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- Question 3, Exercise 10.1
- d $\sin v=\dfrac{4}{5}$ where$u$ and $v$ are between $0$ and $\dfrac{\pi }{2}$, evaluate each of the... sin v=\dfrac{4}{5},$ $0\le v\le \dfrac{\pi }{2}.$ We know $\cos u=\pm \sqrt{1-{{\sin }^{2}}u}$ As $u$... nd $\sin v=\dfrac{4}{5}$ and $u$ and $v$ are between $0$ and $\dfrac{\pi }{2}$, evaluate each of th... },\,\,\,\,\,0\le v\le \dfrac{\pi }{2}.\end{align} We know $\cos u=\pm \sqrt{1-{{\sin }^{2}}u}$ \\ As $
- Question 1, Review Exercise 10
- {3}}{2}$ \\ <btn type="link" collapse="a1">See Answer</btn><collapse id="a1" collapsed="true">(B): $\d... sqrt{3}$ \\ <btn type="link" collapse="a2">See Answer</btn><collapse id="a2" collapsed="true">(B): $\d... c{7}{6}$ \\ <btn type="link" collapse="a3">See Answer</btn><collapse id="a3" collapsed="true">(B): $\d... (d) $1$ \\ <btn type="link" collapse="a4">See Answer</btn><collapse id="a4" collapsed="true">(B): $\d
- Question 2, Exercise 10.1
- e exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }{3}-\d... luate exactly:$\tan {{75}^{\circ }}$ ==Solution== We rewrite ${{75}^{\circ }}$ as ${{45}^{\circ }}+{{3... uate exactly:$\tan {{105}^{\circ }}$ ==Solution== We rewrite ${{105}^{{}^\circ }}$ as ${{60}^{{}^\circ... ate exactly:$\tan \dfrac{5\pi }{12}$ ==Solution== We rewrite $\dfrac{5\pi }{12}$ as $\dfrac{\left( 2+3
- Question 6, Exercise 10.2
- $\dfrac{\theta }{2}=\dfrac{{{30}^{\circ }}}{2}$, we can find $\cos {{15}^{\circ }}$by using half angl... frac{\theta }{2}=\dfrac{{{135}^{\circ }}}{2}$, so we can find $\tan {{67.5}^{\circ }}$by using half an... frac{\theta }{2}=\dfrac{{{225}^{\circ }}}{2}$, so we can find $sin{{112.5}^{\circ }}$by using half ang... dfrac{\theta }{2}=\dfrac{\dfrac{\pi }{4}}{2}$, so we can find $\cos \dfrac{\pi }{8}$by using half angl
- Question 2, Exercise 10.3
- circ }}+\sin {{43}^{\circ }}.$$ ====Solution==== We have an identity: $$\sin \alpha +\sin \beta =2\si... right)\end{align} Since $cos(-\theta)=cos\theta$, we have $$\sin {{37}^{\circ }}+\sin {{43}^{\circ }}=... \circ }}-\cos {{82}^{\circ }}$. ====Solution==== We have an identity: $$\cos \alpha -\cos \beta =-2\s... .\\ &=-2\sin(59^\circ)\sin(-23^\circ) \end{align} We have $\sin(-\theta)=-\sin\theta$, therefore $$\co
- Question 13, Exercise 10.1
- \varphi =\dfrac{3}{5}.\end{align} Thus, from (1), we have \begin{align}&4\sin\theta +3\cos\theta \\ &=... varphi =\dfrac{8}{17}.\end{align} Thus, from (1), we have \begin{align}&15\sin\theta +8\cos\theta \\ &... \dfrac{-5}{\sqrt{29}}.\end{align} Thus, from (1), we have \begin{align}&2\sin\theta -5\cos\theta \\ &=... =\dfrac{1}{\sqrt{2}}.\end{align} Thus, from (1), we have \begin{align}&\sin\theta+\cos\theta \\ &=\sq
- Question 2, Exercise 10.2
- } $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identiti... } $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identiti... } $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identities: Thus, we have the following by using double angle identiti
- Question 3, Exercise 10.2
- 1-kpk-ex10-2-q3.png?nolink |Reference triangle}} We find: $\cos \theta =-\dfrac{3}{5}$. Thus, we have the following by using double angle identity: \beg... 1-kpk-ex10-2-q3.png?nolink |Reference triangle}} We find: $\cos \theta =-\dfrac{3}{5}$. Thus, we have the following by using half angle identities: \be
- Question 1, Exercise 10.3
- or difference $2\sin 6x\sin x$. ====Solution==== We have an identity: $$-2\sin \alpha \sin \beta =\co... \circ }}\cos {{123}^{\circ }}$. ====Solution==== We have an identity: $$2\sin \alpha \cos \beta =\si... c{A+B}{2}\cos \dfrac{A-B}{2}.$$ ====Solution==== We have an identity: $$2\sin \alpha \cos \beta =\s... c{P+Q}{2}\cos \dfrac{P-Q}{2}.$$ ====Solution==== We have an identity: $$2\cos \alpha \cos \beta =\co
- Question, Exercise 10.1
- eta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^... ta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^... ta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^
- Question 5, Exercise 10.3
- {{80}^{\circ }}=\dfrac{1}{16}.$$ ====Solution==== We know that\\ $2\cos \alpha \cos \beta =\cos \left(... dfrac{4\pi }{9}=\dfrac{3}{16}.$$ ====Solution==== We have an identities: $$2\sin \alpha \sin \beta =\c... {{70}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We have an identities: $$2\sin \alpha \sin \beta =\c
- Question 5, Exercise 10.3
- {{80}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We know that\\ $2\cos \alpha \cos \beta =\cos \left(... \dfrac{4\pi }{9}=\dfrac{3}{16}$. ====Solution==== We know that\\ $2\sin \alpha \sin \beta =\cos \left(... {{70}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We know that\\ $2\sin \alpha \sin \beta =\cos \left(
- Question 6, Exercise 10.1
- {\sin }^{2}}\dfrac{\alpha }{2}$ ====Solution==== We start from \begin{align}\cos \alpha &=\cos 2\dfra... \ldots (2)\end{align} Now combining (1) and (2), we get $$\cos \alpha =2{{\cos }^{2}}\dfrac{\alpha }{
- Question 1, Exercise 10.2
- -kpk-ex10-2-q1.png?nolink |reference triangle}} we find $\sin \theta =\dfrac{1}{\sqrt{26}}$ and $\cos \theta =\dfrac{-5}{\sqrt{26}}$ Thus, we have the following by using double angle identiti
- Question 4 and 5, Exercise 10.2
- c{2\pi }{3}$.\\ By using double angle identities, we have \begin{align}\sin 2\theta &=2\sin \theta \co... c{2\pi }{3}$ \\ By using double angle identities, we have \begin{align}\cos 2\theta &=2{{\cos }^{2}}\t