Question

f(x) 7cosx-24sinx
Given that f(x)=Rcos(x+a) where R>=0, 0<=a<=1/2pi and x and a are measured in radians.
Find R and show that a=1.29 to 2 d.p
Hence Write down the minimum value of f(x)
The value of x in the interval 0<=x<=2pi
Find the smallest possible two positive values of x for which 7cosx-24sinx=10

Answer 1

where are you stuck

Answer 2

you have been using `acosx+bsinx` formula right ?

Answer 3

yes i do use this formula @ganeshie8
7cosx-24sinx=Rcosxcosa+RSinxsina
7=Rcosa
-24=Rsina
Tan=-24/7 R=sqrt(7)^2+(-24)^2=25

Answer 4

Yes !

\[7\cos x-24\sin x = 25\cos\left(x - \tan^{-1}(\frac{-24}{7})\right)\]

Answer 5

So how is a=1.29

Answer 6

whats the value of \(\tan^{-1}(\frac{-24}{7})\) ?

Answer 7

oh i get how to do it now tan-1(-24/7)=-73.73979529*pi/180=-1.287 rounded to 2d.p -1.29

Answer 8

Yep !

Answer 9

\[7\cos x-24\sin x = 25\cos\left(x - \tan^{-1}(\frac{-24}{7})\right)\]

\[= 25\cos\left(x - (-1.29)\right)\]

\[= 25\cos\left(x +1.29\right)\]

Answer 10

Thank you @ganeshie8 but how can i solve to find the value of x in the interval 0<=x<=2pi to get this minimum value

Answer 11

\[\LARGE -25 \leq 7cosx x−24\sin x \leq 25\]
clearly minimum value is -25