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

(B) Let $E = \cos^2 10^\circ - \cos 10^\circ \cos 50^\circ + \cos^2 50^\circ$.Multiply and divide by $2$:$E = \frac{1}{2} [2 \cos^2 10^\circ - 2 \cos

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(B) Let $E = \cos^2 10^\circ - \cos 10^\circ \cos 50^\circ + \cos^2 50^\circ$.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 the identities $2 \cos^2 	heta = 1 + \cos 2	heta$ 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 $\cos C - \cos D = 2 \sin(\frac{C+D}{2}) \sin(\frac{D-C}{2})$:$\cos 40^\circ - \cos 100^\circ = 2 \sin(\frac{140^\circ}{2}) \sin(\frac{60^\circ}{2}) = 2 \sin 70^\circ \sin 30^\circ = 2 \sin 70^\circ (\frac{1}{2}) = \sin 70^\circ = \cos 20^\circ$.Substituting back:$E = \frac{1}{2} [\frac{3}{2} + \cos 20^\circ - \cos 20^\circ] = \frac{1}{2} 	imes \frac{3}{2} = \frac{3}{4}$.

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

Similar Questions

If $\frac{\cos \alpha}{\sin \beta} = n$ and $\frac{\cos \alpha}{\cos \beta} = m$,then the value of $\cos^2 \beta$ is:

The value of $\cos 12^\circ + \cos 84^\circ + \cos 156^\circ + \cos 132^\circ$ is

If $\frac{\cos 3A + \sin 3A}{\cos A - \sin A} = 1 - K \sin 2A,$ the value of $K$ is

The value of $(\operatorname{cosec} a - \sin a)(\sec a - \cos a)(\tan a + \cot a)$ is:

If $\cos P = \frac{1}{7}$ and $\cos Q = \frac{13}{14}$,where $P$ and $Q$ both are acute angles,then the value of $P - Q$ is....$^o$

Vedclass Test Series

Exam Paper Generator

Online Exam Module