# Homework Help: Show that 8sinx + 15cosx <= 17

1. Mar 9, 2012

### Bearded Man

1. The problem statement, all variables and given/known data
Show that 8sinx + 15cosx ≤ 17

2. Relevant equations
8^2 + 15^2 = 17^2, or (8,15,17) is a Pythagorean triplet
Identities like sin^2x + cos^2x = 1

3. The attempt at a solution
8sinx + 15cosx <= y
square both sides, clean up
64sin^2x + 240sinxcosx + 225cos^2x ≤y^2
substitute sin^2x for 1-cos^2x and likewise for cos^2x, derived from Pythagorean Identity.
289 - 64cos^2x -225sin^2x +240sinxcosx ≤ y^2

I need to show that the trigonometric functions on the left side equal 0, so I can take the square root of both sides and find 17≤ y, but I don't know how to do that. It is also possible that I'm approaching this from the wrong viewpoint.

2. Mar 9, 2012

### Combsbt

Do you know derivatives yet? Or are you working strictly with trig identities?

If you take the derivative of the function and find the zeroes you will have the maximum/minimum of the original function.

3. Mar 9, 2012

### Bearded Man

I am working strictly with trig identities.

4. Mar 9, 2012

### SammyS

Staff Emeritus
Hello Bearded Man. Welcome to PF !

You demonstrated that a triangle with sides of length 8, 15, and 17 is a right triangle.

Let θ be the larger of the two acute angles of such a triangle.

Then cos(θ) = 8/17 and sin(θ) = 15/17

Get the left hand side of your inequality to have a factor of $\cos(\theta)\sin(x)+\sin(\theta)\cos(x)$.

5. Mar 9, 2012

### Staff: Mentor

This can be done strictly with trig.

6. Mar 10, 2012

### Bearded Man

Now that I have cos(x), sin(x), cos(θ), and sin(θ) defined, can't I just do this:
8sin(x) + 15cos(x) ≤ y
64/17 + 225/17 ≤ y
289/17 ≤ y
17 ≤ y

7. Mar 10, 2012

### LCKurtz

You aren't trying to prove $17\le y$. You are trying to prove $8\sin x + 15\cos x \le 17$. Try writing$$8\sin x + 15\cos x = 17\left( \frac 8 {17}\cos x + \frac{15}{17}\sin x\right )$$and see if you can use the previous hints. Think about an addition formula.

8. Mar 10, 2012

### HallsofIvy

First, I don't see any reason to introduce "y". Second, it is not true that "8 sin(x)= 64/17" and "15 cos(x)= 225/17" which is what you are implying.

Better assert that $sin(x+ y)= sin(x)cos(y)+ cos(x)sin(y)$ and compare that to $17((8/17) sin(x)+ (15/17)cos(x))$. Can you find y so that cos(y)= 8/17 and $sin(y)= 15/17$? Is so, what does that tell you about 8sin(x)+ 15cos(x)?

9. Mar 10, 2012

### Bearded Man

Of course, how silly of me.

Thank you everyone for the help.

10. Mar 11, 2012

### Curious3141

The simplest approach in these cases is to express $a\sin x + b\cos x$ as $R\sin(x + \theta)$, where $R = \sqrt{a^2 + b^2}$ and $tan \theta = \frac{b}{a}$.

You can then make the observation that $\sin(x+\theta) \leq 1$, meaning that the expression has a maximal value of $R$.