C3 exam question

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#1
Can do similar questions but this one has stumped me

QU: The expression T(x) is defined for x in degrees by,
T(x) = 3cos(x - 60) + 2cos(x + 60)

(i) Express T(x) in the form Asinx + Bcosx, giving the exact values of the constants A and B.

(ii) Hence Express T(x) in the form Rsin(x + a), where R>0 and 0<a<90

(iii) Find the smallest positive value of x such that T(x) + 1 = 0
0
11 years ago
#2
Which bit are you stuck with?

i) ?
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#3
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11 years ago
#4
i. Am I mistaken in thinking you just use the addition formula?
I.e. 3cos(x - 60) + 2cos(x + 60) = 3[cosxcos60* + sinxsin60*]... you can do the rest.
ii. Then once you have that value for i, just change it into the form Rsin(x + a) given. You know how to do that right?
iii. You'll know how to do this after you get ii.
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11 years ago
#5
i. 3cos(x - 60) + 2cos(x + 60) = 3(cos x cos 60 + sin x sin 60) + 2(cos x cos 60 - sin x sin 60) = 5 cos 60 cos x + sin 60 sin x = 5/2 cos x + 3^0.5 /2 sin x
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11 years ago
#6
a) Use the exapansion rules, and you end up with 3(cosxcos60 + sinxsin60) + 2(cosxcos60-sinxsin60),
which simplifies to : 5 cosx cos 60 + sinxsin60.

Which simpliefies to give B as 2.5 and A as ROOT 3 / 2

ii ) Expand to Rsinxcosa + Rsinacos x. Look at the above and simply equate co-efficients.
R sin a = 2.5 and R cos a = *ROOT* 3 /2.

squaring and adding all terms leaves to R = *ROOT* 7
tan a = 5/3 a = 59.04 ( 2dp).

Solve :
iii) Sin ( x + 59.04) = -1/*ROOT* 7.
I got x = 202.1
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11 years ago
#7
QU: The expression T(x) is defined for x in degrees by,
T(x) = 3cos(x - 60) + 2cos(x + 60)

(i) Express T(x) in the form Asinx + Bcosx, giving the exact values of the constants A and B.

Using; cos(A+/-B)= cosAcosB-/+sinAsinB
(Given on the formula sheet)
3cos(x-60) = 3 (cosxcos60+sinxsin60)
Seperating the equations
2cos(x + 60) = 2 (cosxcos60-sinxsin60)
3cos(x - 60) + 2cos(x + 60) = 3 (cosxcos60+sinxsin60) + 2 (cosxcos60-sinxsin60)
Expanding the brackets
3cos(x - 60) + 2cos(x + 60) = 3cosxcos60+3sinxsin60+2cosxcos60-2sinxsin60
Simplifying
3cos(x - 60) + 2cos(x + 60) = 5cosxcos60+sinxsin60
Converting cos60 and sin60 to numerical form
3cos(x - 60) + 2cos(x + 60) = 2.5cosx+(squareroot 3/2)sinx

Therefore A=(squareroot 3/2) B=2.5

(ii) Hence Express T(x) in the form Rsin(x + a), where R>0 and 0<a<90
Multiplying each side by R
Rsin(x+a)=R(sinx cosa + cosx sina)
Expanding out the R.H.S
Rsin(x+a)=R sinx cosa + R sina cosx
Equating co-efficients
R sin a = 2.5, R cos a = (squareroot 3/2)
Square and add up all terms
R*2=7, therefore R= squareroot 7
tan a = 5/3
In the exam inverse function may be refferd to as arctan
a= Inverse 5/3 = 59.04

(iii) Find the smallest positive value of x such that T(x) + 1 = 0

squareroot 7 sin(x + 59.04)+1=0
Subtracting one and dividing through by squareroot 7
Sin (x+59.04) = -1/squareroot 7

x=202.1

EDIT: Sorry if I made a mistake i was getting tired and bored of trig tonight lmao... Forfuture referance (im assuming your Edexcel board you can search the exam question on youtube for a detailed step by step walkthrough) I.e searching Youtube for "Edexcel C3 June 2009" will display results from a maths teachers channel with tutorials for each and every question on the exam. I belive this dates back a few years and covers every topic C1-4. However I dont think he covers additional modules or further pure.
0
11 years ago
#8
(Original post by L.McEnaney)
QU: The expression T(x) is defined for x in degrees by,
T(x) = 3cos(x - 60) + 2cos(x + 60)

(i) Express T(x) in the form Asinx + Bcosx, giving the exact values of the constants A and B.

Using; cos(A+/-B)= cosAcosB-/+sinAsinB
(Given on the formula sheet)
3cos(x-60) = 3 (cosxcos60+sinxsin60)
Seperating the equations
2cos(x + 60) = 2 (cosxcos60-sinxsin60)
3cos(x - 60) + 2cos(x + 60) = 3 (cosxcos60+sinxsin60) + 2 (cosxcos60-sinxsin60)
Expanding the brackets
3cos(x - 60) + 2cos(x + 60) = 3cosxcos60+3sinxsin60+2cosxcos60-2sinxsin60
Simplifying
3cos(x - 60) + 2cos(x + 60) = 5cosxcos60+sinxsin60
Converting cos60 and sin60 to numerical form
3cos(x - 60) + 2cos(x + 60) = 2.5cosx+(squareroot 3/2)sinx

Therefore A=(squareroot 3/2) B=2.5

(ii) Hence Express T(x) in the form Rsin(x + a), where R>0 and 0<a<90
Multiplying each side by R
Rsin(x+a)=R(sinx cosa + cosx sina)
Expanding out the R.H.S
Rsin(x+a)=R sinx cosa + R sina cosx
Equating co-efficients
R sin a = 2.5, R cos a = (squareroot 3/2)
Square and add up all terms
R*2=7, therefore R= squareroot 7
tan a = 5/3
In the exam inverse function may be refferd to as arctan
a= Inverse 5/3 = 59.04

(iii) Find the smallest positive value of x such that T(x) + 1 = 0

squareroot 7 sin(x + 59.04)+1=0
Subtracting one and dividing through by squareroot 7
Sin (x+59.04) = -1/squareroot 7

x=202.1
I'm guessing you're new to TSR so I might as well let you know that the maths forum is for nudging only, not full solutions. So don't post full solutions in the future, thanks.
0
11 years ago
#9
Yeah i am sorry. Appologies
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#10
(Original post by L.McEnaney)
QU: The expression T(x) is defined for x in degrees by,
T(x) = 3cos(x - 60) + 2cos(x + 60)

(i) Express T(x) in the form Asinx + Bcosx, giving the exact values of the constants A and B.

Using; cos(A+/-B)= cosAcosB-/+sinAsinB
(Given on the formula sheet)
3cos(x-60) = 3 (cosxcos60+sinxsin60)
Seperating the equations
2cos(x + 60) = 2 (cosxcos60-sinxsin60)
3cos(x - 60) + 2cos(x + 60) = 3 (cosxcos60+sinxsin60) + 2 (cosxcos60-sinxsin60)
Expanding the brackets
3cos(x - 60) + 2cos(x + 60) = 3cosxcos60+3sinxsin60+2cosxcos60-2sinxsin60
Simplifying
3cos(x - 60) + 2cos(x + 60) = 5cosxcos60+sinxsin60
Converting cos60 and sin60 to numerical form
3cos(x - 60) + 2cos(x + 60) = 2.5cosx+(squareroot 3/2)sinx

Therefore A=(squareroot 3/2) B=2.5

(ii) Hence Express T(x) in the form Rsin(x + a), where R>0 and 0<a<90
Multiplying each side by R
Rsin(x+a)=R(sinx cosa + cosx sina)
Expanding out the R.H.S
Rsin(x+a)=R sinx cosa + R sina cosx
Equating co-efficients
R sin a = 2.5, R cos a = (squareroot 3/2)
Square and add up all terms
R*2=7, therefore R= squareroot 7
tan a = 5/3
In the exam inverse function may be refferd to as arctan
a= Inverse 5/3 = 59.04

(iii) Find the smallest positive value of x such that T(x) + 1 = 0

squareroot 7 sin(x + 59.04)+1=0
Subtracting one and dividing through by squareroot 7
Sin (x+59.04) = -1/squareroot 7

x=202.1
Hey, thanks for all your help I really appreciate it. In the mark scheme however for part (ii) they got 70.9, not 59.04 so I dont know whether you know how they've got this value, the exam paper is OCR June 2008 question 8. And of course this has affected your solutions for part (iii)
0
11 years ago
#11
(Original post by L.McEnaney)
Yeah i am sorry. Appologies
No probs buddy.
I wasn't having a go, just letting you know
Welcome to TSR.
0
11 years ago
#12
(Original post by anonR9T2J)
Hey, thanks for all your help I really appreciate it. In the mark scheme however for part (ii) they got 70.9, not 59.04 so I dont know whether you know how they've got this value, the exam paper is OCR June 2008 question 8. And of course this has affected your solutions for part (iii)
t(x) = 5/2 cos x + root 3/2 sinx = root 7 (sin a cos x + cos a sin x)

sin a = 5/(2 root7) => a = 70.89 (2dp)

does this help?
0
2 months ago
#13
3cos (x-60) 2.
0
2 months ago
#14
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