Respuesta :
Answer:
[tex]\begin{aligned}\textsf{1)}\quad\sin(\theta)&=\dfrac{3\sqrt{13}}{13}\qquad\cos(\theta)=\dfrac{2\sqrt{13}}{13}\qquad\tan(\theta)=\dfrac{3}{2}\\\\\csc(\theta)&=\dfrac{\sqrt{13}}{3}\qquad\sec(\theta)=\dfrac{\sqrt{13}}{2}\qquad\cot(\theta)=\dfrac{2}{3}\end{aligned}[/tex]
[tex]\begin{aligned}\textsf{2)}\quad \sin(\theta)&=-\dfrac{\sqrt{2}}{2}\qquad&\cos(\theta)&=\dfrac{\sqrt{2}}{2}\qquad&\tan(\theta)&=-1\\\\\csc(\theta)&=-\sqrt{2}\qquad&\sec(\theta)&=\sqrt{2}\qquad&\cot(\theta)&=-1\end{aligned}[/tex]
Step-by-step explanation:
To find the exact values of the six trigonometric functions for a given angle θ, we can use the coordinates of the point on the terminal side of the angle.
The trigonometric functions are defined as follows:
[tex]\boxed{\begin{array}{l}\underline{\sf Trigonometric\;functions}\\\\\sf \sin(\theta)=\dfrac{O}{H}\qquad\cos(\theta)=\dfrac{A}{H}\qquad\tan(\theta)=\dfrac{O}{A}\\\\\\\sf\csc(\theta)=\dfrac{H}{O}\qquad\sec(\theta)=\dfrac{H}{A}\qquad\cot(\theta)=\dfrac{A}{O}\\\\\sf \textsf{where:}\\\phantom{ww}$\bullet$ $\theta$ is the angle.\\\phantom{ww}$\bullet$ $\sf O$ is the side opposite the angle.\\\phantom{ww}$\bullet$ $\sf A$ is the side adjacent the angle.\\\phantom{ww}$\bullet$ $\sf H$ is the hypotenuse.\end{array}}[/tex]
When working with points in the Cartesian coordinate system, we can form a right triangle by drawing lines from the origin (0, 0) to the point in question. The horizontal distance from the y-axis to the x-coordinate of the point forms the adjacent side of the right triangle, and the vertical distance from the x-axis to the y-coordinate of the point forms the opposite side of the right triangle.
Question 1
Point (2, 3) is in quadrant I.
The side adjacent the angle is 2 units, so A = 2.
The side opposite the angle is 3 units, so O = 3.
The hypotenuse can be found by using the Pythagorean formula:
[tex]H=\sqrt{A^2+O^2}[/tex]
[tex]H=\sqrt{2^2+3^2}[/tex]
[tex]H=\sqrt{4+9}[/tex]
[tex]H=\sqrt{13}[/tex]
Now, substitute A = 2, O = 3 and H = √(13) into the trigonometric functions:
[tex]\begin{aligned}\sin(\theta)&=\dfrac{3}{\sqrt{13}}=\dfrac{3\sqrt{13}}{13}\qquad&\cos(\theta)&=\dfrac{2}{\sqrt{13}}=\dfrac{2\sqrt{13}}{13}\qquad&\tan(\theta)&=\dfrac{3}{2}\\\\\csc(\theta)&=\dfrac{\sqrt{13}}{3}\qquad&\sec(\theta)&=\dfrac{\sqrt{13}}{2}\qquad&\cot(\theta)&=\dfrac{2}{3}\end{aligned}[/tex]
Question 2
Point (5, -5) is in quadrant IV.
The side adjacent the angle is 5 units, so A = 5.
The side opposite the angle is -5 units, so O = -5.
The hypotenuse can be found by using the Pythagorean formula:
[tex]H=\sqrt{A^2+O^2}[/tex]
[tex]H=\sqrt{5^2+(-5)^2}[/tex]
[tex]H=\sqrt{25+25}[/tex]
[tex]H=\sqrt{50}[/tex]
[tex]H=5\sqrt{2}[/tex]
Now, substitute A = 5, O = -5 and H = 5√2 into the trigonometric functions:
[tex]\begin{aligned}\sin(\theta)&=\dfrac{-5}{5\sqrt{2}}=-\dfrac{\sqrt{2}}{2}\qquad&\cos(\theta)&=\dfrac{5}{5\sqrt{2}}=\dfrac{\sqrt{2}}{2}\qquad&\tan(\theta)&=\dfrac{-5}{5}=-1\\\\\csc(\theta)&=\dfrac{5\sqrt{2}}{-5}=-\sqrt{2}\qquad&\sec(\theta)&=\dfrac{5\sqrt{2}}{5}=\sqrt{2}\qquad&\cot(\theta)&=\dfrac{5}{-5}=-1\end{aligned}[/tex]
