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- Question 1, Exercise 8.1
- os \alpha \cos \beta - \sin \alpha \sin \beta \\ \implies \cos (180+60) & = \cos 180 \cos 60 - \sin 180 \sin 60 \\ \implies \cos (180+60) & = (-1)\left(\frac{1}{2}\right) - ... os \alpha \cos \beta + \sin \alpha \sin \beta \\ \implies \cos (180-60) & = \cos 180 \cos 60 + \sin 180 \sin 60 \\ \implies \cos (180-60) & = (-1)\left(\frac{1}{2}\right) +
- Question 4 Exercise 8.2
- frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = \frac{24}{25}}. \end{align... ac{16}{25} \right)\\ \end{align*} \begin{align*} \implies \boxed{\cos 2\theta = -\frac{7}{25}}. \end{align... frac{24/25}{-7/25}\\ \end{align*} \begin{align*} \implies \boxed{\tan 2\theta = -\frac{24}{7}}. \end{align... 1-\cos\theta}{2}}$$ As $0<\theta<\dfrac{\pi}{2}$ implies $0<\dfrac{\theta}{2}<\dfrac{\pi}{4}$, that is, $\
- Question 12, Exercise 8.1
- gin{align*} & \alpha+\beta=180^{\circ}-\gamma \\ \implies & \tan(\alpha+\beta) = \tan(180^{\circ}-\gamma) \\ \implies & \frac{\tan\alpha + \tan\beta}{1-\tan\alpha \tan\beta} = \tan(2(90)-\gamma) \\ \implies & \tan\alpha + \tan\beta = -\tan\gamma[1-\tan\alpha \tan\beta] \\ \implies & \tan\alpha + \tan\beta = -\tan\gamma+\tan\alpha
- Question 5 and 6, Review Exercise
- tan \theta \cdot \tan 45^{\circ}} =\frac{1}{3}\\ \implies & \frac{\tan \theta - 1}{1 + \tan \theta}= \frac{1}{3}\\ \implies & 3 \tan \theta - 3 = 1 + \tan \theta \\ \implies & 2 \tan \theta = 4 \\ \implies & \tan \theta = 2 \end{align*} GOOD =====Question 6(i)===== If $\sin (\a
- Question 2, Review Exercise
- ft(\frac{3}{5}\right)^2\\ & =1-\frac{9}{25} \\ \implies \cos^2 \theta&=\frac{16}{25} \\ \cos \theta&=\pm... \theta$ is obtuse, so $\theta$ lies in II Q. This implies $\cos \theta <0$, thus $$\cos \theta=-\frac{4}{5}... \frac{5}{13}\right)^2\\ & =1-\frac{25}{169} \\ \implies \cos^2 \theta&=\frac{144}{169} \\ \cos \phi&=\pm... } As $\phi$ is acute, so $\phi$ lies in I Q. This implies $\cos\pi >0$, thus $$\cos \phi=\frac{12}{13}$$ As
- Question 13, Exercise 8.1
- 2+(-5)^2=r^2 \cos^2\varphi+r^2 \sin^2 \varphi \\ \implies & 144+25={{r}^{2}}\left( {{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi \right) \\ \implies & 169={{r}^{2}}\left( 1 \right) \\ \implies & r=\sqrt{169}=13 \end{align*} Also \begin{align*} & \frac{-5}{12}=\frac{r\sin \varphi }{r\cos \varphi } \\ \implies & \frac{-5}{12}=\tan \varphi \\ \implies & \varph
- Question 5 Exercise 8.2
- \sqrt{\frac{16}{25}} \end{align*} \begin{align*} \implies \boxed{\sin\theta = \frac{4}{5}} \end{align*} Als... \sqrt{\frac{9}{25}} \end{align*} \begin{align*} \implies \boxed{\cos\theta=\frac{3}{5}} \end{align*} Now \... & = \frac{4/5}{3/5} \end{align*} \begin{align*} \implies \boxed{\tan\theta=\frac{4}{3}} \end{align*} GOOD ... a}{2}}. \] As \(\pi < 2\theta < \frac{3\pi}{2}\) implies \(\frac{\pi}{2} < \theta < \pi\), i.e., \(\theta\
- Question 7, Review Exercise
- in (\alpha - \theta) = \cos (\alpha + \theta) \\ \implies & \sin \alpha \cos \theta - \cos \alpha \sin \the... \alpha \sin \theta}{\cos \alpha \cos \theta} \\ \implies & \tan \alpha -\tan \theta= 1-\tan \alpha \tan \theta\\ \implies & \tan \alpha +\tan \alpha \tan \theta =1+\tan \theta\\ \implies & \tan \alpha(1 + \tan \theta) =1+\tan \theta\\ \
- Question 5 and 6, Exercise 8.1
- n*} \frac{\sin\beta}{\cos\beta} & = \tan\beta \\ \implies \sin\beta & = \tan\beta \cos\beta \\ & = \left(-... c{5}{12} \right) \left(-\frac{12}{13} \right) \\ \implies \sin\beta & = \frac{5}{13}. \end{align*} Now \be... qrt{1-\dfrac{49}{625}}=\sqrt{\dfrac{576}{625}}\\ \implies \sin \alpha&=\dfrac{24}{25}.\end{align*} Also $$\... n*} \frac{\cos\beta}{\sin\beta} & = \cot\beta \\ \implies \cos\beta & = \cot\beta \sin\beta \\ & = \left(\
- Question 14, Exercise 8.1
- n\alpha = \frac{\overline{BC}}{\overline{AB}} \\ \implies &\tan\alpha = \frac{3}{3} = 1 \\ \implies &\alpha = \tan^{-1}(1) = 45^\circ \end{align*} Hence wire will m... rline{AD}^2 = \overline{AB}^2+\overline{BD}^2 \\ \implies \overline{AD}^2 = 3^2 + 7^2 = 58 \\ \implies \overline{AD} = \sqrt{58} \\ \end{align*} Now \begin{align*}
- Question 6 Exercise 8.2
- } \times \frac{1}{2} \end{align*} \begin{align*} \implies \boxed{\sin 15^{\circ} \cos 15^{\circ} = \frac{1}... = \frac{\sqrt{3}}{2} \end{align*} \begin{align*} \implies \boxed{\cos^2 15^\circ - \sin^2 15^\circ = \frac{... rac{\pi}{4}\right)\\ \end{align*} \begin{align*} \implies \boxed{1-2\sin^2 \left(\frac{\pi}{8}\right) = \fr... rac{\pi}{6}\right)\\ \end{align*} \begin{align*} \implies \boxed{2\cos^2 \left(\frac{\pi}{12}\right)-1 = \f
- Question 2, Exercise 8.1
- } & \sin \left(90-15\right) = \cos 15^\circ \\ \implies & \sin 75^\circ = \cos 15^\circ \\ & = \cos \le... frac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}}\\ \implies & \boxed{\sin 75^\circ = \dfrac{\sqrt{3}+1}{2\... qrt{3}-1)/2\sqrt{2}} \end{align*} \begin{align*} \implies \boxed{\tan 75^\circ = \frac{\sqrt{3}+1}{\sqrt{3
- Question 9, Exercise 8.1
- \ &=-\sqrt{1-\dfrac{1}{2}}=\sqrt{\dfrac{1}{2}}\\ \implies \cos \alpha&=-\dfrac{1}{\sqrt{2}}. \end{align*} A... \ &=-\sqrt{1-\dfrac{1}{2}}=\sqrt{\dfrac{1}{2}}\\ \implies \cos \alpha&=-\dfrac{1}{\sqrt{2}}. \end{align*} A... t{2}}\right)}^{2}}}\\ &=-\sqrt{1-\dfrac{1}{2}}\\ \implies \cos \alpha&=-\dfrac{1}{\sqrt{2}}. \end{align*} N
- Question 1, 2 and 3 Exercise 8.2
- } & \sin 2\alpha = 2 \sin \alpha \cos \alpha \\ \implies & \boxed{\sin 2\alpha=- y\sqrt{1-y^2}} \end{align... in{align*} & \cos 2\alpha = 1-2\sin^2 \alpha \\ \implies & \boxed{\cos 2\alpha=1-2y^2} \end{align*} \begin... 2\alpha = \frac{\sin 2\alpha}{\cos 2\alpha} \\ \implies & \boxed{\tan 2\alpha=-\frac{y\sqrt{1-y^2}}{1-2y^
- Question 7, Exercise 8.1
- n*} \frac{\sin\beta}{\cos\beta} & = \tan\beta \\ \implies \sin\beta & = \tan\beta \cdot \cos\beta \\ &= \le... ft(\frac{4}{3}\right)\left(\frac{3}{5}\right) \\ \implies \sin\beta & = \frac{4}{5}. \end{align*} (i) $\si