Question

Five trigonometric functions of theta. Cos0=3/5,Tan0<0. Please explain.

Answer 1

\(\bf cos(\theta)=\cfrac{3}{5}\qquad tan(\theta)<0
\\ \quad \\
tan(\theta)<0\Leftarrow\textit{another way of saying, tangent is negative}
\\ \quad \\
{\color{blue}{ tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad -tan(\theta)=\cfrac{sin(\theta)}{-cos(\theta)}\qquad -tan(\theta)=\cfrac{-sin(\theta)}{cos(\theta)}}}
\\ \quad \\

cos(\theta)=\cfrac{3}{5}\implies \cfrac{adjacent}{hypotenuse}\implies \cfrac{a=3}{c=5}\qquad \textit{what's "b"?}
\\ \quad \\
c^2=a^2+b^2\implies \pm \sqrt{c^2-a^2}={\color{blue}{ b}}\qquad recall\implies sin(\theta)=\cfrac{{\color{blue}{ b}}}{c}\)

so... the pythagorean theorem doesn't tell us if the root will be +/- so what is "b"?

well, notice above, we know the tangent is negative
for that to happen either the cosine or sine has to be negative
we know the cosine isn't, so the sine will be, thus the "b" component is negative

Answer 2

so once you find "b", you can pretty much use all three, "a", "b" and "c" to get the trig functions values

Answer 3

Would b=4???

Answer 4

recall, the tangent is negative, the cosine is positive, or "a" is positive, so "b" have to be negative \(\bf c^2=a^2+b^2\implies \sqrt{c^2-a^2}={\color{blue}{ b}}\implies \pm\sqrt{25-9}={\color{blue}{ b}}\iff -4\)

Answer 5

Thanks but where would five trigonometric functions come from??

Answer 6

http://www.gradeamathhelp.com/image-files/right-triangle-formulas.gif

Answer 7

http://htmartin.myweb.uga.edu/6190/resources/Trig%20Functions.gif <--- this one looks better

Answer 8

Thank you!

Answer 9

yw

Answer 10

Just for clarification sine would be -0.8

Answer 11

well, you'd usually keep it as fraction, but I gather you could use the decimal format as well, \(\bf sin(\theta)=\cfrac{b=-4}{c=5}\implies -\cfrac{4}{5}\iff -0.8\)

Answer 12

How you would you find the exact six trigonometric function at point (-2,6)???