### Sum and Difference of Angles

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The following are trigonometric formulas for the sum and difference of angles:

$\text{sin}(A+B)=\text{sin}A.\text{cos}B+\text{sin}B.\text{cos}A$

$\text{cos}(A+B)=\text{cos}A.\text{cos}B-\text{sin}A.\text{sin}B$

$\text{sin}(A-B)=\text{sin}A.\text{cos}B-\text{sin}B.\text{cos}A$

$\text{cos}(A-B)=\text{cos}A.\text{cos}B+\text{sin}A.\text{sin}B$

$\displaystyle \text{tan}(A+B)=\frac{\text{tan}A+\text{tan}B}{1-\text{tan}A.\text{tan}B}$

$\displaystyle \text{tan}(A-B)=\frac{\text{tan}A-\text{tan}B}{1+\text{tan}A.\text{tan}B}$

Example:

It is known that $\text{sin }A= \frac{2}{5}$and $\text{sin }B = \frac{3}{5}$. Determine the value of:

1. $\text{sin}(A+B)$

2. $\text{cos}(A-B)$

3. $\text{tan}(A+B)$

Solution:

First we have to find the value of cos$A$, cos$B$, tan$A$, and tan$B$. By using the trigonometric identity:

$\displaystyle \text{cos}A=\sqrt{1-\text{sin}^2A}$

$\displaystyle \text{cos}B=\sqrt{1-\text{sin}^2B}$

$\displaystyle \text{tan}A=\frac{\text{sin}A}{\text{cos}A}$

and

$\displaystyle \text{tan}B=\frac{\text{sin}B}{\text{cos}B}$

Than get:

$\displaystyle \text{cos}A=\frac{1}{5}\sqrt{21}$

$\displaystyle \text{cos}B=\frac{4}{5}$

$\displaystyle \text{tan}A=\frac{2}{21}\sqrt{21}$

and

$\displaystyle \text{tan}B=\frac{3}{4}$

So that:

1. $\displaystyle \text{sin}(A+B)=\frac{3}{25}\sqrt{21}+\frac{8}{25}$

2. $\displaystyle \text{cos}(A-B)=\frac{4}{25}\sqrt{21}+\frac{6}{25}$

3. $\displaystyle \text{tan}(A+B)=\frac{\frac{2}{21}\sqrt{21}+\frac{3}{4}}{1-\frac{1}{4}\sqrt{21}}$

Thanks for view, see you and happy learning.