The value of cos 24^{circ} + cos 55^{circ} + | vedclass

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

Question

(A) We use the trigonometric identities: $\cos(180^{\circ} + 	heta) = -\cos 	heta$ and $\cos(360^{\circ} - 	heta) = \cos 	heta$.Given expression:

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) We use the trigonometric identities: $\cos(180^{\circ} + 	heta) = -\cos 	heta$ and $\cos(360^{\circ} - 	heta) = \cos 	heta$.Given expression: $E = \cos 24^{\circ} + \cos 55^{\circ} + \cos 125^{\circ} + \cos 204^{\circ} + \cos 300^{\circ}$.$1$. $\cos 125^{\circ} = \cos(180^{\circ} - 55^{\circ}) = -\cos 55^{\circ}$.$2$. $\cos 204^{\circ} = \cos(180^{\circ} + 24^{\circ}) = -\cos 24^{\circ}$.$3$. $\cos 300^{\circ} = \cos(360^{\circ} - 60^{\circ}) = \cos 60^{\circ} = \frac{1}{2}$.Substituting these values into the expression:$E = \cos 24^{\circ} + \cos 55^{\circ} - \cos 55^{\circ} - \cos 24^{\circ} + \frac{1}{2}$.$E = 0 + 0 + \frac{1}{2} = \frac{1}{2}$.

The value of $\cos 24^{\circ} + \cos 55^{\circ} + \cos 125^{\circ} + \cos 204^{\circ} + \cos 300^{\circ}$ is

Similar Questions

If $f(x) = \cos^2 x + \sec^2 x$,then its value is always:

If $\tan \alpha = \frac{x^2 - x}{x^2 - x + 1}$ and $\tan \beta = \frac{1}{2x^2 - 2x + 1}$ $(x \ne 0, 1)$,where $0 < \alpha, \beta < \frac{\pi}{2}$,then $\tan(\alpha + \beta)$ has the value equal to:

$\frac{\sin 70^\circ + \cos 40^\circ}{\cos 70^\circ + \sin 40^\circ} = $

If $\sec \theta = x + \frac{1}{4x}$ where $0^{\circ} < \theta < 90^{\circ}$,then $\sec \theta + \tan \theta$ is equal to:

If $\tan A = \frac{1}{2}$ and $\tan B = \frac{1}{3}$,then $\cos 2A = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module