Question

1. find the exact value of the trigonometric function given that sin u = 5/13 and u is a qi angle, and cos v=-3/5 and v is a qii angle.

(a) sin(u + v)=
(b) cos(u - v)

1. find the exact value of the trigonometric function given that sin x=-4/5 and cos x > 0, and cos y=-12/13 and sin y > 0.

(a) sin(x - y)=
(b) cos(x + y)

Explanation:

Step1: Find $\cos u$ using $\sin^{2}u+\cos^{2}u = 1$

Since $\sin u=\frac{5}{13}$ and $u$ is in QI, $\cos u=\sqrt{1 - \sin^{2}u}=\sqrt{1-(\frac{5}{13})^{2}}=\sqrt{\frac{169 - 25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}$

Step2: Find $\sin v$ using $\sin^{2}v+\cos^{2}v = 1$

Since $\cos v=-\frac{3}{5}$ and $v$ is in QII, $\sin v=\sqrt{1-\cos^{2}v}=\sqrt{1 - (-\frac{3}{5})^{2}}=\sqrt{\frac{25 - 9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

Step3: Calculate $\sin(u + v)$ using the formula $\sin(A + B)=\sin A\cos B+\cos A\sin B$

$\sin(u + v)=\sin u\cos v+\cos u\sin v=\frac{5}{13}\times(-\frac{3}{5})+\frac{12}{13}\times\frac{4}{5}=\frac{-15 + 48}{65}=\frac{33}{65}$

Step4: Calculate $\cos(u - v)$ using the formula $\cos(A - B)=\cos A\cos B+\sin A\sin B$

$\cos(u - v)=\cos u\cos v+\sin u\sin v=\frac{12}{13}\times(-\frac{3}{5})+\frac{5}{13}\times\frac{4}{5}=\frac{-36+20}{65}=-\frac{16}{65}$

For problem 17:

Step1: Find $\cos x$ using $\sin^{2}x+\cos^{2}x = 1$

Since $\sin x=-\frac{4}{5}$ and $\cos x>0$, $\cos x=\sqrt{1-\sin^{2}x}=\sqrt{1-(-\frac{4}{5})^{2}}=\sqrt{\frac{25 - 16}{25}}=\frac{3}{5}$

Step2: Find $\sin y$ using $\sin^{2}y+\cos^{2}y = 1$

Since $\cos y=-\frac{12}{13}$ and $\sin y>0$, $\sin y=\sqrt{1-\cos^{2}y}=\sqrt{1-(-\frac{12}{13})^{2}}=\sqrt{\frac{169 - 144}{169}}=\frac{5}{13}$

Step3: Calculate $\sin(x - y)$ using the formula $\sin(A - B)=\sin A\cos B-\cos A\sin B$

$\sin(x - y)=\sin x\cos y-\cos x\sin y=(-\frac{4}{5})\times(-\frac{12}{13})-\frac{3}{5}\times\frac{5}{13}=\frac{48 - 15}{65}=\frac{33}{65}$

Step4: Calculate $\cos(x + y)$ using the formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$

$\cos(x + y)=\cos x\cos y-\sin x\sin y=\frac{3}{5}\times(-\frac{12}{13})-(-\frac{4}{5})\times\frac{5}{13}=\frac{-36 + 20}{65}=-\frac{16}{65}$

Answer:

16(a) $\frac{33}{65}$
16(b) $-\frac{16}{65}$
17(a) $\frac{33}{65}$
17(b) $-\frac{16}{65}$