Math / Trigonometry

QuestionEvaluate $\cos (a-b)$ given that $\cos a=\frac{4}{5}$ with $0<a<\frac{\pi}{2}$ and $\sin b=-\frac{15}{17}$ with $\frac{3 \pi}{2}<b<2 \pi$ . $\cos (a-b)=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the cosine of the difference of two angles, knowing the cosine of the first angle and the sine of the second angle, along with their quadrants. Watch out! Don't mix up the cosine and sine values, and make sure to account for the correct signs based on the quadrants of the angles.

STEP 2

1. Find sin a and cos b.
2. Apply the cosine difference formula.

STEP 3

Alright, let's kick things off by finding $\sin{a}$ !
We know that $\cos{a} = \frac{4}{5}$ and a is in the first quadrant (between $0$ and $\frac{\pi}{2}$ ).
In this quadrant, both sine and cosine are positive!

STEP 4

Remember the Pythagorean identity: $(\sin{a})^2 + (\cos{a})^2 = 1$ .
Let's plug in our known value: $(\sin{a})^2 + (\frac{4}{5})^2 = 1$ .

STEP 5

This simplifies to $(\sin{a})^2 + \frac{16}{25} = 1$ .
Subtracting $\frac{16}{25}$ from both sides gives us $(\sin{a})^2 = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25}$ .

STEP 6

Taking the square root of both sides, we get $\sin{a} = \pm\frac{3}{5}$ .
Since a is in the first quadrant, $\sin{a}$ is positive, so $\sin{a} = \frac{3}{5}$ .
Awesome!

STEP 7

Now, let's find $\cos{b}$ .
We know $\sin{b} = -\frac{15}{17}$ and b is in the fourth quadrant (between $\frac{3\pi}{2}$ and $2\pi$ ).
Cosine is positive in the fourth quadrant!

STEP 8

Using the Pythagorean identity again: $(\sin{b})^2 + (\cos{b})^2 = 1$ , we substitute the known value: $(-\frac{15}{17})^2 + (\cos{b})^2 = 1$ .

STEP 9

This simplifies to $\frac{225}{289} + (\cos{b})^2 = 1$ .
Subtracting $\frac{225}{289}$ from both sides, we get $(\cos{b})^2 = 1 - \frac{225}{289} = \frac{289}{289} - \frac{225}{289} = \frac{64}{289}$ .

STEP 10

Taking the square root gives us $\cos{b} = \pm\frac{8}{17}$ .
Since b is in the fourth quadrant, $\cos{b}$ is positive, so $\cos{b} = \frac{8}{17}$ .
Fantastic!

STEP 11

The cosine difference formula is $\cos(a-b) = \cos{a}\cos{b} + \sin{a}\sin{b}$ .
Let's plug in our values: $\cos(a-b) = (\frac{4}{5})(\frac{8}{17}) + (\frac{3}{5})(-\frac{15}{17})$ .

STEP 12

Multiplying the fractions, we get $\cos(a-b) = \frac{32}{85} - \frac{45}{85}$ .

STEP 13

Finally, subtracting the fractions gives us our final result: $\cos(a-b) = -\frac{13}{85}$ .
Boom!

STEP 14

$\cos(a-b) = -\frac{13}{85}$

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QuestionEvaluate $\cos (a-b)$ given that $\cos a=\frac{4}{5}$ with $0<a<\frac{\pi}{2}$ and $\sin b=-\frac{15}{17}$ with $\frac{3 \pi}{2}<b<2 \pi$ . $\cos (a-b)=$ $\square$

Studdy Solution

STEP 1

STEP 2

1. Find sin a and cos b.
2. Apply the cosine difference formula.

STEP 3

Alright, let's kick things off by finding $\sin{a}$ !
We know that $\cos{a} = \frac{4}{5}$ and a is in the first quadrant (between $0$ and $\frac{\pi}{2}$ ).
In this quadrant, both sine and cosine are positive!

STEP 4

Remember the Pythagorean identity: $(\sin{a})^2 + (\cos{a})^2 = 1$ .
Let's plug in our known value: $(\sin{a})^2 + (\frac{4}{5})^2 = 1$ .

STEP 5

This simplifies to $(\sin{a})^2 + \frac{16}{25} = 1$ .
Subtracting $\frac{16}{25}$ from both sides gives us $(\sin{a})^2 = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25}$ .

STEP 6

Taking the square root of both sides, we get $\sin{a} = \pm\frac{3}{5}$ .
Since a is in the first quadrant, $\sin{a}$ is positive, so $\sin{a} = \frac{3}{5}$ .
Awesome!

STEP 7

Now, let's find $\cos{b}$ .
We know $\sin{b} = -\frac{15}{17}$ and b is in the fourth quadrant (between $\frac{3\pi}{2}$ and $2\pi$ ).
Cosine is positive in the fourth quadrant!

STEP 8

Using the Pythagorean identity again: $(\sin{b})^2 + (\cos{b})^2 = 1$ , we substitute the known value: $(-\frac{15}{17})^2 + (\cos{b})^2 = 1$ .

STEP 9

This simplifies to $\frac{225}{289} + (\cos{b})^2 = 1$ .
Subtracting $\frac{225}{289}$ from both sides, we get $(\cos{b})^2 = 1 - \frac{225}{289} = \frac{289}{289} - \frac{225}{289} = \frac{64}{289}$ .

STEP 10

Taking the square root gives us $\cos{b} = \pm\frac{8}{17}$ .
Since b is in the fourth quadrant, $\cos{b}$ is positive, so $\cos{b} = \frac{8}{17}$ .
Fantastic!

STEP 11

The cosine difference formula is $\cos(a-b) = \cos{a}\cos{b} + \sin{a}\sin{b}$ .
Let's plug in our values: $\cos(a-b) = (\frac{4}{5})(\frac{8}{17}) + (\frac{3}{5})(-\frac{15}{17})$ .

STEP 12

Multiplying the fractions, we get $\cos(a-b) = \frac{32}{85} - \frac{45}{85}$ .

STEP 13

Finally, subtracting the fractions gives us our final result: $\cos(a-b) = -\frac{13}{85}$ .
Boom!

STEP 14

$\cos(a-b) = -\frac{13}{85}$

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Studdy Solution

STEP 1

STEP 2

1. Find sin *a* and cos *b*.2. Apply the cosine difference formula.

STEP 3

Alright, let's **kick things off** by finding sin⁡a\sin{a}sina!We know that cos⁡a=45\cos{a} = \frac{4}{5}cosa=54​ and *a* is in the **first quadrant** (between 000 and π2\frac{\pi}{2}2π​).In this quadrant, both sine and cosine are **positive**!

STEP 4

Remember the **Pythagorean identity**: (sin⁡a)2+(cos⁡a)2=1(\sin{a})^2 + (\cos{a})^2 = 1(sina)2+(cosa)2=1.Let's plug in our known value: (sin⁡a)2+(45)2=1(\sin{a})^2 + (\frac{4}{5})^2 = 1(sina)2+(54​)2=1.

STEP 5

STEP 6

Taking the **square root** of both sides, we get sin⁡a=±35\sin{a} = \pm\frac{3}{5}sina=±53​.Since *a* is in the **first quadrant**, sin⁡a\sin{a}sina is **positive**, so sin⁡a=35\sin{a} = \frac{3}{5}sina=53​.Awesome!

STEP 7

Now, let's find cos⁡b\cos{b}cosb.We know sin⁡b=−1517\sin{b} = -\frac{15}{17}sinb=−1715​ and *b* is in the **fourth quadrant** (between 3π2\frac{3\pi}{2}23π​ and 2π2\pi2π).Cosine is **positive** in the fourth quadrant!

STEP 8

Using the **Pythagorean identity** again: (sin⁡b)2+(cos⁡b)2=1(\sin{b})^2 + (\cos{b})^2 = 1(sinb)2+(cosb)2=1, we substitute the known value: (−1517)2+(cos⁡b)2=1(-\frac{15}{17})^2 + (\cos{b})^2 = 1(−1715​)2+(cosb)2=1.

STEP 9

STEP 10

Taking the **square root** gives us cos⁡b=±817\cos{b} = \pm\frac{8}{17}cosb=±178​.Since *b* is in the **fourth quadrant**, cos⁡b\cos{b}cosb is **positive**, so cos⁡b=817\cos{b} = \frac{8}{17}cosb=178​.Fantastic!

STEP 11

STEP 12

Multiplying the fractions, we get cos⁡(a−b)=3285−4585\cos(a-b) = \frac{32}{85} - \frac{45}{85}cos(a−b)=8532​−8545​.

STEP 13

Finally, subtracting the fractions gives us our **final result**: cos⁡(a−b)=−1385\cos(a-b) = -\frac{13}{85}cos(a−b)=−8513​.Boom!

STEP 14

cos⁡(a−b)=−1385\cos(a-b) = -\frac{13}{85}cos(a−b)=−8513​

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1. Find sin a and cos b.
2. Apply the cosine difference formula.

Alright, let's kick things off by finding $\sin{a}$ !
We know that $\cos{a} = \frac{4}{5}$ and a is in the first quadrant (between $0$ and $\frac{\pi}{2}$ ).
In this quadrant, both sine and cosine are positive!

Remember the Pythagorean identity: $(\sin{a})^2 + (\cos{a})^2 = 1$ .
Let's plug in our known value: $(\sin{a})^2 + (\frac{4}{5})^2 = 1$ .

Taking the square root of both sides, we get $\sin{a} = \pm\frac{3}{5}$ .
Since a is in the first quadrant, $\sin{a}$ is positive, so $\sin{a} = \frac{3}{5}$ .
Awesome!

Now, let's find $\cos{b}$ .
We know $\sin{b} = -\frac{15}{17}$ and b is in the fourth quadrant (between $\frac{3\pi}{2}$ and $2\pi$ ).
Cosine is positive in the fourth quadrant!

Using the Pythagorean identity again: $(\sin{b})^2 + (\cos{b})^2 = 1$ , we substitute the known value: $(-\frac{15}{17})^2 + (\cos{b})^2 = 1$ .

Taking the square root gives us $\cos{b} = \pm\frac{8}{17}$ .
Since b is in the fourth quadrant, $\cos{b}$ is positive, so $\cos{b} = \frac{8}{17}$ .
Fantastic!

Multiplying the fractions, we get $\cos(a-b) = \frac{32}{85} - \frac{45}{85}$ .

Finally, subtracting the fractions gives us our final result: $\cos(a-b) = -\frac{13}{85}$ .
Boom!

$\cos(a-b) = -\frac{13}{85}$