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C3 trig/differentiation help

image.jpgHey getting some Easter revision done

I have no idea how to do the last part of this Q and mark scheme alone isn't very revelatory to me

Any guidance would be greatly appreciated, thx
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Avatar for Zacken
Zacken
22
Original post by hi-zen-berg
image.jpgHey getting some Easter revision done

I have no idea how to do the last part of this Q and mark scheme alone isn't very revelatory to me

Any guidance would be greatly appreciated, thx


dydx=2cos⁡y\frac{dy}{dx} = \frac{2}{\cos y} is excellent, now just note that y=sin⁡−12x⇒sin⁡y=2x⇒sin⁡2y=4x2y = \sin^{-1}2x \Rightarrow \sin y = 2x \Rightarrow \sin^2 y = 4x^2

So cos⁡y=1−sin⁡2y=1−4x2\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - 4x^2} and you're done.
Original post by Zacken
dydx=2cos⁡y\frac{dy}{dx} = \frac{2}{\cos y} is excellent, now just note that y=sin⁡−12x⇒sin⁡y=2x⇒sin⁡2y=4x2y = \sin^{-1}2x \Rightarrow \sin y = 2x \Rightarrow \sin^2 y = 4x^2

So cos⁡y=1−sin⁡2y=1−4x2\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - 4x^2} and you're done.


Thanks a lot again bro!
Reply 3
Avatar for Zacken
Zacken
22
Original post by hi-zen-berg
Thanks a lot again bro!


No problem!

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