Circular Functions Questions

[Wingtips]

1. For 3cos(x) + 4sin(x)

a. Express in the form Rcos(x-a), where R>0 and a is acute.

b.Hence, find the maximum value of 3cos(x) + 4 sin(x)

c. Find also, the smallest positive value of x for which 3cos(x) + 4sin(x) takes its minimum value.

2. a Given that sin(x+a)=b*sin(x-a), show that tan(x) = [(b+1)tan(a)]/(b-1)

Urgent help needed. Thanks in advance
                                                                                     

[fred42]

3cos(x)+4sin(x)=Rcos(x-a)
                         =R[cos(x)cos(a)+sin(x)sin(a)]
3cos(x)=Rcos(x)cos(a)               and      4sin(x)=Rsin(x)sin(a)
3=Rcos(a)                                              4=Rsin(a)

Squaring and adding gives
9+16=R^2cos^2(a)+R^2sin^2(a)
25=R^2
R=5                  therefore max value is 5

Back to the 4th line above and divide the equations
4/3=tan(a)   and, since both sin and cos are pos, a must be in Q1
a=tan^-1(4/3)
So expression = 5cos(x-tan^-1(4/3))

c)   Min value occurs at pi + translation = pi+tan^-1(4/3)    (approx = 4.0689)

2.
sin(x+a)=bsin(x-a)
sin(x)cos(a)+cos(x)sin(a)=b[sin(x)cos(a)-cos(x)sin(a)]
Now divide through by cos(x)
tan(x)cos(a)+sin(a)=btan(x)cos(a)-bsin(a)
sin(a)+bsin(a)=btan(x)cos(a)-tan(x)cos(a)
(1+b)sin(a)=(b-1)tan(x)cos(a)
tan(x)=[(1+b)sin(a)]/[(b-1)cos(a)]
         =(1+b)tan(a)/(b-1)

QED

[Special At Specialist]

Do we have to do that rsin(x + a) thing in specialist maths?
That's like the one part of circular functions I don't get.
I've never seen a question like that on the exam but I kind of want to learn it anyway.