Question 635392
Hi, I'm terribly stuck with this question:

The function f is defined on the domain [0,pi] by f(x)=4cos(x)+3sin(x)
a) express in the form Rcos(x-y), where 0 < y < (pi)/2


Thankyou in advance
<pre> 
4cos(x)+3sin(x) = R({{{1/R}}})[4cos(x)+3sin(x)] = R[(cos(x){{{4/R}}}+sin(x){{{3/R}}}] =

 R·cos(x-y) = R·[cos(x)cos(y)+sin(x)sin(y)]

So

R[(cos(x){{{4/R}}}+sin(x){{{3/R}}}] = R·[cos(x)cos(y)+sin(x)sin(y)]

These will be equal if we choose y and R such that 

{{{4/R}}} = cos(y) and {{{3/R}}} = sin(y)

So we draw a right triangle with angle y, adjacent side 4, opposite side 3,
and hypotenuse R:

{{{drawing(100,80,-.5,4.5,-.5,3.5, triangle(0,0,4,0,4,3),locate(1,.9,y),
locate(2,0,4), locate(4.1,1.7,3),locate(1.7,2.5,R)


)}}}

So we choose y = arctan({{{3/4}}}) = 0.6435
and calculate R = {{{sqrt(3^2+4^2)}}} = {{{sqrt(9+16)}}} = {{{sqrt(25)}}} = 5

Therefore 

f(x) = 4·cos(x)+3·sin(x) = R·cos(x-y) = 5·cos(x-0.6435)
</pre>
b) Hence or otherwise, write down the value of x for which f(x) takes its maximum value
<pre>
f(x) will have maximum value when cos(x-0.6435) has maximum value.
cos(<font face="symbol">q</font>) has maximum value of 1 when <font face="symbol">q</font> = 0, so we set

              x-0.6435 = 0
                     x = 0.6435
Edwin</pre>