Math / Trigonometry

Question10 Se arcsen $x=\frac{2}{5} \pi$ , quanto vale arccos $x$ ? (a) $\frac{\pi}{5}$ [c] $\frac{3}{10} \pi$ [b] $\frac{3}{5} \pi$ []. $\frac{\pi}{10}$

Studdy Solution

STEP 1

What is this asking? If $\arcsin(x) = \frac{2}{5}\pi$ , what is $\arccos(x)$ ? Watch out! Remember that $\arcsin(x)$ and $\arccos(x)$ have a specific range, so we can't just assume any angle will work!

STEP 2

1. Rewrite using sine
2. Apply the Pythagorean identity
3. Calculate cosine
4. Find arccos

STEP 3

We are given $\arcsin(x) = \frac{2}{5}\pi$ .
Taking the sine of both sides gives us $x = \sin(\frac{2}{5}\pi)$ .
This isolates our $x$ and lets us work with a familiar trigonometric function.

STEP 4

Remember the Pythagorean identity: $\sin^2(\theta) + \cos^2(\theta) = 1$ .
We'll use this to find $\cos(\frac{2}{5}\pi)$ .

STEP 5

Substituting $\theta = \frac{2}{5}\pi$ , we get $\sin^2(\frac{2}{5}\pi) + \cos^2(\frac{2}{5}\pi) = 1$ .
Since $x = \sin(\frac{2}{5}\pi)$ , we can rewrite this as $x^2 + \cos^2(\frac{2}{5}\pi) = 1$ .
Now, we isolate $\cos^2(\frac{2}{5}\pi)$ to get $\cos^2(\frac{2}{5}\pi) = 1 - x^2$ .

STEP 6

Taking the square root of both sides, we get $\cos(\frac{2}{5}\pi) = \pm\sqrt{1 - x^2}$ .
But since $\frac{2}{5}\pi$ is in the first quadrant (between $0$ and $\frac{\pi}{2}$ ), its cosine is positive.
So, we have $\cos(\frac{2}{5}\pi) = \sqrt{1 - x^2}$ .

STEP 7

We want to find $\arccos(x)$ .
We know that $x = \sin(\frac{2}{5}\pi)$ , and we just found that $\cos(\frac{2}{5}\pi) = \sqrt{1 - x^2}$ .
Let's consider an angle $\alpha$ such that $\cos(\alpha) = x$ .
We are looking for $\alpha = \arccos(x)$ .

STEP 8

We know that $\sin(\theta) = \cos(\frac{\pi}{2} - \theta)$ .
Since $x = \sin(\frac{2}{5}\pi)$ , we have $x = \cos(\frac{\pi}{2} - \frac{2}{5}\pi)$ .
Simplifying the angle, we get $x = \cos(\frac{5\pi - 4\pi}{10}) = \cos(\frac{\pi}{10})$ .

STEP 9

Since $x = \cos(\frac{\pi}{10})$ , we have $\arccos(x) = \frac{\pi}{10}$ .

STEP 10

The value of $\arccos(x)$ is $\frac{\pi}{10}$ .

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Question10 Se arcsen $x=\frac{2}{5} \pi$ , quanto vale arccos $x$ ? (a) $\frac{\pi}{5}$ [c] $\frac{3}{10} \pi$ [b] $\frac{3}{5} \pi$ []. $\frac{\pi}{10}$

Studdy Solution

STEP 1

What is this asking? If $\arcsin(x) = \frac{2}{5}\pi$ , what is $\arccos(x)$ ? Watch out! Remember that $\arcsin(x)$ and $\arccos(x)$ have a specific range, so we can't just assume any angle will work!

STEP 2

1. Rewrite using sine
2. Apply the Pythagorean identity
3. Calculate cosine
4. Find arccos

STEP 3

We are given $\arcsin(x) = \frac{2}{5}\pi$ .
Taking the sine of both sides gives us $x = \sin(\frac{2}{5}\pi)$ .
This isolates our $x$ and lets us work with a familiar trigonometric function.

STEP 4

Remember the Pythagorean identity: $\sin^2(\theta) + \cos^2(\theta) = 1$ .
We'll use this to find $\cos(\frac{2}{5}\pi)$ .

STEP 5

Substituting $\theta = \frac{2}{5}\pi$ , we get $\sin^2(\frac{2}{5}\pi) + \cos^2(\frac{2}{5}\pi) = 1$ .
Since $x = \sin(\frac{2}{5}\pi)$ , we can rewrite this as $x^2 + \cos^2(\frac{2}{5}\pi) = 1$ .
Now, we isolate $\cos^2(\frac{2}{5}\pi)$ to get $\cos^2(\frac{2}{5}\pi) = 1 - x^2$ .

STEP 6

Taking the square root of both sides, we get $\cos(\frac{2}{5}\pi) = \pm\sqrt{1 - x^2}$ .
But since $\frac{2}{5}\pi$ is in the first quadrant (between $0$ and $\frac{\pi}{2}$ ), its cosine is positive.
So, we have $\cos(\frac{2}{5}\pi) = \sqrt{1 - x^2}$ .

STEP 7

We want to find $\arccos(x)$ .
We know that $x = \sin(\frac{2}{5}\pi)$ , and we just found that $\cos(\frac{2}{5}\pi) = \sqrt{1 - x^2}$ .
Let's consider an angle $\alpha$ such that $\cos(\alpha) = x$ .
We are looking for $\alpha = \arccos(x)$ .

STEP 8

We know that $\sin(\theta) = \cos(\frac{\pi}{2} - \theta)$ .
Since $x = \sin(\frac{2}{5}\pi)$ , we have $x = \cos(\frac{\pi}{2} - \frac{2}{5}\pi)$ .
Simplifying the angle, we get $x = \cos(\frac{5\pi - 4\pi}{10}) = \cos(\frac{\pi}{10})$ .

STEP 9

Since $x = \cos(\frac{\pi}{10})$ , we have $\arccos(x) = \frac{\pi}{10}$ .

STEP 10

The value of $\arccos(x)$ is $\frac{\pi}{10}$ .

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Question10 Se arcsen x=25πx=\frac{2}{5} \pix=52​π, quanto vale arccos xxx ? (a) π5\frac{\pi}{5}5π​ [c] 310π\frac{3}{10} \pi103​π [b] 35π\frac{3}{5} \pi53​π []. π10\frac{\pi}{10}10π​

Studdy Solution

STEP 1

What is this asking? If arcsin⁡(x)=25π\arcsin(x) = \frac{2}{5}\piarcsin(x)=52​π, what is arccos⁡(x)\arccos(x)arccos(x)? Watch out! Remember that arcsin⁡(x)\arcsin(x)arcsin(x) and arccos⁡(x)\arccos(x)arccos(x) have a specific range, so we can't just assume any angle will work!

STEP 2

1. Rewrite using sine2. Apply the Pythagorean identity3. Calculate cosine4. Find arccos

STEP 3

We are given arcsin⁡(x)=25π\arcsin(x) = \frac{2}{5}\piarcsin(x)=52​π.Taking the sine of both sides gives us x=sin⁡(25π)x = \sin(\frac{2}{5}\pi)x=sin(52​π).This **isolates** our xxx and lets us work with a familiar trigonometric function.

STEP 4

Remember the **Pythagorean identity**: sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1sin2(θ)+cos2(θ)=1.We'll use this to find cos⁡(25π)\cos(\frac{2}{5}\pi)cos(52​π).

STEP 5

STEP 6

STEP 7

STEP 8

STEP 9

Since x=cos⁡(π10)x = \cos(\frac{\pi}{10})x=cos(10π​), we have arccos⁡(x)=π10\arccos(x) = \frac{\pi}{10}arccos(x)=10π​.

STEP 10

The value of arccos⁡(x)\arccos(x)arccos(x) is π10\frac{\pi}{10}10π​.

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Question10 Se arcsen $x=\frac{2}{5} \pi$ , quanto vale arccos $x$ ? (a) $\frac{\pi}{5}$ [c] $\frac{3}{10} \pi$ [b] $\frac{3}{5} \pi$ []. $\frac{\pi}{10}$

What is this asking? If $\arcsin(x) = \frac{2}{5}\pi$ , what is $\arccos(x)$ ? Watch out! Remember that $\arcsin(x)$ and $\arccos(x)$ have a specific range, so we can't just assume any angle will work!

1. Rewrite using sine
2. Apply the Pythagorean identity
3. Calculate cosine
4. Find arccos

We are given $\arcsin(x) = \frac{2}{5}\pi$ .
Taking the sine of both sides gives us $x = \sin(\frac{2}{5}\pi)$ .
This isolates our $x$ and lets us work with a familiar trigonometric function.

Remember the Pythagorean identity: $\sin^2(\theta) + \cos^2(\theta) = 1$ .
We'll use this to find $\cos(\frac{2}{5}\pi)$ .

Since $x = \cos(\frac{\pi}{10})$ , we have $\arccos(x) = \frac{\pi}{10}$ .

The value of $\arccos(x)$ is $\frac{\pi}{10}$ .