Mathematics
TestU01423475716
9

If tan θ = 3/2 and cos θ < 0, use the fundamental identities to evaluate the other five trig functions of θ.

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(2) Answers
jazzlove

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

kekenunuanitasky2

If you're using the app, try seeing this answer through your browser:   https://brainly.com/question/2804096 _______________ [latex]\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)}[/latex] •   Finding [latex]\mathsf{cos\,\theta:}[/latex] Square both sides of [latex]\mathsf{(i):}[/latex] [latex]\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}[/latex] [latex]\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}[/latex] •   Finding [latex]\mathsf{sin\,\theta:}[/latex] Substitute back the value of [latex]\mathsf{cos\,\theta}[/latex] into the equation [latex]\mathsf{(i):}[/latex] [latex]\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}[/latex] •   Finding [latex]\mathsf{cot\,\theta:}[/latex] [latex]\mathsf{cot\,\theta=\dfrac{1}{tan\,\theta}}\\\\\\ \mathsf{cot\,\theta=\dfrac{1}{~\frac{3}{2}~}}\\\\\\ \mathsf{cot\,\theta=\dfrac{2}{3}\qquad\quad\checkmark}[/latex] •   Finding [latex]\mathsf{sec\,\theta:}[/latex] [latex]\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}[/latex] •   Finding [latex]\mathsf{csc\,\theta:}[/latex] [latex]\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}[/latex] I hope this helps. =) Tags:  trigonometric trig function sine cosine tangent cotangent secant cosecant sin cos tan cot sec csc relation identity trigonometry

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