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Question 144 of 175

Q.(a) For the given base curve y=sin⁡xy=\sin x, draw y=12sin⁡2xy=\dfrac{1}{2}\sin 2x. OR

(b) Write any five different forms of an equation of a straight line.
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2019Subjective· 5mImportance★★★★★
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Figure — Plot y=(1/2)sin 2x over  0,2pi
Figure — Plot y=(1/2)sin 2x over 0,2pi

Starting from y=sin⁡xy=\sin x (amplitude 1, period 2π2\pi), y=12sin⁡2xy=\dfrac{1}{2}\sin2x halves the amplitude to 0.50.5 and halves the period to π\pi, so its graph is a flatter sine wave that completes two cycles in the space of one cycle of sin⁡x\sin x.

The base curve y=sin⁡xy=\sin x has amplitude 11 and period 2π2\pi, starting at (0,0)(0,0), rising to a maximum of 11 at x=π/2x=\pi/2, back to 00 at x=πx=\pi, down to a minimum of −1-1 at x=3π/2x=3\pi/2, and back to 00 at x=2πx=2\pi.

For y=12sin⁡2xy=\dfrac{1}{2}\sin2x: replacing xx by 2x2x inside the sine compresses the period by a factor of 2 (new period =2π/2=π=2\pi/2=\pi), and the factor 12\dfrac12 outside scales the amplitude down to 0.50.5 (new range [−0.5,0.5][-0.5, 0.5]).

A table of values over one period of the base curve, [0,2π][0,2\pi]:

xx00π/4\pi/4π/2\pi/23π/43\pi/4π\pi5π/45\pi/43π/23\pi/27π/47\pi/42π2\pi
2x2x00π/2\pi/2π\pi3π/23\pi/22π2\pi5π/25\pi/23π3\pi7π/27\pi/24π4\pi
sin⁡2x\sin2x001100−1-1001100−1-100

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