This is beta site. https://beta.mathcity.org/ 2026-06-05T23:32:27+00:00 https://beta.mathcity.org/ https://beta.mathcity.org/_media/logo.png text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-2-p2?rev=1737476040&do=diff Question 4 Exercise 8.2 Solutions of Question 4 of Exercise 8.2 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\sin 2 \theta$$\cos 2 \theta$$\tan 2 \theta$$\sin \frac{\theta}{2}$$\cos \frac{\theta}{2}$$\tan \frac{\theta}{2}$$\cos \theta=\frac{3}{5}$$0<\theta<\frac{\pi}{2}$$\cos\theta=\dfrac{3}{5}$$0<\theta<\dfrac{\pi}{2}$$\theta$$$\sin\theta = \pm \sq… text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-2-p3?rev=1737476040&do=diff Question 5 Exercise 8.2 Solutions of Question 5 of Exercise 8.2 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\sin \theta$$\cos \theta$$\tan \theta$$\sin 2 \theta=\frac{24}{25}, 2 \theta$$\sin 2\theta=\dfrac{24}{25}$$2\theta$$$\cos 2\theta = \pm \sqrt{1-\sin^2 2\theta}$$$2\theta$$\cos 2\theta$\begin{align*}\cos 2\theta & = - \sqrt{1-\sin^2 2\theta}\\… text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-2-p4?rev=1737476040&do=diff Question 6 Exercise 8.2 Solutions of Question 6 of Exercise 8.2 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\sin 15^{\circ} \cos 15^{\circ}$$$\sin 2 \theta = 2\sin\theta \cos\theta$$$$\sin\theta \cos\theta = \frac{1}{2}\sin 2\theta$$$\theta = 15^{\circ}$\begin{align*} \sin 15^{\circ} \cos 15^{\circ} & = \frac{1}{2}\sin 2(15^{\circ}) \\ & \frac{1}{2… text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-1-p2?rev=1737476040&do=diff Question 2, Exercise 8.1 Solutions of Question 2 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\cos 15^{\circ}$$\cos \left(45^{\circ}-30^{\circ}\right)$\begin{align*} \cos 15^{\circ} & = \cos \left(45^{\circ}-30^{\circ}\right)\\ &= \cos 45 \cos 30 + \sin 45 \sin 30 \\ &= \dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2… text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-1-p8?rev=1737476040&do=diff Question 9, Exercise 8.1 Solutions of Question 9 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\alpha$$\beta$$\sin \alpha=\dfrac{1}{\sqrt{2}}$$\cos \beta=-\dfrac{3}{5}$$\sin (\alpha \pm \beta)$$\sin \alpha=\dfrac{1}{\sqrt{2}}$$\alpha$$\cos \beta=-\dfrac{3}{5}$$\beta$$$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$$\alpha$$\cos$\begin{align*… text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/re-ex-p2?rev=1737476040&do=diff Question 2, Review Exercise Solutions of Question 2 of Review Exercise of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\sin \theta=\dfrac{3}{5}, \sin \phi=\dfrac{5}{13}$$\theta$$\phi$$\sin (\theta-\phi)$$\sin \theta=\dfrac{3}{5}$$\sin \phi=\dfrac{5}{13}$$\theta$$\phi$\begin{align*} \cos^2 \theta &= 1-\sin^2\theta\\ &= 1-\left(\frac{3}{5}\right)^2\\ & =… text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-1-p5?rev=1737476040&do=diff Question 5 and 6, Exercise 8.1 Solutions of Question 5 and 6 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\sin \alpha=\dfrac{4}{5}, \tan \beta=-\dfrac{5}{12}$$\cos (\alpha+\beta)$$\cos (\alpha-\beta)$$\sin \alpha=\dfrac{4}{5}$$\alpha$$\tan \beta=-\dfrac{5}{12}$$\beta$$$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$$\alpha$$\cos$\begin{alig… text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-1-p13?rev=1737476040&do=diff Question 14, Exercise 8.1 Solutions of Question 14 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\theta$$\sin \theta$$\cos \theta$$\alpha$\begin{align*} &\tan\alpha = \frac{\overline{BC}}{\overline{AB}} \\ \implies &\tan\alpha = \frac{3}{3} = 1 \\ \implies &\alpha = \tan^{-1}(1) = 45^\circ \end{align*}$45^\circ$$\theta$\begin{align*} … text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-1-p3?rev=1737476040&do=diff Question 3, Exercise 8.1 Solutions of Question 3 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\cos 120^{\circ}$$\cos \left(180^{\circ}-60^{\circ}\right)$$\cos \left(90^{\circ}+30^{\circ}\right)$\begin{align*} \cos 120^{\circ} & = \cos \left(180^{\circ}-60^{\circ}\right) \\ &= - \cos 60 ^{\circ}\\ &= -\dfrac{1}{2}. \end{align*}\begin{… text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-1-p6?rev=1737476040&do=diff Question 7, Exercise 8.1 Solutions of Question 7 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\alpha$$\beta$$\sin \alpha=\dfrac{12}{13}$$\tan \beta=\dfrac{4}{3}$$\sin(\alpha+\beta)$$\cos(\alpha+\beta)$$\tan(\alpha+\beta)$$\sin \alpha=\dfrac{12}{13}$$\alpha$$\tan \beta=\dfrac{4}{3}$$\beta$$$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$\(\a… text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-1-p9?rev=1737476040&do=diff Question 10, Exercise 8.1 Solutions of Question 10 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\sin \left(\dfrac{\pi}{2}-\alpha\right)=\cos \alpha$\begin{align*} L.H.S & = \sin \left(\frac{\pi}{2}-\alpha\right) \\ & =\sin\frac{\pi}{2} \cos \alpha - \cos \frac{\pi}{2} \sin\alpha \\ & = 1\times \cos \alpha - 0 \times \sin\alpha \\ & =… text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-2-p1?rev=1737476040&do=diff Question 1, 2 and 3 Exercise 8.2 Solutions of Question 1, 2 and 3 of Exercise 8.2 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $P(-3,4)$$\theta$$\theta$$\cos 2 \theta$$\sin 2 \theta$$2 \theta$$x=-3$$y=4$\begin{align*} r&= \sqrt{(-3)^2+4^2} \\ &=\sqrt{25} = 5. \end{align*}$$\sin\theta = \frac{4}{5} \text{ and } \cos\theta = -\frac{3}{5}.$$\begin{align…