The value of cos 105^circ + sin 105^circ is | vedclass

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

Question

(D) We know that $\cos 105^\circ = \cos(60^\circ + 45^\circ) = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ = \frac{1}{2} \cdot \frac{1}{

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(D) We know that $\cos 105^\circ = \cos(60^\circ + 45^\circ) = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1 - \sqrt{3}}{2\sqrt{2}}$.Similarly,$\sin 105^\circ = \sin(60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.Adding these two values:$\cos 105^\circ + \sin 105^\circ = \frac{1 - \sqrt{3}}{2\sqrt{2}} + \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{1 - \sqrt{3} + \sqrt{3} + 1}{2\sqrt{2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.

The value of $\cos 105^\circ + \sin 105^\circ$ is

Similar Questions

If $\frac{x}{\cos \alpha} = \frac{y}{\cos \left(\frac{2 \pi}{3} - \alpha\right)} = \frac{z}{\cos \left(\frac{2 \pi}{3} + \alpha\right)}$,then the value of $(x + y + z)$ is equal to

If $m \tan (\theta - 30^\circ) = n \tan (\theta + 120^\circ)$,then $\frac{m + n}{m - n} = $

The value of $\frac{3 + \cot 76^{\circ} \cot 16^{\circ}}{\cot 76^{\circ} + \cot 16^{\circ}}$ is:

If $\sin A + \sin B = C$ and $\cos A + \cos B = D$,then the value of $\sin (A + B)$ is:

If $A + C = B$,then $\tan A \tan B \tan C = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module