# Thread: expressing sin(x) in terms of tan(x)

1. ## expressing sin(x) in terms of tan(x)

I saw this in a list of previous examination questions today and became intrigued. It asks:

"The acute angle x radians is such that tan(x) = k. Express, in terms of k, sin(x)."

The obvious answer is k cos(x). But as a 2 mark question I get the feeling that something else is required (and I'm not sure if cos(x) counts as 'in terms of x').

I have feeling there's something deceptively obvious here than I'm missing. Can anyone give me hints, or am I overthinking this?

2. Hints :

$sin^2(x) + cos^2(x) = 1$

$tan^2(x) + 1 = sec^2(x)$

3. Are well known the identities...

$\displaystyle \sin x = \frac{2\ \tan \frac{x}{2}} {1+\tan^{2} \frac{x}{2}}$ (1)

$\displaystyle \cos x = \frac{1- \tan^{2} \frac{x}{2}} {1+\tan^{2} \frac{x}{2}}$ (2)

If You can use $\cos x$ and $\sin x$ as function of $\tan \frac{x}{2}$ instead of $\tan x$ very well!... otherwise You can devide (1) and (2) and obtain...

$\displaystyle \tan x = \frac {2 \tan \frac{x}{2}}{1- \tan^{2} \frac{x}{2}}$ (3)

... the use (3) to find $\tan \frac{x}{2}$ as function of $\tan x$ and then substitute in (1) and (2)... a very tedius job!...

Kind regards

$\chi$ $\sigma$

4. Hello, EvilKitty!

$\text{The acute angle }x\text{ radians is such that: }\:\tan x \,=\, k$

$\text{Express }\sin x\text{ in terms of }k,$

Your answer, $k\cos x$, is in terms of $\,k$ and $\,x.$

We are given: . $\tan x \:=\:\dfrac{k}{1} \:=\:\dfrac{opp}{adj}$

Angle $\,x$ is in a right triangle with: . $opp = k,\; adj = 1$

. . Pythagorus says: . $hyp = \sqrt{k^2+1}$

Therefore: . $\sin x \;=\;\dfrac{opp}{hyp} \;=\;\dfrac{k}{\sqrt{k^2+1}}$

5. Thank you very much, all!

Although I managed to work it out with identities, Soroban's answer may have been the 'intended' one, as neither double angle formulas or sec(x), cosec(x) etc. are part of my syllabus.

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