Using the sum and difference formulas

The formula needs the sine and the cosine of each angle. The missing ones come from $\displaystyle {\sin^2+\cos^2}={1}$ and the quadrant.

$\displaystyle {\sin\alpha}={\frac{3}{5}}$, and the angle $\displaystyle \alpha$ is in the second quadrant.
$\displaystyle {\cos\beta}={-\frac{5}{13}}$, and the angle $\displaystyle \beta$ is in the second quadrant.
Find $\displaystyle \sin(\alpha+\beta)$.

• $\displaystyle {\cos\alpha}={-\frac{4}{5}}$ and $\displaystyle {\sin\beta}={\frac{12}{13}}$ (in the second quadrant the cosine is negative and the sine positive).
• $\displaystyle {\sin(\alpha+\beta)}={\sin\alpha\cos\beta+\cos\alpha\sin\beta}={\frac{3}{5}\cdot\left(-\frac{5}{13}\right)+\left(-\frac{4}{5}\right)\cdot\frac{12}{13}}={-\frac{63}{65}}$.

Such an angle can be written with an inverse function: $\displaystyle \sin^{-1}\frac{4}{5}$ is the acute angle whose sine is $\displaystyle \frac{4}{5}$, and $\displaystyle \cos^{-1}\left(-\frac{5}{13}\right)$ is the angle between $\displaystyle 0$ and $\displaystyle \pi$ whose cosine is $\displaystyle -\frac{5}{13}$.

Divide the numerator and the denominator of $\displaystyle \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}$ by $\displaystyle \cos\alpha\cos\beta$ and you get:

An example: $\displaystyle {\tan\alpha}={\frac{1}{2}}$ and $\displaystyle {\tan\beta}={\frac{1}{3}}$, and both angles are acute.

• $\displaystyle {\tan(\alpha+\beta)}={\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}}={1}$.
• $\displaystyle 0<\alpha+\beta<\pi$, so $\displaystyle {\alpha+\beta}={\frac{\pi}{4}}$.

A sum of a sine and a cosine of the same angle is one sine, shifted and stretched. Take out a common factor so that the brackets hold the $\displaystyle \cos$ and the $\displaystyle \sin$ of an angle from the table:

The factor $\displaystyle 2$ is $\displaystyle \sqrt{1^2+\left(\sqrt{3}\right)^2}$. From this it is plain that the expression’s greatest value is $\displaystyle 2$.

$\displaystyle {\sin\alpha}={\frac{4}{5}}$ and $\displaystyle {\sin\beta}={\frac{5}{13}}$, and both angles are acute: they are $\displaystyle \sin^{-1}\frac{4}{5}$ and $\displaystyle \sin^{-1}\frac{5}{13}$.
Work out $\displaystyle \cos(\alpha-\beta)$: write the formula with the numbers, then the result.

$\displaystyle {\tan\alpha}={2}$ and $\displaystyle {\tan\beta}={3}$, and both angles are acute.
Work out $\displaystyle \tan(\alpha+\beta)$.

What is the angle $\displaystyle \alpha+\beta$? Explain.

Write as one sine, until you reach $\displaystyle 2\sin\left(x-\frac{\pi}{6}\right)$.

Simplify until you reach $\displaystyle 2\sin x\cos y$.

Simplify until you reach $\displaystyle \cos^2x-\sin^2y$.