The Student Room Group
Reply 1
sin^2 x+cos^2 x=1
Reply 2
Are you sure it's January 2006? Cannot find this question.

EDIT:

Found it :smile:
Reply 3
goldsilvy
Are you sure it's January 2006? Cannot find this question.

EDIT:

Found it :smile:


What question. Can't find it.
Reply 4
JTeighty
What question. Can't find it.

It's on the MS. Question 4b.
Reply 5
goldsilvy
It's on the MS. Question 4b.


What a stupid way to answer the question.

Just work out dx/dy and then take the inverse to make it dy/dx
Reply 6
JTeighty
What a stupid way to answer the question.

Just work out dx/dy and then take the inverse to make it dy/dx

That's what they did in the MS. Still there is this one final step,

For those interested,



Reply 7
x = 4 sin(2y + 6) -> x^2 = 16 sin^2(2y+6) -> (1-x^2)^0.5 = 4cos(2y + 6)

you have a 1/8cos(2y+6) so be careful with the constant multiplier on the bottom.
Reply 8
goldsilvy
That's what they did in the MS. Still there is this one final step,

For those interested,





Sorry didn't realise there was another stage.
Reply 9
Am really close. If you square the whole fraction you'll get 1/(64cos^2(arcsinx/4))
use sin^2+cos^2=1
so in the denominator u get 64-sin^2(arcsinx/4) which cancels to 64-x^2/16
Now root it.

Now I'm stuck but I think it's practically done
Reply 10
JTeighty
Am really close. If you square the whole fraction you'll get 1/(64cos^2(arcsinx/4))
use sin^2+cos^2=1
so in the denominator u get 64(1-sin^2(arcsinx/4) which cancels to 64(1-x^2/16)=64/16(16-x^2)=4(16-x^2)
Now root it.

Now I'm stuck but I think it's practically done

fixed
Reply 11
technically, since the answer's in brackets, it's an extra. You won't get marked down for not doing that, the question's not explicit enough.
Reply 12
mathz
fixed


Of course. What a silly mistake.

Thanks for correcting me :-)
how does the (2y+6) translate to arcsin(x/4)
someone pls help ... if so rep+
Because x=4sin(2y+6), you can undo this by dividing by 4, and undoing the sin, so that arcsin(x/4)=2y+6. After differentiation, you're asked to put dy/dx in terms of x, which means you can just substitute in the known value of 2y+6 into it's place in the equation.

Voila.
of course ... duh ... heres you rep
Reply 17
x=4sin(2y+6)x= 4sin(2y+6)

y=12sin1(x4)3y=\frac{1}{2}sin^{-1}(\frac{x}{4})-3 By making y the subject.
dydx=±181x216\frac{dy}{dx}=±\frac{1}{8\sqrt{1-\frac{x^{2}}{16}}}
dydx=±1216x2\frac{dy}{dx}=±\frac{1}{2\sqrt{16-x^{2}}}
dude i'd give you a rep but just used ... 2moro thank btw
morebritishthanindian
how does the (2y+6) translate to arcsin(x/4)


if you use the orginal equation: x =4sin(2y+6)

x/4 = sin(2y+6)

arcsin(x/4) = 2y+6 then sub it back in to dy/dx