begin{vmatrix} cos^2theta & -sin^2theta sin^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The determinant of the matrix is calculated as: $\begin{vmatrix} \cos^2	heta & -\sin^2	heta \ \sin^2	heta & \cos^2	heta \end{vmatrix} = (\cos

Vedclass · Accepted Answer

$\frac{1}{4}(3 + \cos 4	heta)$

Vedclass · Answer

(B) The determinant of the matrix is calculated as: $\begin{vmatrix} \cos^2	heta & -\sin^2	heta \ \sin^2	heta & \cos^2	heta \end{vmatrix} = (\cos^2	heta)(\cos^2	heta) - (-\sin^2	heta)(\sin^2	heta) = \cos^4	heta + \sin^4	heta$. Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$, we get: $\cos^4	heta + \sin^4	heta = (\cos^2	heta + \sin^2	heta)^2 - 2\sin^2	heta\cos^2	heta = 1 - 2\sin^2	heta\cos^2	heta$. Since $\sin 2	heta = 2\sin	heta\cos	heta$, we have $\sin^2 2	heta = 4\sin^2	heta\cos^2	heta$, so $2\sin^2	heta\cos^2	heta = \frac{1}{2}\sin^2 2	heta$. Thus, the expression becomes $1 - \frac{1}{2}\sin^2 2	heta$. Using the identity $\sin^2 2	heta = \frac{1 - \cos 4	heta}{2}$, we substitute: $1 - \frac{1}{2}(\frac{1 - \cos 4	heta}{2}) = 1 - \frac{1}{4} + \frac{1}{4}\cos 4	heta = \frac{3}{4} + \frac{1}{4}\cos 4	heta = \frac{1}{4}(3 + \cos 4	heta)$. Therefore, option $(B)$ is correct.

$\begin{vmatrix} \cos^2\theta & -\sin^2\theta \\ \sin^2\theta & \cos^2\theta \end{vmatrix} = \dots$

Similar Questions

The area of the triangle with vertices $(a, b)$,$(x_1, y_1)$,and $(x_2, y_2)$,where $a, x_1, x_2$ are in $G.P.$ with common ratio $r$ and $b, y_1, y_2$ are in $G.P.$ with common ratio $s$,is given by

If $f(x) = \left| \begin{array}{ccc} 1 & 6+x & 36+x^2 \\ 0 & x-3 & 3x^2-27 \\ 0 & 2x-4 & 8x^2-32 \end{array} \right|$,then $\lim_{x \rightarrow 1} \frac{f(x)}{f(-x)} = $

If $\left| \begin{array}{ccc} x - 1 & 3 & 0 \\ 2 & x - 3 & 4 \\ 3 & 5 & 6 \end{array} \right| = 0$,then $x =$

If $\left[\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$ has no inverse,then the real value of $x$ is

The area of a triangle whose vertices are $(2, -6)$,$(5, 4)$,and $(k, 4)$ is $35$ sq. units. Then,the value of $k$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module