Show that 8sinx + 15cosx <= 17

[Bearded Man]

Homework Statement

Show that 8sinx + 15cosx ≤ 17

Homework Equations

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

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.

 

[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.

 

[Bearded Man]

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.

I am working strictly with trig identities.

 

[SammyS]

Homework Statement

Show that 8sinx + 15cosx ≤ 17

Homework Equations

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

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.

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).

 

[Mark44]

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

This can be done strictly with trig.

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

 

[Bearded Man]

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).

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

 

[LCKurtz]

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

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.

 

[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)?

 

[Bearded Man]

Of course, how silly of me.

Thank you everyone for the help.

 

[Curious3141]

Homework Statement

Show that 8sinx + 15cosx ≤ 17

Homework Equations

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

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.

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.