The value of cos^2 10^circ - cos 10^circ cos | vedclass

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

Question

(C) Given expression: $E = \cos^2 10^\circ - \cos 10^\circ \cos 50^\circ + \cos^2 50^\circ$Multiply and divide by $2$: $E = \frac{1}{2} [2\cos^2 10^\c

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(C) Given expression: $E = \cos^2 10^\circ - \cos 10^\circ \cos 50^\circ + \cos^2 50^\circ$Multiply and divide by $2$: $E = \frac{1}{2} [2\cos^2 10^\circ - 2\cos 10^\circ \cos 50^\circ + 2\cos^2 50^\circ]$Using $2\cos^2 	heta = 1 + \cos 2	heta$ and $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:$E = \frac{1}{2} [(1 + \cos 20^\circ) - (\cos 60^\circ + \cos(-40^\circ)) + (1 + \cos 100^\circ)]$$E = \frac{1}{2} [1 + \cos 20^\circ - \frac{1}{2} - \cos 40^\circ + 1 + \cos 100^\circ]$$E = \frac{1}{2} [\frac{3}{2} + \cos 20^\circ - \cos 40^\circ + \cos 100^\circ]$Using $\cos 100^\circ - \cos 40^\circ = -2 \sin(\frac{100+40}{2}) \sin(\frac{100-40}{2}) = -2 \sin 70^\circ \sin 30^\circ = -2 \cos 20^\circ (\frac{1}{2}) = -\cos 20^\circ$$E = \frac{1}{2} [\frac{3}{2} + \cos 20^\circ - \cos 20^\circ] = \frac{1}{2} 	imes \frac{3}{2} = \frac{3}{4}$

The value of $\cos^2 10^\circ - \cos 10^\circ \cos 50^\circ + \cos^2 50^\circ$ is equal to:

Similar Questions

If $\tan \theta + \cot \theta = 4$,then $\tan^{4} \theta + \cot^{4} \theta = $

If $x = \log \left[ \cot \left( \frac{\pi}{4} + \theta \right) \right]$, then the value of $\sinh x$ is

The value of $\cos 52^\circ + \cos 68^\circ + \cos 172^\circ$ is

If $3 \sin^4 x + 2 \cos^4 x = \frac{6}{5}$ and $x$ is an acute angle,then $\tan 2x =$

The equation $e^{\sin x} - e^{\sin(-x)} - 4 = 0$ has

Vedclass Test Series

Exam Paper Generator

Online Exam Module