Cosine Rule Problem: I Can't Do a(ii) - Get Help Here

AI Thread Summary
To solve part a(ii) of the cosine rule problem, it's essential to express angles θ and φ as inverse cosines of functions of x, ensuring to use radians. The initial approach of taking the cosine of both sides was incorrect; the cosine of a sum does not equal the sum of cosines. Instead, apply the angle difference formula correctly, noting that π - ∠ADB equals ∠ADC. Revisiting the steps and correcting the application of the cosine rule will lead to a clearer solution. Properly utilizing the cosine formulas is crucial for resolving the problem effectively.
CAH
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See the photo attachments of question and marking scheme and my attempt at a solution :)

I've done a(i) but I can't do a(ii).

Thanks
 

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For a(ii) you have, basically ##\theta+\phi=\pi## (always use radians) ... with theta and phi expressed as inverse cosines of functions of x.
Your next step was to take the cosine of both sides - but you did this step incorrectly:
note: ##\cos(A+B)\neq \cos A + \cos B##
 
Still don't see what to do next
 
CAH said:
Still don't see what to do next
You could go back and fix things up as Simon suggested.

I would suggest that there was little point in taking the arccos when you did.

Notice that:

π - ∠ADB = ∠ADC

Take the cosine of both sides of this. As Simon says, use a correct form of angle difference formula for cosine.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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