Which of the following is true?

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

Question

(A) We know that the function $f(x) = \sin\ x$ is increasing in the interval $[0^{\circ}, 90^{\circ}]$.For angles greater than $90^{\circ}$,we use the

Vedclass · Accepted Answer

$\sin\ 95^{\circ} > \sin\ 63^{\circ} > \sin\ 1^{\circ}$

Vedclass · Answer

(A) We know that the function $f(x) = \sin\ x$ is increasing in the interval $[0^{\circ}, 90^{\circ}]$.For angles greater than $90^{\circ}$,we use the identity $\sin\ 	heta = \sin(180^{\circ} - 	heta)$.Thus,$\sin\ 95^{\circ} = \sin(180^{\circ} - 95^{\circ}) = \sin\ 85^{\circ}$.Now we compare $\sin\ 85^{\circ}$,$\sin\ 63^{\circ}$,and $\sin\ 1^{\circ}$.Since $85^{\circ} > 63^{\circ} > 1^{\circ}$ and the sine function is increasing in the first quadrant,we have $\sin\ 85^{\circ} > \sin\ 63^{\circ} > \sin\ 1^{\circ}$.Therefore,$\sin\ 95^{\circ} > \sin\ 63^{\circ} > \sin\ 1^{\circ}$.

Similar Questions

$\tan \left(-\frac{23 \pi}{3}\right)-\cot \left(\theta-\frac{13 \pi}{3}\right) =$

If $\tan \theta = \frac{20}{21}$, then what is the value of $\cos \theta$?

The value of $\cos 1200^{\circ} + \tan 1485^{\circ}$ is

If $\sin x = -\frac{24}{25},$ then the value of $\tan x$ is

$\frac{1}{\sin 250^{\circ}}+\frac{\sqrt{3}}{\cos 290^{\circ}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module