Suppose That $\sin \alpha = \frac{12}{13}$ For A Quadrant II Angle $\alpha$ And $\sin \beta = \frac{3}{5}$ For A Quadrant I Angle $\beta$. Find The Exact Value Of Each Of The Following:a. $\cos \alpha$b.

by ADMIN 204 views

a. cos⁡α\cos \alpha

To find the exact value of cos⁡α\cos \alpha, we can use the Pythagorean identity sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1. Since we are given that sin⁡α=1213\sin \alpha = \frac{12}{13}, we can substitute this value into the equation and solve for cos⁡α\cos \alpha.

sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1

(1213)2+cos⁡2α=1\left(\frac{12}{13}\right)^2 + \cos^2 \alpha = 1

144169+cos⁡2α=1\frac{144}{169} + \cos^2 \alpha = 1

cos⁡2α=1−144169\cos^2 \alpha = 1 - \frac{144}{169}

cos⁡2α=25169\cos^2 \alpha = \frac{25}{169}

Since α\alpha is a quadrant II angle, we know that cos⁡α\cos \alpha is negative. Therefore, we can take the negative square root of both sides of the equation to find the exact value of cos⁡α\cos \alpha.

cos⁡α=−25169\cos \alpha = -\sqrt{\frac{25}{169}}

cos⁡α=−513\cos \alpha = -\frac{5}{13}

b. cos⁡β\cos \beta

To find the exact value of cos⁡β\cos \beta, we can use the Pythagorean identity sin⁡2β+cos⁡2β=1\sin^2 \beta + \cos^2 \beta = 1. Since we are given that sin⁡β=35\sin \beta = \frac{3}{5}, we can substitute this value into the equation and solve for cos⁡β\cos \beta.

sin⁡2β+cos⁡2β=1\sin^2 \beta + \cos^2 \beta = 1

(35)2+cos⁡2β=1\left(\frac{3}{5}\right)^2 + \cos^2 \beta = 1

925+cos⁡2β=1\frac{9}{25} + \cos^2 \beta = 1

cos⁡2β=1−925\cos^2 \beta = 1 - \frac{9}{25}

cos⁡2β=1625\cos^2 \beta = \frac{16}{25}

Since β\beta is a quadrant I angle, we know that cos⁡β\cos \beta is positive. Therefore, we can take the positive square root of both sides of the equation to find the exact value of cos⁡β\cos \beta.

cos⁡β=1625\cos \beta = \sqrt{\frac{16}{25}}

cos⁡β=45\cos \beta = \frac{4}{5}

c. tan⁡α\tan \alpha

To find the exact value of tan⁡α\tan \alpha, we can use the definition of the tangent function: tan⁡α=sin⁡αcos⁡α\tan \alpha = \frac{\sin \alpha}{\cos \alpha}. We are given that sin⁡α=1213\sin \alpha = \frac{12}{13} and cos⁡α=−513\cos \alpha = -\frac{5}{13}, so we can substitute these values into the equation and solve for tan⁡α\tan \alpha.

tan⁡α=sin⁡αcos⁡α\tan \alpha = \frac{\sin \alpha}{\cos \alpha}

tan⁡α=1213−513\tan \alpha = \frac{\frac{12}{13}}{-\frac{5}{13}}

tan⁡α=−125\tan \alpha = -\frac{12}{5}

d. tan⁡β\tan \beta

To find the exact value of tan⁡β\tan \beta, we can use the definition of the tangent function: tan⁡β=sin⁡βcos⁡β\tan \beta = \frac{\sin \beta}{\cos \beta}. We are given that sin⁡β=35\sin \beta = \frac{3}{5} and cos⁡β=45\cos \beta = \frac{4}{5}, so we can substitute these values into the equation and solve for tan⁡β\tan \beta.

tan⁡β=sin⁡βcos⁡β\tan \beta = \frac{\sin \beta}{\cos \beta}

tan⁡β=3545\tan \beta = \frac{\frac{3}{5}}{\frac{4}{5}}

tan⁡β=34\tan \beta = \frac{3}{4}

e. sin⁡α+cos⁡α\sin \alpha + \cos \alpha

To find the exact value of sin⁡α+cos⁡α\sin \alpha + \cos \alpha, we can substitute the values of sin⁡α\sin \alpha and cos⁡α\cos \alpha into the equation.

sin⁡α+cos⁡α=1213−513\sin \alpha + \cos \alpha = \frac{12}{13} - \frac{5}{13}

sin⁡α+cos⁡α=−513\sin \alpha + \cos \alpha = -\frac{5}{13}

f. sin⁡β+cos⁡β\sin \beta + \cos \beta

To find the exact value of sin⁡β+cos⁡β\sin \beta + \cos \beta, we can substitute the values of sin⁡β\sin \beta and cos⁡β\cos \beta into the equation.

sin⁡β+cos⁡β=35+45\sin \beta + \cos \beta = \frac{3}{5} + \frac{4}{5}

sin⁡β+cos⁡β=75\sin \beta + \cos \beta = \frac{7}{5}

g. sin⁡αcos⁡β\sin \alpha \cos \beta

To find the exact value of sin⁡αcos⁡β\sin \alpha \cos \beta, we can substitute the values of sin⁡α\sin \alpha and cos⁡β\cos \beta into the equation.

sin⁡αcos⁡β=1213⋅45\sin \alpha \cos \beta = \frac{12}{13} \cdot \frac{4}{5}

sin⁡αcos⁡β=4865\sin \alpha \cos \beta = \frac{48}{65}

h. cos⁡αsin⁡β\cos \alpha \sin \beta

To find the exact value of cos⁡αsin⁡β\cos \alpha \sin \beta, we can substitute the values of cos⁡α\cos \alpha and sin⁡β\sin \beta into the equation.

cos⁡αsin⁡β=−513⋅35\cos \alpha \sin \beta = -\frac{5}{13} \cdot \frac{3}{5}

cos⁡αsin⁡β=−1565\cos \alpha \sin \beta = -\frac{15}{65}

i. sin⁡αsin⁡β\sin \alpha \sin \beta

To find the exact value of sin⁡αsin⁡β\sin \alpha \sin \beta, we can substitute the values of sin⁡α\sin \alpha and sin⁡β\sin \beta into the equation.

sin⁡αsin⁡β=1213⋅35\sin \alpha \sin \beta = \frac{12}{13} \cdot \frac{3}{5}

sin⁡αsin⁡β=3665\sin \alpha \sin \beta = \frac{36}{65}

j. cos⁡αcos⁡β\cos \alpha \cos \beta

To find the exact value of cos⁡αcos⁡β\cos \alpha \cos \beta, we can substitute the values of cos⁡α\cos \alpha and cos⁡β\cos \beta into the equation.

cos⁡αcos⁡β=−513⋅45\cos \alpha \cos \beta = -\frac{5}{13} \cdot \frac{4}{5}

cos⁡αcos⁡β=−2065\cos \alpha \cos \beta = -\frac{20}{65}

Q&A

Q: What is the value of cos⁡α\cos \alpha?

A: To find the value of cos⁡α\cos \alpha, we can use the Pythagorean identity sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1. Since we are given that sin⁡α=1213\sin \alpha = \frac{12}{13}, we can substitute this value into the equation and solve for cos⁡α\cos \alpha.

sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1

(1213)2+cos⁡2α=1\left(\frac{12}{13}\right)^2 + \cos^2 \alpha = 1

144169+cos⁡2α=1\frac{144}{169} + \cos^2 \alpha = 1

cos⁡2α=1−144169\cos^2 \alpha = 1 - \frac{144}{169}

cos⁡2α=25169\cos^2 \alpha = \frac{25}{169}

Since α\alpha is a quadrant II angle, we know that cos⁡α\cos \alpha is negative. Therefore, we can take the negative square root of both sides of the equation to find the exact value of cos⁡α\cos \alpha.

cos⁡α=−25169\cos \alpha = -\sqrt{\frac{25}{169}}

cos⁡α=−513\cos \alpha = -\frac{5}{13}

Q: What is the value of cos⁡β\cos \beta?

A: To find the value of cos⁡β\cos \beta, we can use the Pythagorean identity sin⁡2β+cos⁡2β=1\sin^2 \beta + \cos^2 \beta = 1. Since we are given that sin⁡β=35\sin \beta = \frac{3}{5}, we can substitute this value into the equation and solve for cos⁡β\cos \beta.

sin⁡2β+cos⁡2β=1\sin^2 \beta + \cos^2 \beta = 1

(35)2+cos⁡2β=1\left(\frac{3}{5}\right)^2 + \cos^2 \beta = 1

925+cos⁡2β=1\frac{9}{25} + \cos^2 \beta = 1

cos⁡2β=1−925\cos^2 \beta = 1 - \frac{9}{25}

cos⁡2β=1625\cos^2 \beta = \frac{16}{25}

Since β\beta is a quadrant I angle, we know that cos⁡β\cos \beta is positive. Therefore, we can take the positive square root of both sides of the equation to find the exact value of cos⁡β\cos \beta.

cos⁡β=1625\cos \beta = \sqrt{\frac{16}{25}}

cos⁡β=45\cos \beta = \frac{4}{5}

Q: What is the value of tan⁡α\tan \alpha?

A: To find the value of tan⁡α\tan \alpha, we can use the definition of the tangent function: tan⁡α=sin⁡αcos⁡α\tan \alpha = \frac{\sin \alpha}{\cos \alpha}. We are given that sin⁡α=1213\sin \alpha = \frac{12}{13} and cos⁡α=−513\cos \alpha = -\frac{5}{13}, so we can substitute these values into the equation and solve for tan⁡α\tan \alpha.

tan⁡α=sin⁡αcos⁡α\tan \alpha = \frac{\sin \alpha}{\cos \alpha}

tan⁡α=1213−513\tan \alpha = \frac{\frac{12}{13}}{-\frac{5}{13}}

tan⁡α=−125\tan \alpha = -\frac{12}{5}

Q: What is the value of tan⁡β\tan \beta?

A: To find the value of tan⁡β\tan \beta, we can use the definition of the tangent function: tan⁡β=sin⁡βcos⁡β\tan \beta = \frac{\sin \beta}{\cos \beta}. We are given that sin⁡β=35\sin \beta = \frac{3}{5} and cos⁡β=45\cos \beta = \frac{4}{5}, so we can substitute these values into the equation and solve for tan⁡β\tan \beta.

tan⁡β=sin⁡βcos⁡β\tan \beta = \frac{\sin \beta}{\cos \beta}

tan⁡β=3545\tan \beta = \frac{\frac{3}{5}}{\frac{4}{5}}

tan⁡β=34\tan \beta = \frac{3}{4}

Q: What is the value of sin⁡α+cos⁡α\sin \alpha + \cos \alpha?

A: To find the value of sin⁡α+cos⁡α\sin \alpha + \cos \alpha, we can substitute the values of sin⁡α\sin \alpha and cos⁡α\cos \alpha into the equation.

sin⁡α+cos⁡α=1213−513\sin \alpha + \cos \alpha = \frac{12}{13} - \frac{5}{13}

sin⁡α+cos⁡α=−513\sin \alpha + \cos \alpha = -\frac{5}{13}

Q: What is the value of sin⁡β+cos⁡β\sin \beta + \cos \beta?

A: To find the value of sin⁡β+cos⁡β\sin \beta + \cos \beta, we can substitute the values of sin⁡β\sin \beta and cos⁡β\cos \beta into the equation.

sin⁡β+cos⁡β=35+45\sin \beta + \cos \beta = \frac{3}{5} + \frac{4}{5}

sin⁡β+cos⁡β=75\sin \beta + \cos \beta = \frac{7}{5}

Q: What is the value of sin⁡αcos⁡β\sin \alpha \cos \beta?

A: To find the value of sin⁡αcos⁡β\sin \alpha \cos \beta, we can substitute the values of sin⁡α\sin \alpha and cos⁡β\cos \beta into the equation.

sin⁡αcos⁡β=1213⋅45\sin \alpha \cos \beta = \frac{12}{13} \cdot \frac{4}{5}

sin⁡αcos⁡β=4865\sin \alpha \cos \beta = \frac{48}{65}

Q: What is the value of cos⁡αsin⁡β\cos \alpha \sin \beta?

A: To find the value of cos⁡αsin⁡β\cos \alpha \sin \beta, we can substitute the values of cos⁡α\cos \alpha and sin⁡β\sin \beta into the equation.

cos⁡αsin⁡β=−513⋅35\cos \alpha \sin \beta = -\frac{5}{13} \cdot \frac{3}{5}

cos⁡αsin⁡β=−1565\cos \alpha \sin \beta = -\frac{15}{65}

Q: What is the value of sin⁡αsin⁡β\sin \alpha \sin \beta?

A: To find the value of sin⁡αsin⁡β\sin \alpha \sin \beta, we can substitute the values of sin⁡α\sin \alpha and sin⁡β\sin \beta into the equation.

sin⁡αsin⁡β=1213⋅35\sin \alpha \sin \beta = \frac{12}{13} \cdot \frac{3}{5}

sin⁡αsin⁡β=3665\sin \alpha \sin \beta = \frac{36}{65}

Q: What is the value of cos⁡αcos⁡β\cos \alpha \cos \beta?

A: To find the value of cos⁡αcos⁡β\cos \alpha \cos \beta, we can substitute the values of cos⁡α\cos \alpha and cos⁡β\cos \beta into the equation.

cos⁡αcos⁡β=−513⋅45\cos \alpha \cos \beta = -\frac{5}{13} \cdot \frac{4}{5}

cos⁡αcos⁡β=−2065\cos \alpha \cos \beta = -\frac{20}{65}