# Question 3, Exercise 10.1

Solutions of Question 3 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

If $\sin u=\dfrac{3}{5}$ and $\sin v=\dfrac{4}{5}$ where$u$ and $v$ are between $0$ and $\dfrac{\pi }{2}$, evaluate each of the following exactly $\cos \left( u+v \right)$

Given $\sin u=\dfrac{3}{5},$ $0\le u\le \dfrac{\pi }{2}.$ $\sin v=\dfrac{4}{5},$ $0\le v\le \dfrac{\pi }{2}.$ We know $\cos u=\pm \sqrt{1-{{\sin }^{2}}u}$ As $u$ lies in first quadrant and $\cos$ is $+ve$ in first quadrant . \begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\sin }^{2}}u}\\ &=\sqrt{1-{{\left( \frac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9}{25}}\,\,=\sqrt{\dfrac{16}{25}}\\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Also $\cos v=\sqrt{1-{{\sin }^{2}}v}$ As $v$ lies in first quadrant and $\cos$ is $+ve$ in first quadrant. \begin{align}\Rightarrow \,\,\,\,\,\cos v&=\sqrt{1-{{\sin }^{2}}v}\\ &=\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{16}{25}}\,\,=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \,\,\,\cos v &=\frac{3}{5}\end{align} Now \begin{align}\cos \left( u+v \right)&=\cos u\cos v-\sin u\sin v\\ &=\frac{4}{5}.\frac{3}{5}-\frac{3}{5}.\frac{4}{5}\\ &=\dfrac{12}{25}-\frac{12}{25}\\ \cos \left( u+v \right)&=0\end{align}

If $\sin u=\dfrac{3}{5}$ and $\sin v=\dfrac{4}{5}$ and $u$ and $v$ are between $0$ and $\dfrac{\pi }{2}$, evaluate each of the following exactly $\tan \left( u-v \right)$

Given \begin{align}\sin u&=\dfrac{3}{5},\,\,\,\, 0\le u\le \dfrac{\pi }{2}.\\ \sin v&=\dfrac{4}{5},\,\,\,\,\,0\le v\le \dfrac{\pi }{2}.\end{align} We know $\cos u=\pm \sqrt{1-{{\sin }^{2}}u}$
As $u$ lies in first quadrant and $\cos$ is $+ve$ in first quadrant.
\begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\sin }^{2}}u}\\ &=\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9}{25}}\,\,=\sqrt{\dfrac{16}{25}}\\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Also $\cos v=\sqrt{1-{{\sin }^{2}}v}$
As $v$ lies in first quadrant and $\cos$ is $+ve$ in first quadrant. \begin{align}\Rightarrow \,\,\,\,\,\cos v&=\sqrt{1-{{\sin }^{2}}v}\\ &=\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{16}{25}}\,\,=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \,\,\,\,\,\cos v&=\dfrac{3}{5}\\ \tan u&=\dfrac{\sin u}{\cos u}\\ &=\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}\\ \tan u&=\dfrac{3}{4}\end{align} Similarly, $\tan v=\dfrac{4}{3}$
Now \begin{align}\tan (u-v)&=\dfrac{\dfrac{3}{4}-\dfrac{4}{3}}{1+\dfrac{3}{4}\dfrac{4}{3}}\\ &=\dfrac{\dfrac{9-16}{12}}{1+\dfrac{12}{12}}\\ &=\dfrac{\dfrac{-7}{12}}{2}\\ &=\dfrac{-7}{24}\end{align}

If $\sin u=\dfrac{3}{5}$ and $\sin v=\dfrac{4}{5}$ and $u$ and $v$ are between $0$ and $\dfrac{\pi }{2}$, evaluate each of the following exactly $\sin \left( u-v \right)$

Given \begin{align}\sin u&=\dfrac{3}{5},\,\,\,\, 0\le u\le \dfrac{\pi }{2}.\\ \sin v&=\dfrac{4}{5},\,\,\,\, 0\le v\le \dfrac{\pi }{2}.\end{align} We know $\cos u=\pm \sqrt{1-{{\sin }^{2}}u}$
As $u$ lies in first quadrant and $\cos$ is $+ve$ in first quadrant.
\begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\sin }^{2}}u}\\ &=\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9}{25}}\,\,=\sqrt{\frac{16}{25}}\\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Also $\cos v=\sqrt{1-{{\sin }^{2}}v}$
As $v$ lies in first quadrant and $\cos$ is $+ve$ in first quadrant. \begin{align}\Rightarrow \,\,\,\,\,\cos v&=\sqrt{1-{{\sin }^{2}}v}\\ &=\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{16}{25}}\,\,=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \,\,\,\,\,\cos v&=\dfrac{3}{5}\end{align} Now \begin{align}\sin (u-v)&=\sin u\cos v-\cos u\sin v\\ &=\dfrac{3}{5}.\dfrac{3}{5}-\dfrac{4}{5}.\dfrac{4}{5}\\ &=\dfrac{9}{25}-\dfrac{16}{25}\\ \sin \left( u-v \right)&=\dfrac{-7}{25}\end{align}

If $\sin u=\dfrac{3}{5}$ and $\sin v=\dfrac{4}{5}$ where$u$ and $v$ are between $0$ and $\dfrac{\pi }{2}$, evaluate each of the following exactly $\cos \left( u+v \right)$

Given $\sin u=\dfrac{3}{5},$ $0\le u\le \dfrac{\pi }{2}.$ $\sin v=\dfrac{4}{5},$ $0\le v\le \dfrac{\pi }{2}.$ We know $\cos u=\pm \sqrt{1-{{\sin }^{2}}u}$ As $u$ lies in first quadrant and $\cos$ is $+ve$ in first quadrant . \begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\sin }^{2}}u}\\ &=\sqrt{1-{{\left( \frac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9}{25}}\,\,=\sqrt{\dfrac{16}{25}}\\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Also $\cos v=\sqrt{1-{{\sin }^{2}}v}$ As $v$ lies in first quadrant and $\cos$ is $+ve$ in first quadrant. \begin{align}\Rightarrow \,\,\,\,\,\cos v&=\sqrt{1-{{\sin }^{2}}v}\\ &=\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{16}{25}}\,\,=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \quad \cos v &=\frac{3}{5}\end{align} Now \begin{align}\cos \left( u-v \right)&=\cos u\cos v+\sin u\sin v\\ &=\frac{4}{5}.\frac{3}{5}+\frac{3}{5}.\frac{4}{5}\\ &=\dfrac{12}{25}+\frac{12}{25}\\ \cos \left( u+v \right)&=\frac{24}{25}\end{align}