Question

Suppose (theta sign) is an angle in standard position whose terminal side lies in the given quadrant. For these functions, find the exact values of the remaining five trigonometric funtctions of (theta)
sin(theta)=-4/5; Quadrant 4

tan(theta)=2; Quadrant 1

Answer 1

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A)  $\mathsf{sin\,\theta=-\,\dfrac{4}{5}\qquad\qquad\theta \in 4th~quadrant.}$

$\mathsf{5\,sin\,\theta=-4\qquad\quad(i)}$


•  Finding $\mathsf{cos\,\theta:}$

Square both sides of $\mathsf{(i):}$

$\mathsf{(5\,sin^2\,\theta)=4^2}\\\\ \mathsf{5^2\,sin^2\,\theta=4^2}\\\\ \mathsf{25\,sin^2\,\theta=16\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\ \mathsf{25\cdot (1-cos^2\,\theta)=16}\\\\ \mathsf{25-25\,cos^2\,\theta=16}$

$\mathsf{25-16=25\,cos^2\,\theta}\\\\ \mathsf{9=25\,cos^2\,\theta}\\\\ \mathsf{cos^2\,\theta=\dfrac{9}{25}}\\\\\\ \mathsf{cos\,\theta=\pm\,\sqrt{\dfrac{9}{25}}}\\\\\\ \mathsf{cos\,\theta=\pm\,\sqrt{\dfrac{3^2}{5^2}}}\\\\\\ \mathsf{cos\,\theta=\pm\,\dfrac{3}{5}}$


But $\mathsf{cos\,\theta}$ is also negative, because $\mathsf{\theta}$ lies in the 4th quadrant. So,

$\mathsf{cos\,\theta=-\,\dfrac{3}{5}\qquad\quad\checkmark}$


•  Finding $\mathsf{tan\,\theta:}$

$\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\ \mathsf{tan\,\theta=\dfrac{-\,\frac{4}{5}}{-\,\frac{3}{5}}}\\\\\\ \mathsf{tan\,\theta=-\,\dfrac{4}{\diagup\hspace{-6}5}\cdot \left(\!-\,\dfrac{\diagup\hspace{-6}5}{3}\right)}\\\\\\ \mathsf{tan\,\theta=\dfrac{4}{3}\qquad\quad\checkmark}$


•  Finding $\mathsf{cot\,\theta:}$

$\mathsf{cot\,\theta=\dfrac{1}{tan\,\theta}}\\\\\\ \mathsf{cot\,\theta=\dfrac{1}{~\frac{4}{3}~}}\\\\\\ \mathsf{cot\,\theta=\dfrac{3}{4}\qquad\quad\checkmark}$


•  Finding $\mathsf{sec\,\theta:}$

$\mathsf{sec\,\theta=\dfrac{1}{cos\,\theta}}\\\\\\ \mathsf{sec\,\theta=\dfrac{1}{-\,\frac{3}{5}}}\\\\\\ \mathsf{sec\,\theta=-\,\dfrac{5}{3}\qquad\quad\checkmark}$


•  Finding $\mathsf{csc\,\theta:}$

$\mathsf{csc\,\theta=\dfrac{1}{sin\,\theta}}\\\\\\ \mathsf{csc\,\theta=\dfrac{1}{-\,\frac{4}{5}}}\\\\\\ \mathsf{csc\,\theta=-\,\dfrac{5}{4}\qquad\quad\checkmark}$

________


B)  $\mathsf{tan\,\theta=2\qquad\qquad\theta \in 1st~quadrant.}$

$\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=2}\\\\\\ \mathsf{sin\,\theta=2\,cos\,\theta\qquad\quad(ii)}$


•  Finding $\mathsf{sin\,\theta:}$

Square both sides of $\mathsf{(ii):}$

$\mathsf{(sin\,\theta)^2=(2\,cos\,\theta)^2}\\\\ \mathsf{sin^2\,\theta=2^2\,cos^2\,\theta}\\\\ \mathsf{sin^2\,\theta=4\,cos^2\,\theta\qquad\qquad(but~cos^2\,\theta=1-sin^2\,\theta)}\\\\ \mathsf{sin^2\,\theta=4\cdot (1-sin^2\,\theta)}\\\\ \mathsf{sin^2\,\theta=4-4\,sin^2\,\theta}$

$\mathsf{sin^2\,\theta+4\,sin^2\,\theta=4}\\\\ \mathsf{5\,sin^2\,\theta=4}\\\\ \mathsf{sin^2\,\theta=\dfrac{4}{5}}\\\\ \mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{4}{5}}}\\\\\\ \mathsf{sin\,\theta=\pm\,\dfrac{2}{\sqrt{5}}}$


But sine is positive because $\theta$ is a 1st quadrant angle:

$\mathsf{sin\,\theta=\dfrac{2}{\sqrt{5}}\qquad\quad\checkmark}$


•  Finding $\mathsf{cos\,\theta:}$

$\mathsf{sin\,\theta=2\,cos\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{1}{2}\,sin\,\theta}\\\\\\ \mathsf{cos\,\theta=\dfrac{1}{\diagup\hspace{-6}2}\cdot \dfrac{\diagup\hspace{-6}2}{\sqrt{5}}}\\\\\\ \mathsf{cos\,\theta=\dfrac{1}{\sqrt{5}}\qquad\quad\checkmark}$


•  Finding $\mathsf{cot\,\theta:}$

$\mathsf{cot\,\theta=\dfrac{1}{tan\,\theta}}\\\\\\ \mathsf{cot\,\theta=\dfrac{1}{2}}\qquad\quad\checkmark}$


•  Finding $\mathsf{sec\,\theta:}$

$\mathsf{sec\,\theta=\dfrac{1}{cos\,\theta}}\\\\\\ \mathsf{sec\,\theta=\dfrac{1}{~\frac{1}{\sqrt{5}}~}}\\\\\\ \mathsf{sec\,\theta=\sqrt{5}\qquad\quad\checkmark}$


•  Finding $\mathsf{csc\,\theta:}$

$\mathsf{csc\,\theta=\dfrac{1}{sin\,\theta}}\\\\\\ \mathsf{csc\,\theta=\dfrac{1}{~\frac{2}{\sqrt{5}}~}}\\\\\\ \mathsf{csc\,\theta=\dfrac{\sqrt{5}}{2}\qquad\quad\checkmark}$


I hope this helps. =)


Tags:  trigonometric trig function sine cosine tangent cotangent secant cosecant sin cos tan cot sec csc trigonometry