# Show that 8sinx + 15cosx <= 17

• Bearded Man
In summary: This is because \sin(x) \gt 0 as \theta \gt 0, and \cos(x) \lt 1 as \theta \lt 0.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.If you take the derivative of the function and find the zeroes you will have the maximum/minimum of the original function.In summary, the problem is to find a function that when graphed has a maximum or minimum at a specific point. Homework Equations8^2 + 15^2 = 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.

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.

Combsbt said:
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.

Bearded Man said:

## 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 $\cos(\theta)\sin(x)+\sin(\theta)\cos(x)$.

Combsbt said:
Do you know derivatives yet? Or are you working strictly with trig identities?
This can be done strictly with trig.
Combsbt said:
If you take the derivative of the function and find the zeroes you will have the maximum/minimum of the original function.

SammyS said:
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

Bearded Man said:
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.

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)?

Of course, how silly of me.

Thank you everyone for the help.

Bearded Man said:

## 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 $R$.

## 1. How do I solve the inequality 8sinx + 15cosx <= 17?

To solve this inequality, you can use the properties of sine and cosine functions to rewrite it as a single trigonometric function. Then, you can use algebraic methods to isolate the variable and determine the range of values that satisfy the inequality.

## 2. What is the general approach to solving trigonometric inequalities?

The general approach to solving trigonometric inequalities is to use the properties of trigonometric functions to rewrite the inequality as a single function, and then use algebraic methods to solve for the variable. It is important to also consider the domain of the trigonometric function to determine the range of valid solutions.

## 3. Is there more than one way to solve the inequality 8sinx + 15cosx <= 17?

Yes, there are multiple ways to solve this inequality. One approach is to use the properties of sine and cosine functions to rewrite the inequality as a single trigonometric function. Another approach is to graph the inequality and visually determine the range of values that satisfy it.

## 4. What is the significance of the inequality 8sinx + 15cosx <= 17 in mathematics?

This inequality is significant in mathematics because it represents a relationship between the sine and cosine functions, two fundamental trigonometric functions. It also demonstrates the use of trigonometric functions in modeling real-world phenomena and solving problems in various fields such as physics, engineering, and finance.

## 5. Can you provide an example of a real-world application of the inequality 8sinx + 15cosx <= 17?

One example of a real-world application of this inequality is in electrical engineering, where it can be used to model the relationship between the voltage and current in an alternating current (AC) circuit. The left side of the inequality represents the amplitude of the resulting voltage, while the right side represents the maximum possible voltage. By solving this inequality, engineers can determine the maximum voltage that the circuit can handle without causing damage.

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