Respuesta :
Use the Pythagorean theorem, tan is y/x so y = -3 and x= -2. Because cos is negative and R is always positive. You need to find R- which is the hypotenuse;
3^2 + 2^2 = r^2
9 + 4 = r^2
13 = r^2
√13 = r.
So you already have tanθ,
cotθ= 2/3
sinθ= -3/√13 BUT
you have to rationalize, so you get -3 √13/ 13
cscθ= √13/ -3
cosθ= -2/ √13 BUT
you have to rationalize, so you get -2√13/ 13
secθ= √13/ -2
3^2 + 2^2 = r^2
9 + 4 = r^2
13 = r^2
√13 = r.
So you already have tanθ,
cotθ= 2/3
sinθ= -3/√13 BUT
you have to rationalize, so you get -3 √13/ 13
cscθ= √13/ -3
cosθ= -2/ √13 BUT
you have to rationalize, so you get -2√13/ 13
secθ= √13/ -2
If you're using the app, try seeing this answer through your browser: https://brainly.com/question/2804096
_______________
[tex]\mathsf{tan\,\theta=\dfrac{3}{2}\qquad\qquad (cos\,\theta<0)}\\\\\\ \mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{3}{2}}\\\\\\ \mathsf{2\,sin\,\theta=3\,cos\,\theta\qquad\quad(i)}[/tex]
• Finding [tex]\mathsf{cos\,\theta:}[/tex]
Square both sides of [tex]\mathsf{(i):}[/tex]
[tex]\mathsf{(2\,sin\,\theta)^2=(3\,cos\,\theta)^2}\\\\ \mathsf{2^2\,sin^2\,\theta=3^2\,cos^2\,\theta}\\\\ \mathsf{4\,sin^2\,\theta=9\,cos^2\,\theta\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\ \mathsf{4\cdot (1-cos^2\,\theta)=9\,cos^2\,\theta}\\\\ \mathsf{4-4\,cos^2\,\theta=9\,cos^2\,\theta}[/tex]
[tex]\mathsf{4=9\,cos^2\,\theta+4\,cos^2\,\theta}\\\\ \mathsf{4=13\,cos^2\,\theta}\\\\ \mathsf{cos^2\,\theta=\dfrac{4}{13}}\\\\\\ \mathsf{cos\,\theta=-\,\sqrt{\dfrac{4}{13}}}\qquad\qquad\textsf{(because }\mathsf{cos\,\theta}\textsf{ is negative)}\\\\\\ \mathsf{cos\,\theta=-\,\dfrac{2}{\sqrt{13}}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{sin\,\theta:}[/tex]
Substitute back the value of [tex]\mathsf{cos\,\theta}[/tex] into the equation [tex]\mathsf{(i):}[/tex]
[tex]\mathsf{2\,sin\,\theta=3\,cos\,\theta}\\\\ \mathsf{sin\,\theta=\dfrac{3}{2}\,cos\,\theta}\\\\\\ \mathsf{sin\,\theta=\dfrac{3}{\diagup\!\!\!\! 2}\cdot \left(\!-\,\dfrac{\diagup\!\!\!\! 2}{\sqrt{13}} \right)}\\\\\\ \mathsf{sin\,\theta=-\,\dfrac{3}{\sqrt{13}}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{cot\,\theta:}[/tex]
[tex]\mathsf{cot\,\theta=\dfrac{1}{tan\,\theta}}\\\\\\ \mathsf{cot\,\theta=\dfrac{1}{~\frac{3}{2}~}}\\\\\\ \mathsf{cot\,\theta=\dfrac{2}{3}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{sec\,\theta:}[/tex]
[tex]\mathsf{sec\,\theta=\dfrac{1}{cos\,\theta}}\\\\\\ \mathsf{sec\,\theta=\dfrac{1}{-\,\frac{2}{\sqrt{13}}}}\\\\\\ \mathsf{sec\,\theta=-\,\dfrac{\sqrt{13}}{2}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{csc\,\theta:}[/tex]
[tex]\mathsf{csc\,\theta=\dfrac{1}{sin\,\theta}}\\\\\\ \mathsf{csc\,\theta=\dfrac{1}{-\,\frac{3}{\sqrt{13}}}}\\\\\\ \mathsf{csc\,\theta=-\,\dfrac{\sqrt{13}}{3}\qquad\quad\checkmark}[/tex]
I hope this helps. =)
Tags: trigonometric trig function sine cosine tangent cotangent secant cosecant sin cos tan cot sec csc relation identity trigonometry
_______________
[tex]\mathsf{tan\,\theta=\dfrac{3}{2}\qquad\qquad (cos\,\theta<0)}\\\\\\ \mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{3}{2}}\\\\\\ \mathsf{2\,sin\,\theta=3\,cos\,\theta\qquad\quad(i)}[/tex]
• Finding [tex]\mathsf{cos\,\theta:}[/tex]
Square both sides of [tex]\mathsf{(i):}[/tex]
[tex]\mathsf{(2\,sin\,\theta)^2=(3\,cos\,\theta)^2}\\\\ \mathsf{2^2\,sin^2\,\theta=3^2\,cos^2\,\theta}\\\\ \mathsf{4\,sin^2\,\theta=9\,cos^2\,\theta\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\ \mathsf{4\cdot (1-cos^2\,\theta)=9\,cos^2\,\theta}\\\\ \mathsf{4-4\,cos^2\,\theta=9\,cos^2\,\theta}[/tex]
[tex]\mathsf{4=9\,cos^2\,\theta+4\,cos^2\,\theta}\\\\ \mathsf{4=13\,cos^2\,\theta}\\\\ \mathsf{cos^2\,\theta=\dfrac{4}{13}}\\\\\\ \mathsf{cos\,\theta=-\,\sqrt{\dfrac{4}{13}}}\qquad\qquad\textsf{(because }\mathsf{cos\,\theta}\textsf{ is negative)}\\\\\\ \mathsf{cos\,\theta=-\,\dfrac{2}{\sqrt{13}}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{sin\,\theta:}[/tex]
Substitute back the value of [tex]\mathsf{cos\,\theta}[/tex] into the equation [tex]\mathsf{(i):}[/tex]
[tex]\mathsf{2\,sin\,\theta=3\,cos\,\theta}\\\\ \mathsf{sin\,\theta=\dfrac{3}{2}\,cos\,\theta}\\\\\\ \mathsf{sin\,\theta=\dfrac{3}{\diagup\!\!\!\! 2}\cdot \left(\!-\,\dfrac{\diagup\!\!\!\! 2}{\sqrt{13}} \right)}\\\\\\ \mathsf{sin\,\theta=-\,\dfrac{3}{\sqrt{13}}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{cot\,\theta:}[/tex]
[tex]\mathsf{cot\,\theta=\dfrac{1}{tan\,\theta}}\\\\\\ \mathsf{cot\,\theta=\dfrac{1}{~\frac{3}{2}~}}\\\\\\ \mathsf{cot\,\theta=\dfrac{2}{3}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{sec\,\theta:}[/tex]
[tex]\mathsf{sec\,\theta=\dfrac{1}{cos\,\theta}}\\\\\\ \mathsf{sec\,\theta=\dfrac{1}{-\,\frac{2}{\sqrt{13}}}}\\\\\\ \mathsf{sec\,\theta=-\,\dfrac{\sqrt{13}}{2}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{csc\,\theta:}[/tex]
[tex]\mathsf{csc\,\theta=\dfrac{1}{sin\,\theta}}\\\\\\ \mathsf{csc\,\theta=\dfrac{1}{-\,\frac{3}{\sqrt{13}}}}\\\\\\ \mathsf{csc\,\theta=-\,\dfrac{\sqrt{13}}{3}\qquad\quad\checkmark}[/tex]
I hope this helps. =)
Tags: trigonometric trig function sine cosine tangent cotangent secant cosecant sin cos tan cot sec csc relation identity trigonometry
