English

Evaluate the Following Determinant: ∣ ∣ ∣ Cos 15 ∘ Sin 15 ∘ Sin 75 ∘ Cos 75 ∘ ∣ ∣ ∣ - Mathematics

Advertisements
Advertisements

Question

Evaluate the following determinant:

\[\begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}\]

Advertisements

Solution

\[∆ = \cos15^\circ\cos75^\circ - \sin15^\circ\sin75^\circ\]
\[ = \cos15^\circ\cos75^\circ - \sin(90^\circ - 75^\circ)\sin(90^\circ - 15^\circ) \left[ \because \sin\left( 90^\circ - \theta \right) = \cos\theta \right]\]
\[ = \cos15^\circ\cos75^\circ - \cos75^\circ\cos15^\circ\]
\[ = \cos15^\circ\cos75^\circ - \cos15^\circ\cos75^\circ = 0\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 6: Determinants - Exercise 6.1 [Page 10]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 6 Determinants
Exercise 6.1 | Q 2.3 | Page 10

RELATED QUESTIONS

If `|[2x,5],[8,x]|=|[6,-2],[7,3]|`, write the value of x.


Solve the system of linear equations using the matrix method.

4x – 3y = 3

3x – 5y = 7


If A = `[(2,-3,5),(3,2,-4),(1,1,-2)]` find A−1. Using A−1 solve the system of equations:

2x – 3y + 5z = 11

3x + 2y – 4z = –5

x + y – 2z = –3


Evaluate

\[\begin{vmatrix}2 & 3 & - 5 \\ 7 & 1 & - 2 \\ - 3 & 4 & 1\end{vmatrix}\] by two methods.

 

Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}\sin\alpha & \cos\alpha & \cos(\alpha + \delta) \\ \sin\beta & \cos\beta & \cos(\beta + \delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma + \delta)\end{vmatrix}\]


Evaluate the following:

\[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix}\]


Prove that:

`[(a, b, c),(a - b, b - c, c - a),(b + c, c + a, a + b)] = a^3 + b^3 + c^3 -3abc`


Prove that

\[\begin{vmatrix}\frac{a^2 + b^2}{c} & c & c \\ a & \frac{b^2 + c^2}{a} & a \\ b & b & \frac{c^2 + a^2}{b}\end{vmatrix} = 4abc\]


\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + a & a \\ 1 & 1 & 1 + a\end{vmatrix} = a^3 + 3 a^2\]


Show that

\[\begin{vmatrix}x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c\end{vmatrix} =\text{ 0 where a, b, c are in A . P .}\]

 


​Solve the following determinant equation:

\[\begin{vmatrix}1 & 1 & x \\ p + 1 & p + 1 & p + x \\ 3 & x + 1 & x + 2\end{vmatrix} = 0\]

If a, b, c are real numbers such that
\[\begin{vmatrix}b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a\end{vmatrix} = 0\] , then show that either
\[a + b + c = 0 \text{ or, } a = b = c\]


If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab.


Prove that :

\[\begin{vmatrix}b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c\end{vmatrix} = 3abc - a^3 - b - c^3\]

 


Prove that :

\[\begin{vmatrix}a^2 & bc & ac + c^2 \\ a^2 + ab & b^2 & ac \\ ab & b^2 + bc & c^2\end{vmatrix} = 4 a^2 b^2 c^2\]

\[\begin{vmatrix}1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1\end{vmatrix} = \left( a^3 - 1 \right)^2\]

If w is an imaginary cube root of unity, find the value of \[\begin{vmatrix}1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w\end{vmatrix}\]


Find the value of the determinant \[\begin{vmatrix}2^2 & 2^3 & 2^4 \\ 2^3 & 2^4 & 2^5 \\ 2^4 & 2^5 & 2^6\end{vmatrix}\].


If \[\begin{vmatrix}x + 1 & x - 1 \\ x - 3 & x + 2\end{vmatrix} = \begin{vmatrix}4 & - 1 \\ 1 & 3\end{vmatrix}\], then write the value of x.

If \[\begin{vmatrix}2x & x + 3 \\ 2\left( x + 1 \right) & x + 1\end{vmatrix} = \begin{vmatrix}1 & 5 \\ 3 & 3\end{vmatrix}\], then write the value of x.

 

 


If \[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix}\] , write the value of x.


Using the factor theorem it is found that a + bb + c and c + a are three factors of the determinant 

\[\begin{vmatrix}- 2a & a + b & a + c \\ b + a & - 2b & b + c \\ c + a & c + b & - 2c\end{vmatrix}\]
The other factor in the value of the determinant is


The value of \[\begin{vmatrix}5^2 & 5^3 & 5^4 \\ 5^3 & 5^4 & 5^5 \\ 5^4 & 5^5 & 5^6\end{vmatrix}\]

 


Let \[f\left( x \right) = \begin{vmatrix}\cos x & x & 1 \\ 2\sin x & x & 2x \\ \sin x & x & x\end{vmatrix}\] \[\lim_{x \to 0} \frac{f\left( x \right)}{x^2}\]  is equal to


Solve the following system of equations by matrix method:
3x + y = 7
5x + 3y = 12


Solve the following system of equations by matrix method:
 2x + 6y = 2
3x − z = −8
2x − y + z = −3


Solve the following system of equations by matrix method:
 x − y + z = 2
2x − y = 0
2y − z = 1


Show that the following systems of linear equations is consistent and also find their solutions:
6x + 4y = 2
9x + 6y = 3


If \[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\] , find A−1. Using A−1, solve the system of linear equations  x − 2y = 10, 2x + y + 3z = 8, −2y + z = 7.

Given \[A = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}, B = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}\] , find BA and use this to solve the system of equations  y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17


A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.


x + y + z = 0
x − y − 5z = 0
x + 2y + 4z = 0


If \[\begin{bmatrix}1 & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ - 1 \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}\] , find x, y and z.


The number of solutions of the system of equations
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
is


x + y = 1
x + z = − 6
x − y − 2z = 3


Prove that (A–1)′ = (A′)–1, where A is an invertible matrix.


`abs ((1, "a"^2 + "bc", "a"^3),(1, "b"^2 + "ca", "b"^3),(1, "c"^2 + "ab", "c"^3))`


The number of real value of 'x satisfying `|(x, 3x + 2, 2x - 1),(2x - 1, 4x, 3x + 1),(7x - 2, 17x + 6, 12x - 1)|` = 0 is


Using the matrix method, solve the following system of linear equations:

`2/x + 3/y + 10/z` = 4, `4/x - 6/y + 5/z` = 1, `6/x + 9/y - 20/z` = 2.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×