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- Question 8 & 9, Review Exercise 10
- n. =====Question 8===== Prove the identity $\sin \left( \dfrac{\pi }{4}-\theta \right)\sin \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{1}{2}\cos 2\theta $.... ==== We know that $2\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ \begin{align}L.H.S.&=\sin \left( \dfrac{\p
- Question 2, Exercise 10.1
- end{align} \begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{3}-\frac{\pi }{4} \right) & =\sin \f... }{4}-\cos \frac{\pi }{3}\sin \frac{\pi }{4} \\ &=\left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{\sqrt{2}} \right)-\left( \frac{1}{2} \right)\left( \frac{1}{\sqrt{2}} \right) \\ & =\frac{\sq
- Question 5, Exercise 10.3
- == We know that\\ $2\cos \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}... c }}\\ &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\df... 0}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\circ }} \rig
- Question 5, Exercise 10.3
- == We know that\\ $2\cos \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}... c }}\\ &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\df... 0}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\circ }} \rig
- Question 5, Exercise 10.1
- $\beta$ in the first Quadrant, then find: $\sin \left( \alpha +\beta \right)$. ====Solution==== Given:... a }\\ \Rightarrow \quad \sec \alpha &=-\sqrt{1+{{\left(\frac{3}{4} \right)}^{2}}}=-\sqrt{1+\frac{9}{16}}... \cos \alpha \\ \Rightarrow \quad \sin \alpha &=\left(\frac{3}{4}\right)\left(-\frac{4}{5} \right)\\ \Rightarrow \quad \sin\alpha &= \frac{3}{5}\end{align}
- Question 7, Exercise 10.2
- &={{\cos }^{4}}\theta -{{\sin }^{4}}\theta \\ &=\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)\\ &=\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)\left( 1 \right)\\ &=\cos 2\theta \quad \text{(By usin
- Question 8, Exercise 10.1
- tan. =====Question 8(i)===== Prove that: $\tan \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{\cos \the... }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi }{4}+\theta \right)\\ &=\dfrac{\sin \left( \dfrac{\pi }{4}+\theta \right)}{\cos \left( \dfrac{\pi }{4}+\theta \right)}\\ &=\dfrac{\sin\dfrac{\
- Question 9 and 10, Exercise 10.1
- \cos 4\theta +\cos \theta \sin 4\theta \\ &=\sin \left( \theta +4\theta \right)\\ &=\sin 5\theta =R.H.S... =====Question 10===== Show that: $\dfrac{\sin \left( {{180}^{\circ }}-\alpha \right)\cos \left( {{270}^{\circ }}-\alpha \right)}{\sin \left( {{180}^{\circ }}+\alpha \right)\cos \left( {{270}^{\cir
- Question, Exercise 10.1
- a $in Quadrant II, find the exact value of $\sin \left( \alpha -\beta \right)$. ====Solution==== Given:... s\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{1-{{\left(-\frac{4}{5} \right)}^{2}}}\\ &=-\sqrt{1-\dfrac{1... s +ive in 2nd quadrant, \begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\\ &=\sqrt{1-\dfrac... sin \alpha\cos \beta+\cos \alpha\sin \beta \\ &=\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)+\l
- Question 2, Exercise 10.3
- ve an identity: $$\sin \alpha +\sin \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right).$$ Put $\alpha ={{37}^{... \sin {{37}^{\circ }}+\sin {{43}^{\circ }}&=2\sin \left( \dfrac{{{37}^{\circ }}+{{43}^{\circ }}}{2} \right)\cos \left( \dfrac{{{37}^{\circ }}-{{43}^{\circ }}}{2} \righ
- Question 3, Exercise 10.1
- }$, evaluate each of the following exactly $\cos \left( u+v \right)$ ====Solution==== Given $\sin u=\df... \,\cos u&=\sqrt{1-{{\sin }^{2}}u}\\ &=\sqrt{1-{{\left( \frac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9}... \,\cos v&=\sqrt{1-{{\sin }^{2}}v}\\ &=\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{1... &=\frac{3}{5}\end{align} Now \begin{align}\cos \left( u+v \right)&=\cos u\cos v-\sin u\sin v\\ &=\fra
- Question 13, Exercise 10.1
- ess each of the following in the form $r\,\,\sin \left( \theta +\phi \right)$ where terminal ray of $\t... varphi+r^2\sin^2\varphi = 4^2+3^2\\ \implies &r^2\left(\cos^2\varphi+\sin^2\varphi\right) = 16+9\\ \impl... ve \begin{align}&4\sin\theta +3\cos\theta \\ &=5 \left(\dfrac{4}{5}\sin\theta +\dfrac{3}{5}\cos\theta\right)\\ &=r\left(\cos\varphi\sin\theta +\sin\varphi\cos\theta\righ
- Question 6 & 7, Review Exercise 10
- ==== \begin{align}L.H.S&=\cos 4\theta \\ &=\cos 2\left( 2\theta \right)\\ &=1-2si{{n}^{2}}2\theta \\ &=1-2{{\left( 2sin\theta \cos \theta \right)}^{2}}\\ &=1-8si{... S.&=\sin 6x\sin x+\cos 4x\cos 3x\\ &=\dfrac{1}{2}\left( 2\sin 6x\sin x+2\cos 4x\cos 3x \right)\\ &=\dfrac{1}{2}\left[ \cos \left( 6x-x \right)-\cos \left( 6x+x \right
- Question 8 and 9, Exercise 10.2
- Solution==== \begin{align}{{\cos}^{4}}\theta &={{\left( {{\cos }^{2}}\theta \right)}^{2}}\\ &={{\left( \dfrac{1+\cos 2\theta }{2} \right)}^{2}}\\ &=\dfrac{1... theta +{{\cos }^{2}}2\theta }{4}\\ &=\dfrac{1}{4}\left[ 1+2\cos 2\theta +{{\cos }^{2}}2\theta \right]\\ &=\dfrac{1}{4}\left[ 1+2\cos 2\theta +\dfrac{1+\cos 4\theta }{2} \rig
- Question 6, Exercise 10.1
- \alpha }{2}\\ &={{\cos }^{2}}\dfrac{\alpha }{2}-\left( 1-{{\cos }^{2}}\dfrac{\alpha }{2} \right)\\ \co... \alpha &=2{{\cos }^{2}}\dfrac{\alpha }{2}-1\\ &=2\left( 1-{{\sin }^{2}}\dfrac{\alpha }{2} \right)-1\\ &=... ign} =====Question 6(ii)===== Show that: $\sin \left( \alpha +\beta \right)\sin \left( \alpha -\beta \right)={{\cos }^{2}}\beta -{{\cos }^{2}}\alpha$ ====