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- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- in 180 \sin 60 \\ \implies \cos (180+60) & = (-1)\left(\frac{1}{2}\right) - (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = -\frac{1}{2} - 0 = -\frac{1}{2} \end{a... in 180 \sin 60 \\ \implies \cos (180-60) & = (-1)\left(\frac{1}{2}\right) + (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = -\frac{1}{2} + 0 = -\frac{1}{2} \end{al
- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- in QI, so \begin{align*} \sin\theta & = \sqrt{1-\left(\frac{3}{5} \right)^2} \\ &= \sqrt{1-\frac{9}{25}... \sin 2\theta & = 2 \sin\theta \cos\theta \\ &= 2\left(\frac{4}{5} \right) \left(\frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\thet... ign*} \cos 2\theta & = 1-2\sin^2\theta \\ &= 1-2\left(\frac{4}{5} \right)^2\\ &= 1-2\left(\frac{16}{25}
- Question 9, Exercise 1.2 @math-11-nbf:sol:unit01
- on 9(iii)==== Find real and imaginary parts of $\left(\dfrac{7+2 i}{3-i}\right)^{-1}$. **Solution.** We use the following formulas: \[Re\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}\right) = \frac{x_1 x_2 + y_1 y_2}{x_1^2 + y_1^2},\] \[Im\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}
- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- ingular and $A^{-1}$ exists. Now \begin{align} & \left[\begin{matrix} 4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{matrix}\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0... d{matrix}\right. \right] \\ \underset{\sim}{R} & \left[\begin{matrix} 1 & 4 & 14 \\ 0 & 5 & 6 \\ -1 & 2 & 3 \end{matrix}\left| \begin{matrix} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0
- Question 8 & 9, Review Exercise 10 @fsc-part1-kpk:sol:unit10
- n. =====Question 8===== Prove the identity $\sin \left( \dfrac{\pi }{4}-\theta \right)\sin \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{1}{2}\cos 2\theta $.... ==== We know that $2\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ \begin{align}L.H.S.&=\sin \left( \dfrac{\p
- Question 8 & 9, Review Exercise 10 @math-11-kpk:sol:unit10
- n. =====Question 8===== Prove the identity $\sin \left( \dfrac{\pi }{4}-\theta \right)\sin \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{1}{2}\cos 2\theta $.... ==== We know that $2\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ \begin{align}L.H.S.&=\sin \left( \dfrac{\p
- Exercise 1.2 (Solutions) @fsc-part1-ptb:sol:ch01
- exists}\; -z\in \mathbb{C} \mbox{ such that}\; z+\left( -z\right) =0 \\ \mbox{In fact if } z=a+bi, \mbox... ray}\nonumber\] - Associative Law for Addition \[\left( z+w\right) +v= z +\left( w+v\right)\nonumber \] </panel> <panel> **Question 2** Verify the multiplic... olutions** $\quad (2,6)\div (3,7)=\dfrac{(2,6)}{\left( 3,7 \right)}$ $=\dfrac{2+6i}{3+7i}=\dfrac{2+6i}
- Exercise 2.4 (Solutions) @matric:9th_science:unit_02
- {\frac{-1}{5}}}{\sqrt(196)^{-1}}$ * (ii) $\left(2x^5y^{-4}\right)\left(-8x^{-3}y^2\right)$ * (iii) $\left(\frac{x^{-2}y^{-1}z^{-4}}{x^4y^{-3}z^0}\right)^{-3}$ * (iv) $\frac{\left(81\right)^n.3^5-\left(3\right)^{4n-1}\left(243\ri
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- egin{bmatrix}x\\y\\z\end{bmatrix}$. Verify that $\left( AB \right)C=A\left( BC \right)$. ====Solution==== Given: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B... bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}\\ &=\left[ x( ax+hy+gz )+y( hx+by+fz )+z( gx+fy+cz ) \right] \\ &=\left[ a{{x}^{2}}+hxy+gxz+hxy+b{{y}^{2}}+fyz+gxz+fyz+c{
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- 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 2, Exercise 10.1 @math-11-kpk:sol:unit10
- 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 @fsc-part1-kpk:sol:unit10
- == 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 @math-11-kpk:sol:unit10
- == 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 1, Exercise 2.5 @math-11-nbf:sol:unit02
- nto echelon form then into reduced echelon form $\left[\begin{array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 &... ight]$. ** Solution. ** \begin{align*} & \quad \left[\begin{array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 26 & 33 \\ 0 ... ad R_2 + 6R_1 \quad R_3 + 4R_1\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 1 & \frac{33}
- Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- == 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