Write the following as an algebriac expression in "x" cot(arccos(3x)), 0 ≤ x ≤ 1/3
HI!!
This problem makes my head hurt :( lol
this is really not that bad at all first step: draw a triangle with some angle \(\theta\) labelled next step, since \[\cos(\theta)=3x\] put a \(3x\) on the "adjacent" side and make the hypotenuse 1
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there is a picture of an angle whose cosine is \(3x\) what you need is the "opposite" side which you find via pythagoras
let me know what you get it should have a square root in it
:o let me try
sqrt(-9x^2 + 1) ?
yes although \[\sqrt{1-9x^2}\] looks neater
haha alright
now you want the cotangent of that angle, which is "adjacent over opposite" so you are pretty much done, just write it i
Yeah I got it form here, you just made that so easy it's ridiculous. Thank you so much
it is ridiculously easy if you know what you are doing hope your head does not hurt anymore \[\color\magenta\heartsuit\]
It's alleviated :)
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