33. -/2 points details my notes larpcalclimaga8 5.4.048. find the exact value of the trigonometric…

Explanation:

Step1: Find cos(u)

Since $\sin^{2}u+\cos^{2}u = 1$ and $u$ is in Quadrant III where $\cos(u)<0$. Given $\sin(u)=-\frac{12}{13}$, then $\cos(u)=-\sqrt{1 - \sin^{2}u}=-\sqrt{1-\left(-\frac{12}{13}\right)^{2}}=-\sqrt{1-\frac{144}{169}}=-\sqrt{\frac{169 - 144}{169}}=-\frac{5}{13}$.

Step2: Find sin(v)

Since $\sin^{2}v+\cos^{2}v = 1$ and $v$ is in Quadrant III where $\sin(v)<0$. Given $\cos(v)=-\frac{4}{5}$, then $\sin(v)=-\sqrt{1-\cos^{2}v}=-\sqrt{1-\left(-\frac{4}{5}\right)^{2}}=-\sqrt{1 - \frac{16}{25}}=-\sqrt{\frac{25-16}{25}}=-\frac{3}{5}$.

Step3: Use the cosine - sum formula for $\cos(u + v)$

The formula is $\cos(u + v)=\cos(u)\cos(v)-\sin(u)\sin(v)$. Substitute $\cos(u)=-\frac{5}{13}$, $\sin(u)=-\frac{12}{13}$, $\cos(v)=-\frac{4}{5}$, and $\sin(v)=-\frac{3}{5}$ into the formula: [ \begin{align*} \cos(u + v)&=\left(-\frac{5}{13}\right)\times\left(-\frac{4}{5}\right)-\left(-\frac{12}{13}\right)\times\left(-\frac{3}{5}\right)\ &=\frac{20}{65}-\frac{36}{65}\ &=-\frac{16}{65} \end{align*} ]

Step4: Use the cosine - difference formula for $\cos(v - u)$

The formula is $\cos(v - u)=\cos(v)\cos(u)+\sin(v)\sin(u)$. Substitute the values: [ \begin{align*} \cos(v - u)&=\left(-\frac{4}{5}\right)\times\left(-\frac{5}{13}\right)+\left(-\frac{3}{5}\right)\times\left(-\frac{12}{13}\right)\ &=\frac{20}{65}+\frac{36}{65}\ &=\frac{56}{65} \end{align*} ]

Answer:

For $\cos(u + v)=-\frac{16}{65}$ For $\cos(v - u)=\frac{56}{65}$