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[Core 3] Inverse Trig Functions


Haven't got a clue what to do for part ii. I was going to try dividing both sides by arccosx but then I realised that arcsinx/arccosx =/= arctanx.

Here's what the mark scheme says, I don't understand it though.

Spoiler

Original post by DarkEnergy

Haven't got a clue what to do for part ii. I was going to try dividing both sides by arccosx but then I realised that arcsinx/arccosx =/= arctanx.

Here's what the mark scheme says, I don't understand it though.

Spoiler




does sin(90-x)=cosx help at all?
Original post by DarkEnergy

Haven't got a clue what to do for part ii. I was going to try dividing both sides by arccosx but then I realised that arcsinx/arccosx =/= arctanx.

Here's what the mark scheme says, I don't understand it though.

Spoiler




You can use the fact that arcsin(x) = pi/2 - arccos(x)
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Avatar for Notnek
Notnek
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Original post by DarkEnergy

Haven't got a clue what to do for part ii. I was going to try dividing both sides by arccosx but then I realised that arcsinx/arccosx =/= arctanx.

Here's what the mark scheme says, I don't understand it though.

Spoiler



A different method:

Since they're equal, you can let arcsin⁡x\arcsin x and arccos⁡x\arccos x both equal yy.

arcsin⁡x=y⇒x=siny\arcsin x = y \Rightarrow x = sin y

arccos⁡x=y⇒x=cos⁡y\arccos x = y \Rightarrow x = \cos y

Then use cos⁡2y+sin⁡2y≡1\cos^2 y + \sin^2 y \equiv 1 and you'll end up with an equation in xx.

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