Suppose $\alpha$ is interval $$\pi/2 \leq \alpha \leq \pi$$ and $$ \cos(\alpha) = - 1/3 $$ and $\beta$ is in the interval $$0 \leq \beta \leq \pi/2$$ and $$ \sin\beta = 2/5. $$ Use the facts above to find the exact value of
$$\cos(\alpha + \beta) $$
I think that I should be able to use the identity to solve this but the information that it gave has confused me. How would I solve it and what's the right answer?
$\endgroup$ 53 Answers
$\begingroup$HINT, we have that:
$$\cos\left(\alpha\right)=-\frac{1}{3}\space\Longleftrightarrow\space\alpha=2\pi\text{n}\pm\arccos\left(-\frac{1}{3}\right)\tag1$$
Where $\text{n}\in\mathbb{Z}$
Now, we also know that:
$$\frac{\pi}{2}\space<\space\alpha\space<\space\pi\space\Longleftrightarrow\space\frac{\pi}{2}\space<\space2\pi\text{n}\pm\arccos\left(-\frac{1}{3}\right)\space<\space\pi\space\Longleftrightarrow\space\text{n}=0\tag2$$
Another way:
$$\cos\left(\alpha+\beta\right)=\cos\left(\alpha\right)\cos\left(\beta\right)-\sin\left(\alpha\right)\sin\left(\beta\right)=-\frac{1}{3}\cdot\cos\left(\beta\right)-\sin\left(\alpha\right)\cdot\frac{2}{5}\tag3$$
Now, you can use:
$$\cos^2\left(\alpha\right)+\sin^2\left(\alpha\right)=\left(-\frac{1}{3}\right)^2+\sin^2\left(\alpha\right)=1\space\Longleftrightarrow\space\sin^2\left(\alpha\right)=\frac{8}{9}\tag4$$
And:
$$\cos^2\left(\beta\right)+\sin^2\left(\beta\right)=\cos^2\left(\beta\right)+\left(\frac{2}{5}\right)^2=1\space\Longleftrightarrow\space\cos^2\left(\beta\right)=\frac{21}{25}\tag5$$
$\endgroup$ 3 $\begingroup$Hint: Form $2$ triangle with angle $\alpha$ and $\beta$ and put the values in this $$\cos{(A+B)}=\cos A\cos B-\sin A\sin B$$
$\endgroup$ $\begingroup$Since $\alpha \in Q2$, then $\sin (\alpha) = \frac{2 \sqrt{2}}{3}$ and since $\beta \in Q1$, then $\cos (\beta)= \frac{\sqrt{21}}{5}$.
Hence,
\begin{eqnarray} \cos (\alpha + \beta) &=& \cos (\alpha) \cos (\beta) - \sin (\alpha) \sin (\beta) \\ &=& \left( - \frac{1}{3} \right) \left( \frac{\sqrt{21}}{5} \right) - \left( \frac{2 \sqrt{2}}{3} \right) \left( \frac{2}{5} \right) \\ &=&- \frac{\sqrt{21} + 4 \sqrt{2}}{15} \end{eqnarray}
$\endgroup$ 3
