Andiio's Questions Thread :)

[pi]

You sure this is the right image, lol?

Vitamin C must be code for something algebraically...

[brightsky]

Could anyone please provide the proof for this?

I understand the a = tan(theta 2) <-- compound angle derivation, but not sure about why m = tan (theta) and why a = theta₂ - theta₁.

Thanks!


EDIT: MY BAD LOL DIDN'T NOTICE

Look at the red line. Draw a vertical line from the red line to the x-axis. Let the point at which this vertical line touches the red line be (x,y). By trig, we know that tan(theta) = y/x = gradient m.

a + theta_1 + (180 - theta_2) = 180
a + theta_1 - theta_2 = 0
a = theta_2 - theta_1
tan(a) = tan(theta_2 - theta_1)

[Andiio]

Thanks!

How would you approach part (d) of this question?

[kamil9876]

Linear approximation ie: gradient of the line segment joining and is approximately at

[brightsky]

f'(x) = (f(x+h) - f(x))/h
hf'(x) = f(x+h) - f(x)
f(x) = f(x+h) - hf'(x)

So..let f(x) = x^(-1/2)
f'(x) = -1/2x^(-3/2) = -1/(2sqrt(x^3))

Substitute a suitable number for x and h.

[Andiio]

For a linear approximation question like this:

We want y when x = 0.9, so would we just 'approximate' and 'guess' and just let x = 1 as it is a whole number close to 0.9?

Thanks!

[vgardiy]

Yeah, i think you just use the closest number that fits into the formula well, so your x would be 1 and your h would be -0.1

[Andiio]

How should one approach this question?

[evaever]

you don,t need to use lin app
A(new)=(1.02)^2 x A(old)=1.0404 x A(old)
% change in A = 4.04%

[Andiio]

When sketching circular/trigonometric functions with transformations, is it best to apply each dilation/reflection/translation etc separately?

Also, does anyone know any 'special ways' of applying translations? (Not sure how everyone else does them.)

Thanks!

EDIT: Didn't mean the order of translations >_> I know my question is quite obscure but mm

[forumguy]

reflection first, horizontal and then vertical. That's how i do it

[Andiio]

If 2^x + 2^-x = k^2 + 2, where k E R, then 8^x + 8^-x = ?

Thanks!

[kamil9876]

Try to cube both sides.

[Andiio]

I thought of that, but was daunted by the fact that there'd be more variables and everything. I'll try and see if I can simplify it!

[brightsky]

Don't know if you've worked it out already but cubing it works because 2^x * 2^(-2x) and vice versa gives you 2^x and 2^-x respectively:

(2^x + 2^(-x))^3
= 2^3x + 3*2^x*2^(-2x) + 3*2^(2x)*2^(-x) + 2^(-3x)
= 2^3x + 2^(-3x) + 3(2^x + 2^(-x))
Substitute 2^x + 2^(-x) = k^2 + 2 in.
(k^2 + 2)^3 = 8^x + 8^(-x) + 3(k^2 + 2)
So 8^x + 8^(-x) = (k^2 + 2)^3 - 3(k^2 + 2)

Now simplify.