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High School Math Algebra Algebra 2: New York Edition (holt Mcdougal Larson Algebra 2)

The value of expression sin a − b . The value of sin a − b is − 87 425 . Given information: The given value is sin a = 8 17 with 0 < a < π 2 and cos b = − 24 25 with π < b < 3 π 2 . Formula used: The difference formula for cosine is given by: sin a − b = sin a cos b − cos a sin b Calculation: Use Pythagorean identity to find the value of cosine function. sin 2 a + cos 2 a = 1 Substitute 8 17 for sin a in the above equation. 8 17 2 + cos 2 a = 1 cos 2 a = 1 − 64 289 cos 2 a = 225 289 cos a = ± 15 17 The angle a lies in the first quadrant. Cosine function is positive in the first quadrant. cos a = 15 17 Use Pythagorean identity to find the value of cosine function. sin 2 b + cos 2 b = 1 Substitute − 24 25 for cos b in the above equation. sin 2 b + − 24 25 2 = 1 sin 2 b = 1 − 576 625 sin 2 b = 49 625 sin b = ± 7 25 The angle b lies in the third quadrant. Sine function is negative in the third quadrant. sin b = − 7 25 Substitute 8 17 for sin a , − 7 25 for sin b , 15 17 for cos a and − 24 25 for cos b in the formula. sin a − b = 8 17 ⋅ − 24 25 − 15 17 ⋅ − 7 25 = − 192 425 + 105 425 = − 87 425 Therefore, the value of sin a − b is − 87 425 .

The value of expression sin a − b . The value of sin a − b is − 87 425 . Given information: The given value is sin a = 8 17 with 0 < a < π 2 and cos b = − 24 25 with π < b < 3 π 2 . Formula used: The difference formula for cosine is given by: sin a − b = sin a cos b − cos a sin b Calculation: Use Pythagorean identity to find the value of cosine function. sin 2 a + cos 2 a = 1 Substitute 8 17 for sin a in the above equation. 8 17 2 + cos 2 a = 1 cos 2 a = 1 − 64 289 cos 2 a = 225 289 cos a = ± 15 17 The angle a lies in the first quadrant. Cosine function is positive in the first quadrant. cos a = 15 17 Use Pythagorean identity to find the value of cosine function. sin 2 b + cos 2 b = 1 Substitute − 24 25 for cos b in the above equation. sin 2 b + − 24 25 2 = 1 sin 2 b = 1 − 576 625 sin 2 b = 49 625 sin b = ± 7 25 The angle b lies in the third quadrant. Sine function is negative in the third quadrant. sin b = − 7 25 Substitute 8 17 for sin a , − 7 25 for sin b , 15 17 for cos a and − 24 25 for cos b in the formula. sin a − b = 8 17 ⋅ − 24 25 − 15 17 ⋅ − 7 25 = − 192 425 + 105 425 = − 87 425 Therefore, the value of sin a − b is − 87 425 .

Algebra 2: New York Edition (holt Mcdougal Larson Algebra 2)

Algebra 2: New York Edition (holt Mcdougal Larson Algebra 2)

2nd Edition

ISBN: 9780618912407

Author: MCDOUGAL LITTEL

Publisher: Mcdougal Littel

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Expert Solution & Answer

Chapter 10.6, Problem 5GP

Solution

To find: The value of expression sina−b .

The value of sina−b is −87425 .

Given information: The given value is sina=817 with 0<a<π2 and cosb=−2425 with π<b<3π2 .

Formula used: The difference formula for cosine is given by:

sina−b=sinacosb−cosasinb

Calculation:

Use Pythagorean identity to find the value of cosine function.

sin2a+cos2a=1

Substitute 817 for sina in the above equation.

8172+cos2a=1cos2a=1−64289cos2a=225289cosa=±1517

The angle a lies in the first quadrant. Cosine function is positive in the first quadrant.

cosa=1517

Use Pythagorean identity to find the value of cosine function.

sin2b+cos2b=1

Substitute −2425 for cosb in the above equation.

sin2b+−24252=1sin2b=1−576625sin2b=49625sinb=±725

The angle b lies in the third quadrant. Sine function is negative in the third quadrant.

sinb=−725

Substitute 817 for sina , −725 for sinb , 1517 for cosa and −2425 for cosb in the formula.

sina−b=817⋅−2425−1517⋅−725=−192425+105425=−87425

Therefore, the value of sina−b is −87425 .

Chapter 10.6, Problem 4GP

Chapter 10.6, Problem 6GP

Chapter 10 Solutions

Algebra 2: New York Edition (holt Mcdougal Larson Algebra 2)

Ch. 10.1 - Prob. 1GP Ch. 10.1 - Prob. 2GP Ch. 10.1 - Prob. 3GP Ch. 10.1 - Prob. 4GP Ch. 10.1 - Prob. 5GP Ch. 10.1 - Prob. 6GP Ch. 10.1 - Prob. 7GP Ch. 10.1 - Prob. 8GP Ch. 10.1 - Prob. 9GP Ch. 10.1 - Prob. 10GP

Ch. 10.1 - Prob. 11GP Ch. 10.1 - Prob. 12GP Ch. 10.1 - Prob. 13GP Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Prob. 13E Ch. 10.1 - Prob. 14E Ch. 10.1 - Prob. 15E Ch. 10.1 - Prob. 16E Ch. 10.1 - Prob. 17E Ch. 10.1 - Prob. 18E Ch. 10.1 - Prob. 19E Ch. 10.1 - Prob. 20E Ch. 10.1 - Prob. 21E Ch. 10.1 - Prob. 22E Ch. 10.1 - Prob. 23E Ch. 10.1 - Prob. 24E Ch. 10.1 - Prob. 25E Ch. 10.1 - Prob. 26E Ch. 10.1 - Prob. 27E Ch. 10.1 - Prob. 28E Ch. 10.1 - Prob. 29PS Ch. 10.1 - Prob. 30PS Ch. 10.1 - Prob. 31PS Ch. 10.1 - Prob. 32PS Ch. 10.1 - Prob. 33PS Ch. 10.2 - Prob. 1GP Ch. 10.2 - Prob. 2GP Ch. 10.2 - Prob. 3GP Ch. 10.2 - Prob. 4GP Ch. 10.2 - Prob. 5GP Ch. 10.2 - Prob. 6GP Ch. 10.2 - Prob. 7GP Ch. 10.2 - Prob. 1E Ch. 10.2 - Prob. 2E Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - Prob. 5E Ch. 10.2 - 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