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

• Oct 12th 2010, 04:24 AM
EvilKitty
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?
• Oct 12th 2010, 04:38 AM
Traveller
Hints :

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

$\displaystyle tan^2(x) + 1 = sec^2(x)$
• Oct 12th 2010, 06:55 AM
chisigma
Are well known the identities...

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Oct 12th 2010, 07:08 AM
Soroban
Hello, EvilKitty!

Quote:

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

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

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

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

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

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

Therefore: .$\displaystyle \sin x \;=\;\dfrac{opp}{hyp} \;=\;\dfrac{k}{\sqrt{k^2+1}}$
• Oct 14th 2010, 01:32 AM
EvilKitty
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.