cos ^2 45^{circ}+cos ^2 135^{circ}+cos ^2 225 | vedclass

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

Question

(B) We use the property $\cos^2 	heta = \cos^2(180^{\circ} \pm 	heta) = \cos^2(360^{\circ} \pm 	heta)$.$\cos 135^{\circ} = \cos(180^{\circ} - 45^{\

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) We use the property $\cos^2 	heta = \cos^2(180^{\circ} \pm 	heta) = \cos^2(360^{\circ} \pm 	heta)$.$\cos 135^{\circ} = \cos(180^{\circ} - 45^{\circ}) = -\cos 45^{\circ} \implies \cos^2 135^{\circ} = \cos^2 45^{\circ}$.$\cos 225^{\circ} = \cos(180^{\circ} + 45^{\circ}) = -\cos 45^{\circ} \implies \cos^2 225^{\circ} = \cos^2 45^{\circ}$.$\cos 315^{\circ} = \cos(360^{\circ} - 45^{\circ}) = \cos 45^{\circ} \implies \cos^2 315^{\circ} = \cos^2 45^{\circ}$.Thus,the expression becomes:$\cos^2 45^{\circ} + \cos^2 45^{\circ} + \cos^2 45^{\circ} + \cos^2 45^{\circ} = 4 \cos^2 45^{\circ}$.Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have $\cos^2 45^{\circ} = \frac{1}{2}$.Therefore,$4 	imes \frac{1}{2} = 2$.

$\cos ^2 45^{\circ}+\cos ^2 135^{\circ}+\cos ^2 225^{\circ}+\cos ^2 315^{\circ} = $

Similar Questions

Which of the following is correct?

$\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta} = $

$\cos 1^\circ + \cos 2^\circ + \cos 3^\circ + \dots + \cos 180^\circ = $

If $\tan \theta = - \frac{1}{\sqrt{10}}$ and $\theta$ lies in the fourth quadrant,then $\cos \theta = $

$\sin \left(\frac{5 \pi}{3}\right) + \sec \left(\frac{13 \pi}{3}\right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module