If $\sin \alpha = 3/5, \cos \beta = 12/13$, what's $\cos(\alpha + \beta)$?

If $\sin \alpha = 3/5, \cos \beta = 12/13$, what's $\cos(\alpha + \beta)$?

Question

Given that $\sin \alpha = \tfrac{3}{5}$ and $\cos \beta = \tfrac{12}{13}$, prove that one possible value of $\cos(\alpha + \beta)$ is $\tfrac{33}{65}$ and find all the other possible values.

Suggestion

Given that $\sin \alpha = \dfrac{3}{5}$ and $\cos \beta = \dfrac{12}{13}$, prove that one possible value of $\cos(\alpha + \beta)$ is $\dfrac{33}{65}$…

Could we draw a diagram with right-angle triangles and the angles $\alpha$, $\beta$ and $\alpha+\beta$?

If two sides of a right-angle triangle are $3$ and $5$, what is the third side?

Solution

Given that $\sin \alpha = \dfrac{3}{5}$ and $\cos \beta = \dfrac{12}{13}$, prove that one possible value of $\cos(\alpha + \beta)$ is $\tfrac{33}{65}$ and find all the other possible values.

The diagram tells us that $\cos(\alpha+\beta) = \cos\alpha\cos\beta-\sin\alpha\sin\beta$ for $\alpha+\beta < 90^\circ$. This is, in fact, true for all angles $\alpha$ and $\beta$.

Now $\cos^2 \theta + \sin^2 \theta = 1$ for all $\theta$, which means $\cos \alpha = \pm\dfrac{4}{5}$, and $\sin \beta = \pm \dfrac{5}{13}$.

Thus $\cos(\alpha+\beta) = \cos\alpha\cos\beta-\sin\alpha\sin\beta = \pm \dfrac{48}{65}\pm \dfrac{15}{65},$ where the signs are independent of each other.

We therefore have four solutions, $\cos(\alpha+\beta) = \pm \dfrac{63}{65}$, or $\pm \dfrac{33}{65}$.