wildareal's questions thread

[wildareal]

Here's another Trig Question:

(sec(x)-cos(x))(cosec(x)-sin(x))

Thanks.

[kamil9876]

That's not a question it is an expression, what do you want me to do with it? differentiate? integrate? sketch? simplify?

I'll assume the latter:

It is

[wildareal]

Simplify sin(tan^-1(-2))

Thanks.

[Pixon]

let tan^-1(-2)=x
therefore, tan(x)=-2
for tan(x)=Opposite/Adjacent, O=2, A=1 in the 4th quadrant (since tan^1(x) has a range (-pi/2,pi/2) )
thus, hypotenuse=root(2^2+1^2)=root5

sin(x)=opposite/hypotenuse=-2/root5  (sin(x) is negative in the 4th quadrant)

remember, tan^-1(-2)=x

sin(tan^-1(-2))=sin(x)
                     =-2/root5

(sorry about not using latex..it's late...there may also be a better way to do it)

[wildareal]

Use a compound angle formula to show than tan^-1(3)-tan^-1(1/2)=TT/4

Hence show that tan^-1(x)-tan^-1((x-1)/(x+1))=TT/4, x>-1

Thanks.

[wildareal]

Given that the domain of sin(x) and cos(x) are restricted to [-TT/2,TT/2] and [0,TT] respectively, define the implied domain and range of tan^-1(cos(x)).

Thanks.

[brightsky]

Let x = tan^(-1)(3)
3 = tan(x)
Let y = tan^(-1)(1/2)
1/2 = tan(y)
tan(x-y) = (tanx - tany)/(1 + tanx.tany)
tan(x-y) = (5/2)/(1 + 3/2) = 1
x - y = tan^(-1)(1) = pi/4 given assumed restrictions.

I'll do this the long way for the sake of clarity:
Let a = tan^(-1)x
tan(a) = x
Let b = tan^(-1)((x-1)/(x+1))
Then tan(b) = (x-1)/(x+1)
Doing what we did before:
tan(a-b) = (tan(a) - tan(b))/(1 + tan(a) tan(b))
= (x - (x-1)/(x+1))/(1+ x(x-1)/(x+1))
With a bit of algebra you get:
= 1
Which means:
a-b = tan^(-1)(1) = pi/4

[brightsky]

Let y = tan^(-1)(cos(x))
tan(y) = cos(x)
Since cos(x) is restricted [-1,1], so tan(y) is likewise restricted to [-1,1]. y is then restricted [tan^(-1)(-1), tan^(-1)(1)] = [-pi/4, pi/4].

(But not too sure on this, someone might want to verify.)

[Mao]

Let y = tan^(-1)(cos(x))
tan(y) = cos(x)
Since cos(x) is restricted [-1,1], so tan(y) is likewise restricted to [-1,1]. y is then restricted [tan^(-1)(-1), tan^(-1)(1)] = [-pi/4, pi/4].

(But not too sure on this, someone might want to verify.)

Half way there. I'm here going to use the definition that , in which case has a domain of (intersection of cos and sin) and a range of . This implies has a maximal domain of and a range of .

Since the argument of is with a domain of , we need to restrict this domain such that its result fits in the domain of . This requires which subsequently implies . With this domain, we obtain the range of

Domain:
Range:

[wildareal]

Here's one vectors question:

Thanks

[wildareal]

Here's another vectors question:

Thanks

[pHysiX]

Your first question/post (@12:27:10 AM):

Given:







Question a)

i:




ii:





Question b)
Equating and components of both equations:

Equation 1
Equation 2

-1 x Equation 2 =

By substitution into Equation 1:





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Your second question/post (@12:27:30 AM):

Given:




============================

Question a)
i:

Given that Q is the midpoint of AB






--------------------------------------------
ii:

Given





--------------------------------------------
iii:





--------------------------------------------
iv: \overrightarrow{\text{RP}}





--------------------------------------------
=============================

Question b)

This is in essence going to be a collinearity proof:
Two parallel vectors with a common point.







Thus, is parallel to with a common point A.

Therefore, it is shown that points A, R & P are collinear; i.e. R lies on the line AP.

Ratio of AR:RP


=============================

Question c)

Given that

Find such that








For




&
*We just equated the component and the component since we said the two vectors were equal.



You can check by substituting back into vector and find that it is a multiple of ; meaning that it is parallel as required.

Enjoy
*Please correct me immediately if I am wrong for anything :uglystupid2: sorry i cant do under-tildes for vectors*

[wildareal]

Here's an interesting one:

Using exactly four 4's and as many +,-,divide, sqr root signs, make the numbers from -1 to -10.

[luken93]

Can we use multiplication?














TBC

[iNerd]

Using exactly three even numbers in any shape or form make an odd number.



ANSWER:
2 and 2/2 = 2 + 1 = 3
Cant' do Latex so its a mixed fraction 2 and 2/2.