4 votes 4 votes For $0\leq{x}\leq{2\pi}$, $\sin x \text{ and } \cos x$ are both decreasing functions in the interval _________ . $\left(0,\dfrac{\pi}{2}\right)$ $\left(\dfrac{\pi}{2},\pi\right)$ $\left(\pi,\dfrac{3\pi}{2}\right)$ $\left(\dfrac{3\pi}{2},2\pi\right)$ Quantitative Aptitude gate2018-ch quantitative-aptitude functions trigonometry + – Lakshman Bhaiya asked Feb 20, 2018 • edited Jun 2, 2019 by Lakshman Bhaiya Lakshman Bhaiya 2.7k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 6 votes 6 votes Both functions are decreasing in the interval $(\pi/2, \pi )$ pankaj_vir answered Mar 4, 2018 • edited Jun 19, 2019 by ajaysoni1924 pankaj_vir comment Share Follow See 1 comment See all 1 1 comment reply This_is_Nimishka commented Dec 25, 2023 reply Follow Share And it is increasing in the interval $\left ( \frac{3\pi }{2},2\pi \right )$, right? 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes Answer will be C) As, Sin is -ve in the interval 3rd and 4th quadrant of axis and Cos -ve for 2nd and 3rd quadrant of axis srestha answered Feb 20, 2018 srestha comment Share Follow See all 2 Comments See all 2 2 Comments reply Nikhil gate 2020 commented Jan 21, 2020 reply Follow Share here not ask nigative value edit your answer 0 votes 0 votes Rutvik2900 commented Aug 21, 2020 reply Follow Share That’s right, but if we solve it using definition, which is f’(x)<0 then both function will be -Ve in the interval $(\prod, 3\prod/2)$ 0 votes 0 votes Please log in or register to add a comment.