Question

If $\cos A=\frac{24}{25}$ and $\cos B=\frac{3}{5}$ and $A$ lies in first quadrant,find $\cos (A+B)$ and $\cos (A-B)$

Answer 1

As angles $A$ and $B$ are acute angles all trigonometric ratios are positive, hence

as $\cos A=\frac{24}{25}$

$\sin A=\sqrt{1-{\left(\frac{24}{25}\right)}^2}=\sqrt{1-\frac{576}{625}}=\frac{49}{625}=\frac{7}{25}$

and as $\cos B=\frac{3}{5}$

$\sin B=\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

Hence, $\cos (A+B)=\cos A\cos B-\sin A\sin B$

$=\frac{24}{25}\times \frac{3}{5}-\frac{7}{25}\times \frac{4}{5}$

$=\frac{72-28}{125}=\frac{44}{125}$

and $\cos (A-B)=\cos A\cos B+\sin A\sin B$

$=\frac{24}{25}\times \frac{3}{5}+\frac{7}{25}\times \frac{4}{5}$

$=\frac{72+28}{125}=\frac{100}{125}=\frac{4}{5}$