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
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.