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    Hi,

    could someone please give me a detailed description and working out for this question as I dont seem to understand how you do them....Iv got the basics of expanding the formulas etc but im really quite confused...!!


    (i)Express 4 cosθ − sinθ in the form Rcos(θ + α ), where R > 0 and 0◦ < α < 90◦. [3]

    (ii) Hence solve the equation 4 cosθ − sinθ = 2, giving all solutions for which −180◦ < θ < 180◦.
    [5]




    WILL REP THE MOST CONCISE ANSWER!!!
    thank you
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    ok

    i) R is the square root of 4 squared + 2 squared
    To find the angle you do cos/sin as in 4/1 then do inverse tan of this answer

    ii) you use the previuos answer to work out the angles
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    First expand the RHS:
    4cosx-sinx=Rcosxcosa - Rsinxsina

    [Mod edit: Please don't post full solutions. Read the posting guide sticky at the top of the thread list.]

    (ii)use your Rcos(x+a) formula from (i) and solve!
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    I didn't help which is a good reason to rep me.
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    (Original post by SK-mar)
    Hi,

    could someone please give me a detailed description and working out for this question as I dont seem to understand how you do them....Iv got the basics of expanding the formulas etc but im really quite confused...!!


    (i)Express 4 cosθ − sinθ in the form Rcos(θ + α ), where R > 0 and 0◦ < α < 90◦. [3]

    (ii) Hence solve the equation 4 cosθ − sinθ = 2, giving all solutions for which −180◦ < θ < 180◦.
    [5]




    WILL REP THE MOST CONCISE ANSWER!!!
    thank you
    i) Use the cos(A+B) rule and expand. Remember to write 'R' in front. A = θ and B = α

    Next step. You must group the terms together where you will find Rcosα and Rcosθ and that will be equal to a number.

    You know that Rcosα = x and Rcosθ = y (whatever number you get, in this case 4 & 1).

    Then, you draw a right angled triangle, fill in the lengths, O and A and work out R by using pythag. Work out what tana = O/A and do inverse tan to find out what a is. Now you have R and a. Re-write.

    ii) From part i, you have formed an equation in the form of Rcos(θ + α ) = 2 (2's given in question)

    Divide by the value of R by bringing it across to the RHS, so you get: 2/R

    Solve like normal trig.

    Hope it helps.
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    (Original post by *MJ*)
    It works so why do I care.

    You guys go back and do sina / cosa to obtain tana and then do tan^-1 right?

    I find it's easier, just stick the R back in and obtain the angle.

    :yeah:
    yeah. But, it's funner to make it tan! Honest.
    I don't like dealing with R, since it's usually a surd, and I hate surds!
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    (Original post by SK-mar)
    sweet thanks......just wandering tho, strictly speaking, in case i get another type, should it be Rsina = -1? as its 4cosa - Sina.....

    and could u actually do part ii.) for me so I fully understand, i will then rep you!!! thankssss
    it's not -1 because it's also -Rsinxsina on the RHS. If that made any sense.
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    (Original post by *MJ*)
    It works so why do I care.

    You guys go back and do sina / cosa to obtain tana and then do tan^-1 right?

    I find it's easier, just stick the R back in and obtain the angle.

    :yeah:
    I draw a right angled triangle. Label cos and sin, work out R using pythag. Then work out tana by O/A. :yeah:
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    (Original post by unamed)
    yeah. But, it's funner to make it tan! Honest.
    I don't like dealing with R, since it's usually a surd, and I hate surds!
    True mathematicians love surds. :love:

    I even eat them for breakfast. :mmm:
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    (Original post by *MJ*)
    True mathematicians love surds. :love:

    I even eat them for breakfast. :mmm:
    I'm not a true mathematician, then. I'm a physicist - and we like our calculus.
    We do it in our sleep.

    [by the way, i swear you were a girl the other day! :eek:]
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    (Original post by *MJ*)
    True mathematicians love surds. :love:

    I even eat them for breakfast. :mmm:
    right that was verrrrry heplful I completely get it now thankyou so much.....not just you but everyone who posted....Im gonna rep MJ but wish I could rep you all, sorry to those who I havnt!! but thanks anyway!!! ......... btw in these question will it most likely always be in the form
    show that xcosa - xsina = Rcos(x-a) ....basicaly do they only use COS and SINE in C3, on the left hand side??
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    (Original post by SK-mar)
    right that was verrrrry heplful I completely get it now thankyou so much.....not just you but everyone who posted....Im gonna rep MJ but wish I could rep you all, sorry to those who I havnt!! but thanks anyway!!! ......... btw in these question will it most likely always be in the form
    show that xcosa - xsina = Rcos(x-a) ....basicaly do they only use COS and SINE in C3 on the left hand side??
    Thanks and yes.

    You may get 2sinx + 5cosx for example.

    Solving 2secx = 3 is a totally different question.

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    Just a reminder people, posting full solutions here is not permitted. Please see the posting guide at the top of the thread list.
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    (Original post by gottastudy)
    I draw a right angled triangle. Label cos and sin, work out R using pythag. Then work out tana by O/A. :yeah:
    That is one long method, it is just an excuse for you to draw triangles??
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    (Original post by Illusionary)
    Just a reminder people, posting full solutions here is not permitted. Please see the posting guide at the top of the thread list.
    I actually never knew this! :eek: I shall obey the rule from now on.
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    (Original post by crazedmonkey09)
    That is one long method, it is just an excuse for you to draw triangles??
    :awesome: Yes. I have finally mastered drawing perfect right angled triangles. :yep:

    Spoiler:
    Show
    Actually, I just learnt the method in the book. :p:
 
 
 
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