cos ^2 10^{circ}+cos ^2 50^{circ}-sin 40^{cir | vedclass

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

Question

(D) We have the expression: $\cos ^2 10^{\circ}+\cos ^2 50^{\circ}-\sin 40^{\circ} \sin 80^{\circ}$ Using the identity $2 \cos ^2 A = 1 + \cos 2A$ and

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(D) We have the expression: $\cos ^2 10^{\circ}+\cos ^2 50^{\circ}-\sin 40^{\circ} \sin 80^{\circ}$ Using the identity $2 \cos ^2 A = 1 + \cos 2A$ and $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$: $= \frac{1}{2} [ (1 + \cos 20^{\circ}) + (1 + \cos 100^{\circ}) - (\cos(40^{\circ}-80^{\circ}) - \cos(40^{\circ}+80^{\circ})) ]$ $= \frac{1}{2} [ 2 + \cos 20^{\circ} + \cos 100^{\circ} - \cos(-40^{\circ}) + \cos 120^{\circ} ]$ Since $\cos(-	heta) = \cos 	heta$ and $\cos 120^{\circ} = -\frac{1}{2}$: $= \frac{1}{2} [ 2 + \cos 20^{\circ} + \cos 100^{\circ} - \cos 40^{\circ} - \frac{1}{2} ]$ $= \frac{1}{2} [ \frac{3}{2} + (\cos 100^{\circ} + \cos 20^{\circ}) - \cos 40^{\circ} ]$ Using $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$: $= \frac{1}{2} [ \frac{3}{2} + 2 \cos 60^{\circ} \cos 40^{\circ} - \cos 40^{\circ} ]$ Since $\cos 60^{\circ} = \frac{1}{2}$: $= \frac{1}{2} [ \frac{3}{2} + 2(\frac{1}{2}) \cos 40^{\circ} - \cos 40^{\circ} ]$ $= \frac{1}{2} [ \frac{3}{2} + \cos 40^{\circ} - \cos 40^{\circ} ] = \frac{1}{2} 	imes \frac{3}{2} = \frac{3}{4}$

$\cos ^2 10^{\circ}+\cos ^2 50^{\circ}-\sin 40^{\circ} \sin 80^{\circ}$ is equal to

Similar Questions

If $\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$,then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=$

If $\alpha, \beta$ are acute angles such that $\frac{\sin \alpha}{\sin \beta} = \frac{6}{5}$ and $\frac{\cos \alpha}{\cos \beta} = \frac{9}{5 \sqrt{5}}$,then $\sin \alpha = $

If $\theta$ is an acute angle,$\cosh x = K$ and $\sinh x = \tan \theta$,then $\sin \theta =$

If $y^2+z^2=a y z$,$z^2+x^2=b x z$,and $x^2+y^2=c x y$,then the value of $\frac{x z}{y^2}+\frac{y^2}{z x}$ is

If $x: y: z = \tan \left(\frac{\pi}{15}+\alpha\right): \tan \left(\frac{\pi}{15}+\beta\right): \tan \left(\frac{\pi}{15}+\gamma\right)$,then find the value of $\frac{z+x}{z-x} \sin ^2(\gamma-\alpha)+\frac{x+y}{x-y} \sin ^2(\alpha-\beta)+\frac{y+z}{y-z} \sin ^2(\beta-\gamma)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module