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Core 3 Maths Question

Hi I'm currently doing a question in the Jan 07 Core 3 OCR, and I'm kind of a bit confused about a question.
It goes:
Express 4cos? – sin? in the form Rcos(?+a), where R is greater than 0 and a is between 0 and 90 degrees.

For me its a very simple q- for R i get sq root 17, but for ? I get a different answer to the mark scheme, and I want to know why.

Usually I do cos-^1 or sin-^1 of the other number divided by R.
Ie Cos-^1 (1/sq.r.17) or sin^-1 (4/sq.r.17)
Normally I get the right answer- but I don't- i got 76 degrees, the answer is 14 degrees- so you obviously have to do cos-^1 of (4/sq.r17)

Please can anybody tell me why? I think its because on this one its a 4c. MINUS instead of plus sin? but im not too sure.
Thankyou!

can't recognise theta so ? means theta symbol
(edited 13 years ago)
Reply 1
Avatar for Noble.
Noble.
17
f(θ)=4cosθ−sinθf(\theta) = 4cos\theta - sin\theta

Rcos(θ+α)=R(cosθcosα−sinθsinα)Rcos(\theta + \alpha) = R(cos\theta cos\alpha - sin\theta sin\alpha)

You've already worked out that

R=17R=\sqrt{17}

Now compare the two expressions.

4cosθ=17cosθcosα4cos\theta = \sqrt{17}cos\theta cos\alpha

417=cosα\frac{4}{\sqrt{17}} = cos\alpha

Now do the same for the other part of the expression, and you'll be able to find the value of α\alpha
Reply 2
Avatar for Goldfishy
Goldfishy
8
Original post by Rokaf0
Usually I do cos-^1 or sin-^1 of the other number divided by R.
Ie Cos-^1 (1/sq.r.17) or sin^-1 (4/sq.r.17)


Since this wants you to write it as Rcos(θ+a) Rcos(\theta + a) , you don't take the other 'number' (coefficient) but instead take the same one.

So it is actually Rcosa=4 Rcosa = 4 and Rsina=1 Rsina = 1 .

If the question said to rewrite it in the form Rsin(θ−a)Rsin(\theta - a) then you would take the other number.

Spoiler

Reply 3
Avatar for Rokaf0
Rokaf0
OP
Thankyou! I understand it now :smile:

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