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A) [tex]\mathsf{sin\,\theta=-\,\dfrac{4}{5}\qquad\qquad\theta \in 4th~quadrant.}[/tex]
[tex]\mathsf{5\,sin\,\theta=-4\qquad\quad(i)}[/tex]
• Finding [tex]\mathsf{cos\,\theta:}[/tex]
Square both sides of [tex]\mathsf{(i):}[/tex]
[tex]\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}[/tex]
[tex]\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}}[/tex]
But [tex]\mathsf{cos\,\theta}[/tex] is also negative, because [tex]\mathsf{\theta}[/tex] lies in the 4th quadrant. So,
[tex]\mathsf{cos\,\theta=-\,\dfrac{3}{5}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{tan\,\theta:}[/tex]
[tex]\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}[/tex]
• Finding [tex]\mathsf{cot\,\theta:}[/tex]
[tex]\mathsf{cot\,\theta=\dfrac{1}{tan\,\theta}}\\\\\\ \mathsf{cot\,\theta=\dfrac{1}{~\frac{4}{3}~}}\\\\\\ \mathsf{cot\,\theta=\dfrac{3}{4}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{sec\,\theta:}[/tex]
[tex]\mathsf{sec\,\theta=\dfrac{1}{cos\,\theta}}\\\\\\ \mathsf{sec\,\theta=\dfrac{1}{-\,\frac{3}{5}}}\\\\\\ \mathsf{sec\,\theta=-\,\dfrac{5}{3}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{csc\,\theta:}[/tex]
[tex]\mathsf{csc\,\theta=\dfrac{1}{sin\,\theta}}\\\\\\ \mathsf{csc\,\theta=\dfrac{1}{-\,\frac{4}{5}}}\\\\\\ \mathsf{csc\,\theta=-\,\dfrac{5}{4}\qquad\quad\checkmark}[/tex]
________
B) [tex]\mathsf{tan\,\theta=2\qquad\qquad\theta \in 1st~quadrant.}[/tex]
[tex]\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=2}\\\\\\ \mathsf{sin\,\theta=2\,cos\,\theta\qquad\quad(ii)}[/tex]
• Finding [tex]\mathsf{sin\,\theta:}[/tex]
Square both sides of [tex]\mathsf{(ii):}[/tex]
[tex]\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}[/tex]
[tex]\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}}}[/tex]
But sine is positive because [tex]\theta[/tex] is a 1st quadrant angle:
[tex]\mathsf{sin\,\theta=\dfrac{2}{\sqrt{5}}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{cos\,\theta:}[/tex]
[tex]\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}[/tex]
• Finding [tex]\mathsf{cot\,\theta:}[/tex]
[tex]\mathsf{cot\,\theta=\dfrac{1}{tan\,\theta}}\\\\\\ \mathsf{cot\,\theta=\dfrac{1}{2}}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{sec\,\theta:}[/tex]
[tex]\mathsf{sec\,\theta=\dfrac{1}{cos\,\theta}}\\\\\\ \mathsf{sec\,\theta=\dfrac{1}{~\frac{1}{\sqrt{5}}~}}\\\\\\ \mathsf{sec\,\theta=\sqrt{5}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{csc\,\theta:}[/tex]
[tex]\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}[/tex]
I hope this helps. =)
Tags: trigonometric trig function sine cosine tangent cotangent secant cosecant sin cos tan cot sec csc trigonometry
_______________
A) [tex]\mathsf{sin\,\theta=-\,\dfrac{4}{5}\qquad\qquad\theta \in 4th~quadrant.}[/tex]
[tex]\mathsf{5\,sin\,\theta=-4\qquad\quad(i)}[/tex]
• Finding [tex]\mathsf{cos\,\theta:}[/tex]
Square both sides of [tex]\mathsf{(i):}[/tex]
[tex]\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}[/tex]
[tex]\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}}[/tex]
But [tex]\mathsf{cos\,\theta}[/tex] is also negative, because [tex]\mathsf{\theta}[/tex] lies in the 4th quadrant. So,
[tex]\mathsf{cos\,\theta=-\,\dfrac{3}{5}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{tan\,\theta:}[/tex]
[tex]\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}[/tex]
• Finding [tex]\mathsf{cot\,\theta:}[/tex]
[tex]\mathsf{cot\,\theta=\dfrac{1}{tan\,\theta}}\\\\\\ \mathsf{cot\,\theta=\dfrac{1}{~\frac{4}{3}~}}\\\\\\ \mathsf{cot\,\theta=\dfrac{3}{4}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{sec\,\theta:}[/tex]
[tex]\mathsf{sec\,\theta=\dfrac{1}{cos\,\theta}}\\\\\\ \mathsf{sec\,\theta=\dfrac{1}{-\,\frac{3}{5}}}\\\\\\ \mathsf{sec\,\theta=-\,\dfrac{5}{3}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{csc\,\theta:}[/tex]
[tex]\mathsf{csc\,\theta=\dfrac{1}{sin\,\theta}}\\\\\\ \mathsf{csc\,\theta=\dfrac{1}{-\,\frac{4}{5}}}\\\\\\ \mathsf{csc\,\theta=-\,\dfrac{5}{4}\qquad\quad\checkmark}[/tex]
________
B) [tex]\mathsf{tan\,\theta=2\qquad\qquad\theta \in 1st~quadrant.}[/tex]
[tex]\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=2}\\\\\\ \mathsf{sin\,\theta=2\,cos\,\theta\qquad\quad(ii)}[/tex]
• Finding [tex]\mathsf{sin\,\theta:}[/tex]
Square both sides of [tex]\mathsf{(ii):}[/tex]
[tex]\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}[/tex]
[tex]\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}}}[/tex]
But sine is positive because [tex]\theta[/tex] is a 1st quadrant angle:
[tex]\mathsf{sin\,\theta=\dfrac{2}{\sqrt{5}}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{cos\,\theta:}[/tex]
[tex]\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}[/tex]
• Finding [tex]\mathsf{cot\,\theta:}[/tex]
[tex]\mathsf{cot\,\theta=\dfrac{1}{tan\,\theta}}\\\\\\ \mathsf{cot\,\theta=\dfrac{1}{2}}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{sec\,\theta:}[/tex]
[tex]\mathsf{sec\,\theta=\dfrac{1}{cos\,\theta}}\\\\\\ \mathsf{sec\,\theta=\dfrac{1}{~\frac{1}{\sqrt{5}}~}}\\\\\\ \mathsf{sec\,\theta=\sqrt{5}\qquad\quad\checkmark}[/tex]
• Finding [tex]\mathsf{csc\,\theta:}[/tex]
[tex]\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}[/tex]
I hope this helps. =)
Tags: trigonometric trig function sine cosine tangent cotangent secant cosecant sin cos tan cot sec csc trigonometry
