Question

If sin $\theta =\frac{3}{5}$ and cos $\varphi =\frac{-12}{13}$ where $\theta$and $\varphi$ both lie in the second quadrant, find the values of

(i) sin $(\theta -\varphi )$, (ii) cos $(\theta +\varphi )$, (iii) tan $(\theta -\varphi )$.

Answer 1

Given : sin $\theta =\frac{3}{5}$ and cos $\varphi =\frac{-12}{13}$

Since $\theta$ lies in the second quadrant, we have

sin $\theta$ > 0, cos $\theta$ <0 and tan $\theta$ <0.

Again, since $\varphi$ lies in the second quadrant, we have

sin $\varphi >0$ , cos $\varphi <0$ and tan $\varphi <0$ Now, cos $\theta$ = -$\sqrt{1-{\text{sin}}^2\theta }=-\sqrt{1-\frac{9}{25}}=-\sqrt{\frac{16}{25}}=\frac{-4}{5}$

$\text{tan}\theta =\frac{\text{sin}\theta }{\text{cos}\theta }=\frac{3}{5}\times (-\frac{5}{4})=\frac{-3}{4}$ $\text{sin}\varphi =+\sqrt{1-{\text{cos}}^2\varphi }=+\sqrt{1-\frac{144}{169}}=+\sqrt{\frac{25}{169}}=+\frac{5}{13}$

$\text{tan}\varphi =\frac{\text{sin}\varphi }{\text{cos}\varphi }=\frac{5}{13}\times \left(\frac{-13}{12}\right)=\frac{-5}{12}$

$\therefore \left(i\right)\text{sin}(\theta -\varphi )=\text{sin}\theta \text{cos}\varphi -cos\theta \text{sin}\varphi$

=$\{\frac{3}{5}\times \left(\frac{-12}{13}\right)\}-\left\{(\frac{-4}{5}\times \frac{5}{13})\right\}=(\frac{-36}{65}+\frac{20}{65})=\frac{-16}{65}$

(ii)$\text{cos}(\theta +\varphi )=\text{cos}\theta \text{cos}\varphi -\text{sin}\theta \text{sin}\varphi$

=$\left\{(\frac{-4}{5}\times \left(\frac{-12}{13}\right))\right\}-\{\frac{3}{5}\times \frac{5}{13}\}=(\frac{48}{65}-\frac{15}{65})=\frac{33}{65}$