# C3 exam question

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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

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

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#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.

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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#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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#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

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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#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

3cos(x-60) = 3 (cosxcos60+sinxsin60)

2cos(x + 60) = 2 (cosxcos60-sinxsin60)

3cos(x - 60) + 2cos(x + 60) = 3 (cosxcos60+sinxsin60) + 2 (cosxcos60-sinxsin60)

3cos(x - 60) + 2cos(x + 60) = 3cosxcos60+3sinxsin60+2cosxcos60-2sinxsin60

3cos(x - 60) + 2cos(x + 60) = 5cosxcos60+sinxsin60

3cos(x - 60) + 2cos(x + 60) = 2.5cosx+(squareroot 3/2)sinx

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

Rsin(x+a)=R(sinx cosa + cosx sina)

Rsin(x+a)=R sinx cosa + R sina cosx

R sin a = 2.5, R cos a = (squareroot 3/2)

tan a = 5/3

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

Sin (x+59.04) = -1/squareroot 7

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.

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)

**Adding the equations**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

**Using your answer for T(x) from part 2**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.

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#8

(Original post by

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

3cos(x-60) = 3 (cosxcos60+sinxsin60)

2cos(x + 60) = 2 (cosxcos60-sinxsin60)

3cos(x - 60) + 2cos(x + 60) = 3 (cosxcos60+sinxsin60) + 2 (cosxcos60-sinxsin60)

3cos(x - 60) + 2cos(x + 60) = 3cosxcos60+3sinxsin60+2cosxcos60-2sinxsin60

3cos(x - 60) + 2cos(x + 60) = 5cosxcos60+sinxsin60

3cos(x - 60) + 2cos(x + 60) = 2.5cosx+(squareroot 3/2)sinx

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

Rsin(x+a)=R(sinx cosa + cosx sina)

Rsin(x+a)=R sinx cosa + R sina cosx

R sin a = 2.5, R cos a = (squareroot 3/2)

tan a = 5/3

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

Sin (x+59.04) = -1/squareroot 7

**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)

**Adding the equations**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

**Using your answer for T(x) from part 2**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**
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**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)

**Adding the equations**

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

**Using your answer for T(x) from part 2**

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**

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#11

(Original post by

Yeah i am sorry. Appologies

**L.McEnaney**)Yeah i am sorry. Appologies

I wasn't having a go, just letting you know

Welcome to TSR.

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#12

(Original post by

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)

**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)

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

does this help?

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