ssNake's Methods 3/4 Q's.

[REBORN]

3/4 time.

Q - How come  [ Cos(2x) ] ^2 + [ Sin(2x) ] ^2 is equivalent to 1?

[LOLs99]

Sorry if I'm not answering your question.
But isn't that the formula for the circle thingy?

[Panicmode]

Ok, the trig identity states that cos^2(x) + sin^2(x) = 1 (I'll go into the proof of this later)

let x = 2u

cos^2(2u) + sin^2(2u) = 1

let x=y/2

cos^2(y/2) + sin^2(y/2) = 1

You getting my drift? As the unknowns are arbitrary, it doesn't matter what they are as long as they are the same.


EDIT:

Proof of Trig identity:

Take a right angled triangle with angle x. Pythagoras' theorem states that a^2 + b^2 = c^2 (side lengths)

cos(x) = adjacent / hypotenuse = a/c
sin(x) = opposite / hypotenuse = b/c

[cos(x)]^2 + [sin(x)]^2 = (a/c)^2 + (b/c)^2= a^2/c^2 + b^2/c^2 = (a^2 + b^2) / c^2 [by factorisation]

since a^2 +b^2 = c^2

[cos(x)]^2 + [sin(x)]^2 = c^2/c^2 =1

Q.E.D.

[REBORN]

Yep I knew the trig identity although I always thought it was for 'x' only.

Now I understand. Thanks Panic

[abd123]

i assume it might be of the general rule of .
if both sine and cosine have cycles that still goes more than 2 or 3 or more cycles periodically,
so if  both have the exponents of 2, not matter if there is a cycle that is over 2 oscillations then
it will always end being equated to 1.

[REBORN]

Q - Differentiate 2 times Cos(3-2x)

I know cos(g(x)) goes to - diff g(x) x sin (g(x))

but it isn't working :S

[pi]

Q - Differentiate 2 times Cos(3-2x)

I know cos(g(x)) goes to - diff g(x) x sin (g(x))

but it isn't working :S

I remember d/dx(f(g(x))) = g'(x)f'(g(x))

Yeh, so:
g'(x) = -2, f'(x) = -2sin(x), g(x) = 3-2x
therefore using above formula: 4sin(3-2x)


edit: assuming '2 times' means 2cos(...) not to find the second derivative

I know cos(g(x)) goes to - diff g(x) x sin (g(x))

but it isn't working :S

Actually, g'(x)(-sin(g(x)))

derivative of cos(x) is -sin(x)

[REBORN]

yeah it's right. never seen that formula. oh well, time to memorise it -.-

[pi]

yeah it's right. never seen that formula. oh well, time to memorise it -.-

It's just the chain rule written nicer, your way was fine, it's just that you forgot the negative when diffing cos

[REBORN]

I'll just use the nice way.

[REBORN]

For f: [-1,inf) --> R, f(x) = x^2 + 2x find its inverse stating dom and range?

My working.

y = x^2+2x
x = y^2+2y

and now I'm stuck.

[brightsky]

y^2 + 2y - x = 0
y = (-2 +- sqrt(4 +4x))/2
= (-2 +- 2sqrt(1 + x))/2
= -1 +- sqrt(1+x)
since dom f = [-1,inf), ran f = [-1, inf)
hence dom f^(-1) = [-1, inf), ran f^(-1) = [-1,inf)
so we only want -1 + sqrt(1+x)
so inverse function is f^(-1)(x) = sqrt(1+x) - 1, x E [-1,inf)

[REBORN]

Q - Solve, algebraically, [ (1-|1-2x|) over 5] greater than 1/2

1-2x is in abs value lines.

[brightsky]

assuming you mean (1-|1-2x|)/5 > 1/2
1-|1-2x|>5/2
|1-2x|<-3/2
hence no solutions

[pi]

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