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- Question, Exercise 10.1
- n \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in ... lpha$ is in 3rd quadrant, \\ $\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. We have an ide... quadrant, \begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{144}{169}}... \beta \\ &=\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)+\left(-\frac{3}{5}\right)\left(\frac{
- Question 2, Exercise 10.2
- {169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using doubl... \\ &=2\left( -\dfrac{5}{13} \right)\left( \dfrac{12}{13} \right)\end{align} $$\implies \bbox[4px,border:2px solid black]{\sin 2\theta=-\dfrac{120}{169}.}$$ =====Question 2(ii)===== If $\sin \t... {169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using doubl
- Question 2, Exercise 10.1
- on 2(i)==== Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }{3}-\dfrac{\pi }{4}$ and using the ide... ion 2(iv)=== Evaluate exactly:$\tan \dfrac{5\pi }{12}$ ==Solution== We rewrite $\dfrac{5\pi }{12}$ as $\dfrac{\left( 2+3 \right)\pi }{12}$ or $\dfrac{\pi
- Question 3, Exercise 10.1
- 5}.\frac{3}{5}-\frac{3}{5}.\frac{4}{5}\\ &=\dfrac{12}{25}-\frac{12}{25}\\ \cos \left( u+v \right)&=0\end{align} =====Question 3(ii)===== If $\sin u=\df... 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} =====Question 3(iii
- Question 5, Exercise 10.1
- 4}{169}} \\ \Rightarrow \quad \sin\beta&=-\dfrac{12}{13} \end{align} Now \begin{align} \sin(\alph... {13}\right)+\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)\\ &=-\frac{3}{13}+\frac{48}{65}\end{a... 4}{169}} \\ \Rightarrow \quad \sin\beta&=-\dfrac{12}{13} \end{align} Now \begin{align} \cos(\alph... }{13}\right)-\left(\frac{3}{5}\right)\left(-\frac{12}{13}\right)\\ &=-\frac{20}{65}+\frac{36}{65}\end{
- Question 1, Exercise 10.3
- as sum or difference $\sin {{55}^{\circ }}\cos {{123}^{\circ }}$. ====Solution==== We have an identi... $ Put $\alpha ={{55}^{\circ }}$ and $\beta ={{123}^{\circ }}$ \begin{align}2\sin {{55}^{\circ }}\cos {{123}^{\circ }}&=\sin ({{55}^{\circ }}+{{123}^{\circ }})+\sin ({{55}^{\circ }}-{{123}^{\circ }})\\ &=\sin
- Question 1, Exercise 10.1
- single angle. $\dfrac{\tan {{35}^{\circ }}-\tan {{12}^{\circ }}}{1+\tan {{35}^{\circ }}\tan {{12}^{\circ }}}$ ==== Solution ==== As \begin{align}\tan (\a... \begin{align}\dfrac{\tan {{35}^{\circ }}-\tan {{12}^{\circ }}}{1+\tan {{35}^{\circ }}\tan {{12}^{\circ }}}&=\tan \left( 35-12 \right)\\ &=\tan {{23}^{\c
- Question 6, Exercise 10.2
- half angle identities to evaluate exactly $sin{{112.5}^{\circ }}$. ====Solution==== Because ${{112.5}^{\circ }}=\dfrac{{{225}^{\circ }}}{2}$, the $\dfrac... frac{{{225}^{\circ }}}{2}$, so we can find $sin{{112.5}^{\circ }}$by using half angle identity as, \begin{align} sin{{112.5}^{\circ }}&=\sin \dfrac{{{225}^{\circ }}}{2}=\s
- Question 3, Exercise 10.3
- Prove that $$\dfrac{\sin {{135}^{\circ }}-\cos {{120}^{\circ }}}{\sin {{135}^{\circ }}+\cos {{120}^{\circ }}}=3+2\sqrt{2}.$$ ====Solution==== \begin{align}L.H.S.&=\dfrac{\sin {{135}^{\circ }}-\cos {{120}^{\circ }}}{\sin {{135}^{\circ }}+\cos {{120}^{\circ }}}\\ &=\dfrac{\dfrac{1}{\sqrt{2}}-\left( -\df
- Question11 and 12, Exercise 10.1
- ====== Question11 and 12, Exercise 10.1 ====== Solutions of Question 11 and 12 of Exercise 10.1 of Unit 10: Trigonometric Identi... cot\dfrac{\gamma}{2}.\end{align} =====Question 12===== If $\alpha +\beta +\gamma ={{180}^{\circ}}$,
- Question 1, Exercise 10.2
- ox[4px,border:2px solid black]{\cos2\theta=\dfrac{12}{13}}$$ \begin{align}\tan 2\theta &=\dfrac{\sin 2... heta }{\cos 2\theta }=\dfrac{\frac{-5}{13}}{\frac{12}{13}}\end{align} $$ \implies \bbox[4px,border:2px solid black]{\tan 2\theta=-\dfrac{5}{12}}$$ ====Go to==== <text align="right"><btn type="
- Question 5, Exercise 10.3
- cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( \cos {{120}^{\circ }}+\cos {{40}^{\circ }} \right)\cos {{20... using (1)}\\ &=-\dfrac{\sqrt{3}}{4}\left( \cos {{120}^{\circ }}-\cos {{40}^{\circ }} \right)\sin {{20
- Question 5, Exercise 10.3
- cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( \cos {{120}^{\circ }}+\cos {{40}^{\circ }} \right)\cos {{20... }^{\circ }}\\ &=-\dfrac{\sqrt{3}}{4}\left( \cos {{120}^{\circ }}-\cos {{40}^{\circ }} \right)\sin {{20
- Question 9 and 10, Exercise 10.1
- lign="right"><btn type="success">[[math-11-kpk:sol:unit10:ex10-1-p10|Question 11,12 >]]</btn></text>
- Question 13, Exercise 10.1
- align="left"><btn type="primary">[[math-11-kpk:sol:unit10:ex10-1-p10|< Question 11,12]]</btn></text>