\left( \frac{37 \pi}{6}\right)

We use the periodicity of cos:

\cos \left(\frac{37\pi }{6}\right)\ =\ \cos \left(\frac{36\ \pi +\pi }{6}\right)=\cos \left(6 \pi +\frac{\pi }{6}\right)\ =\ \cos \left(\frac{\pi }{6}\right)\ =\ \frac{\sqrt{3}}{2}

Example 2
Solve in ]-π,π[ the following equation:

2\sin(x) + \sqrt 2 = 0

Let's start by simplifying the expression

\begin{array}{ll}&2\sin (x)+\sqrt{2}=0\ \\ \iff& 2\sin (x)=-\sqrt{2}\\ \iff& \sin (x)= -\frac{\sqrt{2}}{2}\end{array}

Next, let's look at the trigonometric circle:

circle trig equation

Graphically we see that we have 2 solutions. Let's finish the resolution.

\begin{array}{ll}&\sin\left(x\right)= -\dfrac{\sqrt{2}}{2}\\ \Leftrightarrow& -\sin\left(x\right)= \dfrac{ \sqrt{2}}{2}\\ \Leftrightarrow& \sin\left(-x\right) =\dfrac{\sqrt{2}}{2}\\ \Leftrightarrow& -x\in\left\{\dfrac {\pi}{4};\pi-\dfrac{\pi}{4}\right\}\\ \Leftrightarrow& - x \in\left\{\dfrac{\pi}{4}; \dfrac{3\pi}{4}\right\}\\ \Leftrightarrow& x \in\left\{-\dfrac{3\pi}{4},\dfrac{\pi}{4}\right\} \end{array}

Example 3
Solve the following equation in ]-π,π]:

2 \cos^2(x) - 3\cos(x) - 2 =0

We set X = cos(x) and we solve the equation:

\begin{array}{l}2X^{2}-3X -2 = 0\\ 
\Delta = (-3)^2-4\times(-2)\times2= 9-8 =1\\ 
X_1=\dfrac{3-1}{2}=1\\ 
X_2\ =\dfrac{3+1}{2}=2\end{array}

We then have 2 solutions: 1 and 2. But we can eliminate one of them.
Indeed, cos(x)=X = 2 has no solution. We are then reduced to solving cos(x) = 1. On the considered interval, 0 is the unique solution.

Exercices

Exercise 1

\begin{array}{l}\text{Let f be the function defined by } f\left(x\right) =\dfrac{5}{5+\cos\left(x\right)}\\ 1)\ text{ Determine the definition set of f}\\ 2)\text{ Show that the function f is even and determine its period}\\ 3)\text{ Calculate the derivative } f^{\prime}\text{ and give its interval on } [0;\pi]\\ 4)\text{ Draw the variation table on } \left[-\pi;\pi\right]\text{ and draw the function on }\left[ -\pi;3\pi\right]\end{array}

Exercise 2
Solve in R the following equations on the given interval I:

\begin{array}{ll}\sin^2\left(x\right) -\frac{3}{2}\sin\left(x\right) -1 = 0 &I = \left[-\pi, \pi\right]\\ \sin\left(x\right)=-\frac{1}{2}&I = \left[-2\pi,2\pi\right]\\ \cos\left(x \right)= \sin\left(x\right)& I = \left[-\pi,\pi\right] \end{array}

Exercise 3

\begin{array}{l}\text{Knowing that } \cos\left(-\dfrac{9\pi}{5}\right) = \dfrac{\sqrt{5}+1}{4}\: \\ 1)\text{ Calculate the value of }\sin\left(\dfrac{9\pi}{\sin}\right)\\ 2)\text{ Deduce the value of } \cos\left(\ dfrac{\pi}{5}\right)\text{ and } \sin\left(\dfrac{\pi}{5}\right)\end{array}

Exercise 4
Solve in ]-π,π] the following inequalities:

\begin{array}{l}\frac{1}{2}\le \cos\left(x\right) <1\\ \sin\left(x\right) <\frac{\sqrt{3}} {2}\\ \cos\left(x\right) \ge\frac{1}{\sqrt{2}}\\ -\frac{1}{2}\le \sin\left(x\right) \le1\end{array}

Do you want to continue revising the baccalaureate? Here are our last 5 contents on this theme:

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