Given:
The point on the terminal side of theta is (4,-7).
To find:
The exact values of sin theta, secant theta, and tangent theta.
Solution:
If a point [tex]P(x,y)[/tex] is on the terminal side of theta and [tex]r=\sqrt{x^2+y^2}[/tex], then
[tex]\sin \theta=\dfrac{y}{r}[/tex]
[tex]\cos \theta=\dfrac{x}{r}[/tex]
[tex]\sec \theta=\dfrac{r}{x}[/tex]
[tex]\tan \theta=\dfrac{y}{x}[/tex]
The point on the terminal side of theta is (4,-7). Here, [tex]x=4[/tex] and [tex]y=-7[/tex].
[tex]r=\sqrt{(4)^2+(-7)^2}[/tex]
[tex]r=\sqrt{16+49}[/tex]
[tex]r=\sqrt{65}[/tex]
Now,
[tex]\sin \theta=\dfrac{-7}{\sqrt{65}}[/tex]
[tex]\cos \theta=\dfrac{4}{\sqrt{65}}[/tex]
[tex]\sec \theta=\dfrac{\sqrt{65}}{4}[/tex]
[tex]\tan \theta=\dfrac{-7}{4}[/tex]
Therefore, [tex]\sin \theta=\dfrac{-7}{\sqrt{65}},\cos \theta=\dfrac{4}{\sqrt{65}},\sec \theta=\dfrac{\sqrt{65}}{4},\tan \theta=\dfrac{-7}{4}[/tex].
