Find the remaining trigonometric ratios.
tan (a) = 4, 0<a<pi/2
sin(α)
=cos(α)
=cot(α)
=sec(α)
=csc(α)
=
I would use the Pythagorean identities, keeping in mind all roots will be positive since the angle given is in the first quadrant. For example:
$\displaystyle \tan(a)=4$
$\displaystyle \frac{\sin(a)}{\cos(a)}=4$
$\displaystyle \sin(a)=4\cos(a)$
$\displaystyle \sin^2(a)=16\cos^2(a)$
$\displaystyle \sin^2(a)=16(1-\sin^2(a))$
$\displaystyle 17\sin^2(a)=16$
$\displaystyle \sin^2(a)=\frac{16}{17}$
$\displaystyle \sin(a)=\frac{4}{\sqrt{17}}$
You can now easily find $\displaystyle \cos(a)$ from $\displaystyle \sin(a)=4\cos(a)$. Now that you have the three primary functions, finding the rest is straightforward...
Hello, Oldspice1212!
Given: .$\displaystyle \tan\theta \,=\,4,\;0 < \theta < \tfrac{\pi}{2}$
Find the remaining trigonometric ratios.
We are given: .$\displaystyle \tan\theta\:=\:\frac{4}{1}\:=\:\frac{opp}{adj}$
$\displaystyle \theta$ is in a right triangle with: $\displaystyle opp = 4,\:adj = 1$
Pythagorus tells us that: $\displaystyle hyp \:=\:\sqrt{4^2+1^2} \:=\:\sqrt{17}$
So we have: .$\displaystyle \begin{Bmatrix}opp &=& 4 \\ adj &=& 1 \\ hyp &=& \sqrt{17} \end{Bmatrix}$
And we can write all six trig ratios:
. . $\displaystyle \begin{array}{cccccccccc}\sin\theta &=& \frac{4}{\sqrt{17}} &\;\;& \csc\theta &=& \frac{\sqrt{17}}{4} \\ \\[-3mm] \cos\theta &=& \frac{1}{\sqrt{17}} && \sec\theta &=& \sqrt{17} \\ \\[-3mm] \tan\theta &=& 4 && \cot\theta &=& \frac{1}{4} \end{array}$