cos(x- y)= cos(x)cos(y)- sin(x)sin(y). So you want cos(y)= 11, sin(y)= -3. Of course, that's impossible because sine and cosine are always less than 1, and, of course, \(\displaystyle sin^2(y)+ cos^2(y)\) must equal 1. That's where "R" comes in. If we take \(\displaystyle R= \sqrt{11^2+ 3^2}= \sqrt{121+ 9}= \sqrt{130}\) and then we can write \(\displaystyle \sqrt{130}cos(x- y)= \sqrt{130}\left(\frac{11}{\sqrt{130}}cos(x)- \frac{-3}{\sqrt{130}}sin(x)\right)\) so that we must have \(\displaystyle R= \sqrt{130}\), as you say, and \(\displaystyle cos(y)= \frac{11}{\sqrt{130}}\). Unfortunately, \(\displaystyle cos(15.3)= .26\) while \(\displaystyle \frac{11}{\sqrt{130}}= .96\). (But \(\displaystyle .96^2+ .26^2= 1\) (approximately) so I suspect you have just made a trivial mistake, perhaps using sine when you should have used cosine.)Express 11cosx + 3sinx in the for R cos(x-y), where R and y are constants. Hence find the maximum and minimum values of (11cosx + 3sinx)^3, and the corresponding values of x where 0<x<360 [degrees].
[I calculated that R = square root of 130 while y = 15.3 degrees, do correct me if I'm wrong.] Thank you!
\(\displaystyle \text{Express }11\cos x + 3\sin x\text{ in the form }R\cos(x-y)\text{, where }R\text{ and }y\text{ are constants.}\)
\(\displaystyle \text{Hence, find the maximum and minimum values of }(11\cos x + 3\sin x)^3,\),
\(\displaystyle \text{and the corresponding values of }x\text{, where }0^o <x<360^o.\)
\(\displaystyle \text{I calculated that }R = \sqrt{130}\text{, while }y = 15.3^o.\) .Good!