If A+B=C,then the value of cos ^2 A+cos ^2 B+ | vedclass

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

Question

(A) Given that $A+B=C$. We need to evaluate the expression $E = \cos ^2 A+\cos ^2 B+\cos ^2 C-2 \cos A \cos B \cos C$. Using the identity $\cos ^2 	h

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given that $A+B=C$. We need to evaluate the expression $E = \cos ^2 A+\cos ^2 B+\cos ^2 C-2 \cos A \cos B \cos C$. Using the identity $\cos ^2 	heta = \frac{1+\cos 2	heta}{2}$,we have: $E = \frac{1+\cos 2A}{2} + \frac{1+\cos 2B}{2} + \cos ^2 C - 2 \cos A \cos B \cos C$ $E = 1 + \frac{1}{2}(\cos 2A + \cos 2B) + \cos ^2 C - 2 \cos A \cos B \cos C$ Using $\cos 2A + \cos 2B = 2 \cos(A+B) \cos(A-B)$: $E = 1 + \cos(A+B) \cos(A-B) + \cos ^2 C - 2 \cos A \cos B \cos C$ Since $A+B=C$,substitute $\cos(A+B) = \cos C$: $E = 1 + \cos C \cos(A-B) + \cos ^2 C - 2 \cos A \cos B \cos C$ $E = 1 + \cos C [\cos(A-B) + \cos C] - 2 \cos A \cos B \cos C$ Substitute $\cos C = \cos(A+B) = \cos A \cos B - \sin A \sin B$: $E = 1 + \cos C [\cos A \cos B + \sin A \sin B + \cos A \cos B - \sin A \sin B] - 2 \cos A \cos B \cos C$ $E = 1 + \cos C [2 \cos A \cos B] - 2 \cos A \cos B \cos C$ $E = 1 + 2 \cos A \cos B \cos C - 2 \cos A \cos B \cos C = 1$.

If $A+B=C$,then the value of $\cos ^2 A+\cos ^2 B+\cos ^2 C-2 \cos A \cos B \cos C$ is equal to:

Similar Questions

The minimum value of $(8 \sec^2 \theta + 2 \cos^2 \theta)$ is equal to :-

Which of the following functions have the maximum value unity?

If $f(x) = \sin^6 x + \cos^6 x$ for $x \in R$,then $f(x)$ lies in the interval

The minimum and maximum values of $\cos \left(x+\frac{\pi}{3}\right)+2 \sqrt{2} \sin \left(x+\frac{\pi}{3}\right)$ are respectively

If $A+B+C=2S$,then $\sin(S-A)+\sin(S-B)-\sin C=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module