Problem 34 Find angles $u$ and $v$ such... [FREE SOLUTION]

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Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

1 Edition

Chapter 5: Problem 34

Find angles $u$ and $v$ such that $\sin u=\sin v$ but $\cos u \neq \cos v$.

Short Answer

Expert verified

A pair of angles that satisfy the given conditions are $u = \frac{\pi}{2}$ and $v = \frac{3\pi}{2}$. Their sine values are equal ($\sin u = \sin v = 1$), but their cosine values are different ($\cos u = 0$ and $\cos v = -1$).

Step by step solution

Get the sine property

For angles with equal sines, we can use the property that states if $\sin u = \sin v$, then $u = n \pi + (-1)^n v$, where $n$ is an integer value. In other words, for angles to have the same sine value, they will either be the same angle or be separated by an integer multiple of $\pi$.

Get the cosine property

Now let's focus on the condition $\cos u \neq \cos v$. For angles with not equal cosines, we know from the property that if $\cos u = \cos v$, then $u = 2n\pi \pm v$, where $n$ is an integer value. Consequently, in order for the cosine values to be different, this equation should not hold.

Find angles with equal sines but different cosines

Let's take the integer value $n=1$ in our sine property equation, which gives us $u = \pi + (-1)^1 v = \pi - v$. So, we have an angle pair $u$ and $\pi - v$ with the same sine values. To satisfy the $cos(u) \neq cos(v)$ condition, we need to ensure that $u = 2m\pi \pm v$ does not hold, where $m$ is an integer value. Let's check using the obtained value for $u$: \[\pi - v \neq 2m\pi \pm v\] For $u$ to satisfy this inequality, the only possibility is when $m=0$: \[\pi - v \neq \pm v\] Now, we can choose angles $u$ and $v$ such that these conditions hold. A simple choice would be $u = \frac{\pi}{2}$ and $v = \pi - u = \frac{\pi}{2}$.

Verify the solution

Finally, let's verify our solution using the obtained angles: \[\sin u = \sin(\frac{\pi}{2}) = 1\] \[\sin v = \sin(\frac{\pi}{2}) = 1\] \[\cos u = \cos(\frac{\pi}{2}) = 0\] \[\cos v = \cos(\pi - \frac{\pi}{2}) = -1\] Since $\sin u = \sin v$ and $\cos u \neq \cos v$, our solution is correct. Angles $u = \frac{\pi}{2}$ and $v = \frac{\pi}{2}$ satisfy the given conditions.

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91影视

Find angles \(u\) and \(v\) such that \(\sin u=\sin v\) but \(\cos u \neq \cos v\).

Short Answer

Step by step solution

Get the sine property

Get the cosine property

Find angles with equal sines but different cosines

Verify the solution

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