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A-LEVEL MATHS HELP! Asap

Hello, I was just wondering how on earth I am getting extra solutions when there is only 2 because I though you had to look the graph in order to fund other values.
(edited 12 months ago)
Reply 1
"Squaring both sides" creates extraneous solutions.
Reply 2
Original post by skyeforster15
Hello, I was just wondering how on earth I am getting extra solutions when there is only 2 because I though you had to look the graph in order to fund other values.

you might find harmonic form a "safer" way to approach this question :smile:
Original post by tonyiptony
"Squaring both sides" creates extraneous solutions.

How are you meant to do the question btw?
Reply 4
Original post by toxicgamage56
How are you meant to do the question btw?


not clear from the OP's post, but my suggestion in post #3 would be one way to go...
Reply 5
Original post by toxicgamage56
How are you meant to do the question btw?


As davros suggests, transforming the equation into a harmonic form, i.e. Rcos(θϕ)=constantR\cos(\theta - \phi) = \text{constant} with R and phi are constants to be found, would be the safe way*.
If you choose to square the equation like skye does, just make sure to check your answers at the end.

*How? Well if you expand the harmonic form and compare coefficients...
Also sidenote, I use cosine mostly out of habit. Some uses sine instead, and that works too. After all, cosine is just sine with a phase shift (i.e. the value of phi is different).
(edited 12 months ago)
Original post by tonyiptony
As davros suggests, transforming the equation into a harmonic form, i.e. Rcos(θϕ)=constantR\cos(\theta - \phi) = \text{constant} with R and phi are constants to be found, would be the safe way*.
If you choose to square the equation like skye does, just make sure to check your answers at the end.

*How? Well if you expand the harmonic form and compare coefficients...

Ah ok I'll try this rn and tell you how it goes, thanks.
Original post by tonyiptony
As davros suggests, transforming the equation into a harmonic form, i.e. Rcos(θϕ)=constantR\cos(\theta - \phi) = \text{constant} with R and phi are constants to be found, would be the safe way*.
If you choose to square the equation like skye does, just make sure to check your answers at the end.

*How? Well if you expand the harmonic form and compare coefficients...
Also sidenote, I use cosine mostly out of habit. Some uses sine instead, and that works too. After all, cosine is just sine with a phase shift (i.e. the value of phi is different).

Yep, it worked, thanks. I don't know if I would've thought to have used harmonic form in the exam though lol.
Original post by davros
not clear from the OP's post, but my suggestion in post #3 would be one way to go...

Yes, this is probably the way you're meant to do it, as I got the correct answer.
Reply 9
Original post by davros
you might find harmonic form a "safer" way to approach this question :smile:


Oh okayy thank you I will try now!

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