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11TH · MATH

Question 1

menu_bookSolutionmenu_bookTheorymenu_book•(i)y=−sin⁡2xy = -\sin 2xy=−sin2x, $x \in [-2\pi, 2\pi]$menu_book•(ii)y=2cos⁡2xy = 2\cos 2xy=2cos2x, $x \in [-2\pi, 2\pi]$menu_book•(iii)y=tan⁡2xy = \tan 2xy=tan2x, $x \in [-\pi, \pi]$menu_book•(iv)y=tan⁡x2y = \tan \dfrac{x}{2}y=tan2x​, $x \in [-2\pi, 2\pi]$menu_book•(v)y=sin⁡π2xy = \sin \dfrac{\pi}{2}xy=sin2π​x, $x \in [0, 2\pi]$menu_book•(vi)y=cos⁡π2xy = \cos \dfrac{\pi}{2}xy=cos2π​x, $x \in [-\pi, \pi]$
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school

11TH · MATH

Question 1

menu_bookSolutionmenu_bookTheorymenu_book•(i)y=−sin⁡2xy = -\sin 2xy=−sin2x, $x \in [-2\pi, 2\pi]$menu_book•(ii)y=2cos⁡2xy = 2\cos 2xy=2cos2x, $x \in [-2\pi, 2\pi]$menu_book•(iii)y=tan⁡2xy = \tan 2xy=tan2x, $x \in [-\pi, \pi]$menu_book•(iv)y=tan⁡x2y = \tan \dfrac{x}{2}y=tan2x​, $x \in [-2\pi, 2\pi]$menu_book•(v)y=sin⁡π2xy = \sin \dfrac{\pi}{2}xy=sin2π​x, $x \in [0, 2\pi]$menu_book•(vi)y=cos⁡π2xy = \cos \dfrac{\pi}{2}xy=cos2π​x, $x \in [-\pi, \pi]$
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Accedmychevron_right11thchevron_rightmathchevron_rightTrigonometric Functions And Their Graphschevron_rightExercise 11.2
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