Trig Ratios

**Oldspice1212** · September 19th 2012, 06:04 PM

Find the remaining trigonometric ratios.

tan (a) = 4, 0<a<pi/2

sin(α) =		cos(α) =
cot(α) =		sec(α) =
csc(α) =

**MarkFL** · September 19th 2012, 06:36 PM

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:

$\tan(a)=4$

$\frac{\sin(a)}{\cos(a)}=4$

$\sin(a)=4\cos(a)$

$\sin^2(a)=16\cos^2(a)$

$\sin^2(a)=16(1-\sin^2(a))$

$17\sin^2(a)=16$

$\sin^2(a)=\frac{16}{17}$

$\sin(a)=\frac{4}{\sqrt{17}}$

You can now easily find $\cos(a)$ from $\sin(a)=4\cos(a)$ . Now that you have the three primary functions, finding the rest is straightforward...

**Oldspice1212** · September 19th 2012, 06:43 PM

So would cos(a) be the same as sin (a), and the rest would be just reciprocals? Sorry I haven't done this in a long time :S

**MarkFL** · September 19th 2012, 06:52 PM

No, from the equation relating sin(a) and cos(a), we see that cos(a) is 1/4 the value of sin(a). Then yes, the secondary functions would be the reciprocals of the corresponding primary functions.

**Soroban** · September 19th 2012, 09:00 PM

Hello, Oldspice1212!

Given: . $\tan\theta \,=\,4,\;0 < \theta < \tfrac{\pi}{2}$
Find the remaining trigonometric ratios.

We are given: . $\tan\theta\:=\:\frac{4}{1}\:=\:\frac{opp}{adj}$

$\theta$ is in a right triangle with: $opp = 4,\:adj = 1$
Pythagorus tells us that: $hyp \:=\:\sqrt{4^2+1^2} \:=\:\sqrt{17}$

So we have: . $\begin{Bmatrix}opp &=& 4 \\ adj &=& 1 \\ hyp &=& \sqrt{17} \end{Bmatrix}$

And we can write all six trig ratios:

. . $\begin{array}{cccccccccc}\sin\theta &=& \frac{4}{\sqrt{17}} &\;\;& \csc\theta &=& \frac{\sqrt{17}}{4} \\ \\[-3mm] \cos\theta &=& \frac{1}{\sqrt{17}} && \sec\theta &=& \sqrt{17} \\ \\[-3mm] \tan\theta &=& 4 && \cot\theta &=& \frac{1}{4} \end{array}$

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