If the roots of x^3 + ax^2 + bx + c = 0 are t | vedclass

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

Question

(B) Let the roots be $\cos A, \cos B, \cos C$ where $A, B, C$ are angles of an acute triangle.From Vieta's formulas,we have:$\cos A + \cos B + \cos C

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let the roots be $\cos A, \cos B, \cos C$ where $A, B, C$ are angles of an acute triangle.From Vieta's formulas,we have:$\cos A + \cos B + \cos C = -a$$\cos A \cos B + \cos B \cos C + \cos C \cos A = b$$\cos A \cos B \cos C = -c$We need to evaluate $a^2 - 2b - 2c$.$a^2 - 2b - 2c = (-a)^2 - 2b - 2c = (\cos A + \cos B + \cos C)^2 - 2(\cos A \cos B + \cos B \cos C + \cos C \cos A) - 2(\cos A \cos B \cos C)$.Expanding the square: $(\cos A + \cos B + \cos C)^2 = \cos^2 A + \cos^2 B + \cos^2 C + 2(\cos A \cos B + \cos B \cos C + \cos C \cos A)$.Substituting this into the expression:$a^2 - 2b - 2c = \cos^2 A + \cos^2 B + \cos^2 C + 2(\cos A \cos B + \cos B \cos C + \cos C \cos A) - 2(\cos A \cos B + \cos B \cos C + \cos C \cos A) - 2(\cos A \cos B \cos C)$.$a^2 - 2b - 2c = \cos^2 A + \cos^2 B + \cos^2 C - 2 \cos A \cos B \cos C$.Using the identity for angles of a triangle: $\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2 \cos A \cos B \cos C$.Therefore,$a^2 - 2b - 2c = 1 - 2 \cos A \cos B \cos C - 2 \cos A \cos B \cos C$ is incorrect in the prompt's logic. Re-evaluating: The identity is $\cos^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C = 1$.Thus,$a^2 - 2b = \cos^2 A + \cos^2 B + \cos^2 C$.$a^2 - 2b - 2c = \cos^2 A + \cos^2 B + \cos^2 C - 2(-c) = \cos^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C = 1$.

If the roots of $x^3 + ax^2 + bx + c = 0$ are the cosines of the angles of an acute triangle,then the value of $a^2 - 2b - 2c$ is:

Similar Questions

If $x \neq -y$ and $\sin x + \sin y = 3(\cos y - \cos x)$,then $\tan(x - y) =$

If $\sin A, \cos A$ and $\tan A$ are in $G.P.$,then $\cos^3 A + \cos^2 A$ is equal to

If $2 \sec 2\alpha = \tan \beta + \cot \beta$,then one of the values of $\alpha + \beta$ is

If $\frac{3\pi}{4} < \alpha < \pi,$ then $\sqrt{\csc^2 \alpha + 2\cot \alpha}$ is equal to

If $\sin x+\sin ^2 x=1$,where $x \in\left(0, \frac{\pi}{2}\right)$,then the value of $(\cos ^{12} x+\tan ^{12} x)+3(\cos ^{10} x+\tan ^{10} x+\cos ^8 x+\tan ^8 x)+(\cos ^6 x+\tan ^6 x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module