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[Group5][Trig][HL]Exam Preparation Question


tutorinseoul

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I find part 2 ridiculously difficult. I gave up, but I think the logic is something like have a quadratic in sin x or cos x, then find the other root using Viete's theorem. I looked up the answer in a solver and it's not a familiar angle like pi/4 or pi/3. Thus it's also very difficult to guess and check. I graduated already but it is interesting to see if anyone in the program can figure this out!

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21 minutes ago, kw0573 said:

I find part 2 ridiculously difficult. I gave up, but I think the logic is something like have a quadratic in sin x or cos x, then find the other root using Viete's theorem. I looked up the answer in a solver and it's not a familiar angle like pi/4 or pi/3. Thus it's also very difficult to guess and check. I graduated already but it is interesting to see if anyone in the program can figure this out!

Hey thanks for the reply :)!. It gets simple if you use factor formula from compound angle. Beauty of compound angle is that you can make whatever the sum of two trig equations become a product,which is nice to solve due to null factor theorem. Give it a try again :)!

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9 minutes ago, SC2Player said:

My answer's in the attached file.  Part 2 was pretty interesting; forgot about expressing something in the form rsin(x+a) until I stumbled upon it on a web page.  Pretty easy tbh once you realized it - the trick is remembering this technique.  

Other answer.pdf

Very well done :)!! With this technique, we can pretty much calculate EVERY trig equation; when they come in product of two different trig, we can use null factor law to work out the x. Even if we get an equation of sum of trig, we can combine them into product of trig functions to apply null factor law again. Good job :)!

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