Delger

Posted on Friday, 26 December, 2003 - 08:34 pm:

Find the range of the function f, where

f(x) = 8 cosx + 15 sin x, x belongs to R

It is P2, so can anyone help me to solve this problem to P2 sylablis.

Delger

Posted on Friday, 26 December, 2003 - 08:39 pm:

It is possible to find it by finding maximum value if f(x) but it is P3, and I want to find it by a simpler way.

Mark Lobo

Posted on Friday, 26 December, 2003 - 09:01 pm:

Use the fact that the maximum value of cosx and sinx is 1 and the minimum value of cosx and sinx is -1.

Mark

Arne Smeets

Posted on Friday, 26 December, 2003 - 09:35 pm:

Be careful. Transform the given expression into an expression of the form k sin x. [You will find k = 17.]

Delger

Posted on Friday, 26 December, 2003 - 10:18 pm:

Arne how do u do it? Sorry, can u write the whole solution.

Shu Cao

Posted on Friday, 26 December, 2003 - 10:22 pm:

acosx+bsinx=sqrt(a^2+b^2)sin(x+c)

Arne Smeets

Posted on Friday, 26 December, 2003 - 10:47 pm:

We have 8 cos x + 15 sin x = 15(sin x + 8/15 cos x).
Now suppose t = arctan 8/15, with t in [0, pi/2]. Then
15(sin x + 8/15 cos x)
= 15(sin x + tan t cos x)
= 15(sin x cos t + sin t cos x)/(cos t)
= (15/cos t) sin(x + t).
As tan t = 8/15, cos t = LaTeX Image

.
Hence 8 cos x + 15 sin x = 17 sin(x + t).
Since LaTeX Image

for all x, we deduce that the minimum (maximum) of the given expression equals -17 (17) respectively.

Mark Lobo

Posted on Friday, 26 December, 2003 - 11:17 pm:

I should have made my post clearer:

You can express $8 \cos x + 15 \sin x$ in the form $R \cos (x - α)$ or $R \sin (x + α)$ and then note that the maximum and minimum values of $\cos x$ and $\sin x$ are $1$ and $- 1$ respectively.