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- Question 7, Exercise 10.2
- \sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align}L.H.S&={{\cos }^{4}}\theta -{{\s... }\dfrac{\theta }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{align}L.H.S&=\tan \dfrac{\theta }{2}+c... eta }{1-\cos 2\theta }={{\cot }^{2}}\theta $. ====Solution==== \begin{align}L.H.S&=\dfrac{1+\cos 2\theta }{1... y $\cos ec2\theta -\cot 2\theta =tan\theta $. ====Solution==== \begin{align}L.H.S.&=\cos ec2\theta -\cot 2\t
- Question 1, Exercise 10.1
- }}+\cos {{37}^{\circ }}\sin {{22}^{\circ }}$ ==== Solution ==== As \begin{align} \sin (\alpha +\beta )=\si... }}+\sin {{83}^{\circ }}\sin {{53}^{\circ }}$ ====Solution==== As \begin{align}\cos (\alpha -\beta )&=\cos... }}-\sin {{19}^{\circ }}\sin {{5}^{\circ }}$ ==== Solution ==== As \begin{align}\cos (\alpha +\beta )&=\cos... }}-\cos {{40}^{\circ }}\sin {{15}^{\circ }}$ ==== Solution ==== As \begin{align}\sin (\alpha -\beta )=\sin
- Question 2, Exercise 10.1
- === Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }... ii)=== Evaluate exactly:$\tan {{75}^{\circ }}$ ==Solution== We rewrite ${{75}^{\circ }}$ as ${{45}^{\circ }... i)=== Evaluate exactly:$\tan {{105}^{\circ }}$ ==Solution== We rewrite ${{105}^{{}^\circ }}$ as ${{60}^{{}^... v)=== Evaluate exactly:$\tan \dfrac{5\pi }{12}$ ==Solution== We rewrite $\dfrac{5\pi }{12}$ as $\dfrac{\left
- Question 6, Exercise 10.2
- s to evaluate exactly $\cos {{15}^{\circ }}$. ====Solution==== Because ${{15}^{\circ }}=\dfrac{{{30}^{\circ ... to evaluate exactly $\tan {{67.5}^{\circ }}$. ====Solution==== Because ${{67.5}^{\circ }}=\dfrac{{{135}^{\ci... to evaluate exactly $sin{{112.5}^{\circ }}$. ====Solution==== Because ${{112.5}^{\circ }}=\dfrac{{{225}^{\c... s to evaluate exactly $\cos \dfrac{\pi }{8}$. ====Solution==== Because $\dfrac{\pi }{8}=\dfrac{\dfrac{\pi }{
- Question 3, Exercise 10.1
- following exactly $\cos \left( u+v \right)$ ====Solution==== Given $\sin u=\dfrac{3}{5},$ $0\le u\le \d... e following exactly $\tan \left( u-v \right)$ ====Solution==== Given \begin{align}\sin u&=\dfrac{3}{5},\,\,... following exactly $\sin \left( u-v \right)$ ====Solution==== Given \begin{align}\sin u&=\dfrac{3}{5},\,\,\... following exactly $\cos \left( u+v \right)$ ====Solution==== Given $\sin u=\dfrac{3}{5},$ $0\le u\le \d
- Question 8, Exercise 10.1
- a +\sin \theta }{\cos \theta -\sin \theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi ... \right)=\dfrac{1-tan\theta }{1+tan\theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi ... {1-{{\tan }^{2}}\alpha {{\tan }^{2}}\beta }$ ====Solution==== \begin{align}\tan (\alpha +\beta )&=\dfrac{\... \right)}{\cos \left( \theta -\phi \right)}$ ====Solution==== \begin{align}L.H.S.&=\dfrac{1-\tan \theta \t
- Question 8 and 9, Exercise 10.2
- first power of one or more cosine functions. ====Solution==== \begin{align}{{\cos}^{4}}\theta &={{\left( {{... os }^{3}}\theta -4\sin \theta \cos \theta $ . ====Solution==== \begin{align}L.H.S.&=\sin 4\theta \\ &=\sin 2... c{1-{{\tan }^{2}}2\theta }{2\tan 2\theta }$ . ====Solution==== \begin{align}L.H.S.&=\cot 4\theta =\dfrac{\co... cot^3\theta-3\cot\theta}{3\cot^2\theta -1}$ . ====Solution==== \begin{align}L.H.S.&=\cot 3\theta =\dfrac{1}{
- Question 1, Exercise 10.3
- roduct as sum or difference $2\sin 6x\sin x$. ====Solution==== We have an identity: $$-2\sin \alpha \sin \b... $\sin {{55}^{\circ }}\cos {{123}^{\circ }}$. ====Solution==== We have an identity: $$2\sin \alpha \cos \b... : $$\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}.$$ ====Solution==== We have an identity: $$2\sin \alpha \cos \... : $$\cos \dfrac{P+Q}{2}\cos \dfrac{P-Q}{2}.$$ ====Solution==== We have an identity: $$2\cos \alpha \cos \b
- Question 2, Exercise 10.3
- $\sin {{37}^{\circ }}+\sin {{43}^{\circ }}.$$ ====Solution==== We have an identity: $$\sin \alpha +\sin \be... $\cos {{36}^{\circ }}-\cos {{82}^{\circ }}$. ====Solution==== We have an identity: $$\cos \alpha -\cos \be... $$\sin \dfrac{P+Q}{2}-\sin \dfrac{P-Q}{2}.$$ ====Solution==== We have an identity: $$\sin \alpha -\sin \be... $$\cos \dfrac{A+B}{2}+\cos \dfrac{A-B}{2}.$$ ====Solution==== We have an identity: $$\cos \alpha +\cos \b
- Question, Exercise 10.1
- alue of $\sin \left( \alpha -\beta \right)$. ====Solution==== Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$ ... alue of $\cos \left( \alpha +\beta \right)$. ====Solution==== Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$ ... ue of $\tan \left( \alpha +\beta \right)$ . ====Solution==== Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$
- Question 5, Exercise 10.1
- find: $\sin \left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha =\dfrac{3}{4}$. As $\tan... find: $\cos \left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha =\dfrac{3}{4}$. As $\tan... find: $\tan \left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha =\dfrac{3}{4}$. As $\tan
- Question 2, Exercise 10.2
- e second quadrant, then find $\sin 2\theta $. ====Solution==== Given: $\sin \theta =\dfrac{5}{13}$. Using ... e second quadrant, then find $\cos 2\theta $. ====Solution==== Given: $\sin \theta =\dfrac{5}{13}$. Using ... e second quadrant, then find $\tan 2\theta $. ====Solution==== Given: $\sin \theta =\dfrac{5}{13}$. Using
- Question 4 and 5, Exercise 10.2
- uadrant, then find $\sin \dfrac{\theta }{2}$. ====Solution==== Given: $\cos \theta =-\dfrac{3}{7}$ and termi... to evaluate exactly $\sin \dfrac{2\pi }{3}$. ====Solution==== Given: $\sin \dfrac{2\pi }{3}$.\\ By using do... to evaluate exactly $\cos \dfrac{2\pi }{3}$. ====Solution==== Given: $\cos \dfrac{2\pi }{3}$ \\ By using do
- Question 5, Exercise 10.3
- \circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}.$$ ====Solution==== We know that\\ $2\cos \alpha \cos \beta =\cos... pi }{3}\sin \dfrac{4\pi }{9}=\dfrac{3}{16}.$$ ====Solution==== We have an identities: $$2\sin \alpha \sin \b... {\circ }}\sin {{70}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We have an identities: $$2\sin \alpha \sin \b
- Question 5, Exercise 10.3
- {\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We know that\\ $2\cos \alpha \cos \beta =\cos... \pi }{3}\sin \dfrac{4\pi }{9}=\dfrac{3}{16}$. ====Solution==== We know that\\ $2\sin \alpha \sin \beta =\cos... {\circ }}\sin {{70}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We know that\\ $2\sin \alpha \sin \beta =\cos