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# C3 watch

1. Hi need help with this c3 question:

Given that f(x)=5sinx+12cosx

find f(x) in the form of Rsin(x+alpha) where R and alpha are positive constants to be found. (0<alpha<90degrees)

I found R which is equal to 13, but how do i find alpha?

and how do i find the minimum value of f(x)+4??? thank you
2. I'm actually not sure why but this works for me (this is a bit silly to not understand, but we've got exams to pass...!). If you're converting to a sin graph I do and if converting to a cos graph .

Consider the minimum value of a sin graph, for the last part...
3. (Original post by QED)
I'm actually not sure why but this works for me (this is a bit silly to not understand, but we've got exams to pass...!). If you're converting to a sin graph I do and if converting to a cos graph .

Consider the minimum value of a sin graph, for the last part...
haha thats true, anwyways many thanks it still doesn't make sense to my because how can cos/sin=tan , but thanks anyways REP+ remind me to give u one tomorow ok I already gave one

edit: in the markin scheme for this question it gives tan(alpha)=12/5 >> alpha=67.38 degrees

i didn't get this i got tan(alpha)=5/12 it's so confusin isn't tan=Sin/cos??
4. Well Rsin(x+alpha)= R(sinx. cosalpha+cosx. sinalpha), so equate coeffiecients, sinx: 5=Rcosalpha and cosx: 12=Rsinalpha. divide so that Rsinalpha/Rcosalpha= 5/12, therefore tanalpha= 5/12 then just get rid of the tan to find alpha. sorry but not sure how to find the minimum value
5. (Original post by Remarqable M)
Hi need help with this c3 question:

Given that f(x)=5sinx+12cosx

find f(x) in the form of Rsin(x+alpha) where R and alpha are positive constants to be found. (0<alpha<90degrees)

I found R which is equal to 13, but how do i find alpha?

and how do i find the minimum value of f(x)+4??? thank you
You find alpha by drawing a right angled triangle. If you know what O and A are, and you know tan = O/A sub in the values of O and A.

Then, tan a (alpha) = O/A

Do inverse tan to find what alpha is.

6. You use the special angle formula:

nd you take either the cosx or sinx part of the equation:

Divide by cosx or sinx (which ever you've chosen beforehand)

and solve for alpha

Hope it helped.
7. (Original post by dinkyness_blob)
Well Rsin(x+alpha)= R(sinx. cosalpha+cosx. sinalpha), so equate coeffiecients, sinx: 5=Rcosalpha and cosx: 12=Rsinalpha. divide so that Rsinalpha/Rcosalpha= 5/12, therefore tanalpha= 5/12 then just get rid of the tan to find alpha. sorry but not sure how to find the minimum value
I did that, but in the mark scheme it says tan(alpha)=12/5 not 5/12 and this is confusing because how can tan=cos/sin??
8. QED is right (I think ) about finding the value of alpha using tan, if you want to understand how it works, using the sine addition formula:

and compare coefficients of both:

and

You should then see how you can eliminate R, and find in one easy step

EDIT: Too slow, damn me and my slow latex hands!
9. (Original post by maxfire)

You use the special angle formula:

nd you take either the cosx or sinx part of the equation:

Divide by cosx or sinx (which ever you've chosen beforehand)

and solve for alpha

Hope it helped.
aah i see thanks for the help remind me after the exams to rep+

edit: actually you know what remind me tomorrow to rep+++ you lol your explanation is wicked
10. Thanks to everyone for your input, but can someone answer the last question?? find the minimum?
11. The minimum is always the R value.
12. Max or min is found by setting differential of equation to zero, i.e. solve

or solve in the other form

Either should work... Then sub in value of x you get from this into the original equation - I get 17 as answer!

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