Trigonometry : corrected 1st grade math exercises in PDF.

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Answers to math exercises in 1st grade on trigonometry.

Exercise 1:

Solve in $%5D-\pi;\pi%5D$ the following equations.

1. $cosx=\frac{\sqrt{2}}{2}.$

$S=\,\{-\frac{\pi}{4};\frac{\pi}{4}\,\,\}$

2. $sinx=-\frac{\sqrt{3}}{2}.$

$S=\,\{-\frac{\pi}{6};-\frac{5\pi}{6}\,\,\}$

Exercise 2:

In this exercise, we give:

$cos(\frac{\pi}{5})=\frac{1+\sqrt{5}}{4}.$

Calculate the exact value of $cos(\frac{2\pi}{5})$ and then of $cos(\frac{3\pi}{5}).$

$cos\,\,\frac{2\pi}{5}=\,cos^2\frac{\pi}{5}-1$

$cos\,\,\frac{2\pi}{5}=\,\,(\,\frac{1+\sqrt{5}}{4}\,\,)^2-1$

$cos\,\,\frac{2\pi}{5}=\,\frac{1+2\sqrt{5}+5}{16}\,-1$

$cos\,\,\frac{2\pi}{5}=\,\frac{1+2\sqrt{5}+5-16}{16}$

$cos\,\,\frac{2\pi}{5}=\,\frac{2\sqrt{5}-10}{16}$

${\color{DarkRed}\,cos\,\,\frac{2\pi}{5}=\,\frac{\sqrt{5}-5}{8}\,}$

Hint: for $cos\,\,\frac{3\pi}{5}$ , use the addition formula with $x=\frac{\pi}{5}$ and $y=\frac{2\pi}{5}$ .

Exercise 3:

In this exercise, we have the following data: $tan(\frac{\pi}{12})=2-\sqrt{3}.$

1. Let $x\in%5D0;\frac{\pi}{2}%5B$ . Prove that $tan(\frac{\pi}{2}-x)=\frac{1}{tanx}.$

$tan(\frac{\pi}{2}-x)=\frac{sin(\frac{\pi}{2}-x)}{cos(\frac{\pi}{2}-x)}=\frac{cosx}{sinx}=\frac{1}{tanx}$

2. Deduce that :

$tan(\frac{5\pi}{12})=2+\sqrt{3}.$

Use the equality of 1. by taking $x=\frac{\pi}{12}$

$tan(\frac{\pi}{2}-\frac{\pi}{12})=\frac{1}{tan\frac{\pi}{12}}$

so $tan(\frac{6\pi}{12}-\frac{\pi}{12})=\frac{1}{tan\frac{\pi}{12}}$

$tan(\frac{5\pi}{12})=\frac{1}{tan\frac{\pi}{12}}$

and using the data from the statement :

$tan(\frac{5\pi}{12})=\frac{1}{2-\sqrt{3}}$

Let’s multiply by the conjugate quantity and use the remarkable identity .

$tan(\frac{5\pi}{12})=\frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}$

$tan(\frac{5\pi}{12})=\frac{2+\sqrt{3}}{2^2-(\sqrt{3})^2}$

$tan(\frac{5\pi}{12})=\frac{2+\sqrt{3}}{4-3}$

$tan(\frac{5\pi}{12})=\frac{2+\sqrt{3}}{1}$

Conclusion: $tan(\frac{5\pi}{12})=2+\sqrt{3}.$

Exercise 4:

Solve in $%5D-\pi;\pi%5D$ the equation: sin(2x) = cos(x).

It’s a product equation.

A product of factors is zero if and only if at least one of the factors is zero.

$cosx=0\,ou\,sinx=\frac{1}{2}$

$S=\,\,\{\,-\frac{\pi}{2};\frac{\pi}{2};\frac{\pi}{6};\frac{5\pi}{6}\,\}$

Exercise 5:

Solve in $%5D-\pi;\pi%5B$ the following equations:

Hint: make changes of variable and use trigonometric formulas.

For the 1, put X=cosx.

$1.2cos^3x-7cos^2x+2cosx+3=0.\\2.2sin^3x+cos^2x5sinx-3=0.$

Exercise 6:

Solve in $\mathbb{R}$ the equation :

Let’s put

We have to solve :

An obvious root is X = 1.

a = 2

-c=8 so c= -8

c-b=7 so -8-b=7 so b=-8-7=-15

Let’s calculate the value of the discriminant :

$\Delta\,=(-15)^2+4\times \,2\times \,8=289$

The discriminant is positive, there are two distinct real roots.

$X_1=\frac{15+17}{4}=8\,,\,X_2=\frac{15-17}{4}=-\frac{1}{2}$

$2X^3-17X^2+7X+8=(X-1)(X-8)(x+\frac{1}{2})$

so we get :

$sinx=-1\,ou\,sinx=8\,ou\,sinx=-\frac{1}{2}$

or $-1\leq\,\,sinx\,\leq\,\,1$

so either

$sinx=-1\,ou\,\,sinx=-\frac{1}{2}$

Conclusion:
$x=-\frac{\pi}{2}+2k\pi\,ou\,\,x=-\frac{\pi}{6}+2k\pi\,ou\,\,x=-\frac{5\pi}{6}+2k\pi$

Exercise 7:

Exercise 8:

Exercise 9:

Exercise 10:

ABC is a triangle with $BC=4,\widehat{B}=\frac{\pi}{4};\widehat{C}=\frac{\pi}{3}$ .

1. Prove that $sin\,\,\,\widehat{A}=\frac{\sqrt{6}+\sqrt{2}}{4}$ .

2. Calculate the exact values of AB and AC .

Indications:

Use the Al-Kashi formulas and the generalized Pythagorean theorem.

Exercise 11:

Show that the graphical representation of the function defined on $\mathbb{R}$ by :

is located between the lines of equation y = – 3 and y = 1 .

All this comes from the fact that $-1\leq\,\,sinx\leq\,\,1$ and $-1\leq\,\,cosx\leq\,\,1$ .

Just add up each member.

Exercise 12:

Show that, for any real :

Let’s use the remarkable identity

$cos^4x-sin^4x=cos(2x)\times \,1$ because and

Conclusion: ${\color{Blue}\,cos^4x-sin^4x=cos(2x).}$

Exercise 13:

Using the addition formulas, calculate the exact value of $sin(\frac{7\pi}{12})\,et\,cos(\frac{7\pi}{12}).$

$sin(\frac{7\pi}{12})=sin(\frac{\pi}{2}+\frac{\pi}{12})=cos(\frac{\pi}{12})$

$cos(\frac{7\pi}{12})=cos(\frac{\pi}{2}+\frac{\pi}{12})=-sin(\frac{\pi}{12})$

Exercise 14:

Let ABC be any triangle.

In the triangles AEB and BEC rectangles in E ,

by applying the Pythagorean theorem :

$\,\{\,h^2=c^2-x^2\\h^2=a^2-(b-x)^2\,.$

so by equality, we deduce that :

(*)

In the triangle AEB right-angled at E, using right-angled trigonometry:

$cos(\alpha\,)=\frac{x}{c}$

$x=c\times \,cos(\alpha\,)$

Let’s take the equality ( *)

${\color{DarkRed}\,c^2=a^2-b^2+2bc\times \,cos(\alpha\,)}$

Remark: these are the same demonstrations for the other formulas of Al-Kashi,

which are also called generalized Pythagorean formulas.

Exercise 15:

$\frac{sin3x}{sinx}+\frac{cos3x}{cox}=\frac{sin3x\times \,cosx}{sinx\times \,cosx}+\frac{cos3x\times \,sinx}{cox\times \,sinx}$

$=\frac{sin3x\times \,cosx+cos3x\times \,sinx}{sinx\times \,cosx}$

$=\frac{sin(3x+x)}{sinx\times \,cosx}$

$=\frac{sin(4x)}{\frac{sin(2x)}{2}}$

$=2\frac{sin(4x)}{sin(2x)}$

$=2\frac{2sin(2x)cos(2x))}{sin(2x)}$

${\color{DarkRed}=4cos(2x)}$

Exercise 16:

Exercise 17:

Exercise 18:

Exercise 19:

Answers to the exercises on trigonometry in 1st grade.

After having consulted the answers to these exercises on trigonometry in 1ère, you can return to the exercises in première.

The exercises in the first year.

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