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Symbol Defintion
Example : Sqr(x) = square root of x
Q01. Sine Curve : y = cos(x)
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Q02. Sketch y = cos(x) using five points
Solution
- The period of sine finction is 2*pi
- Hence we can sketch the function from 0 to 2*pi
- Five points
- x ...... 0 ...... pi/2 ..... pi ...... 1.5*pi ...... 2*pi
- y ...... 1 ...... 0 ........ -1 ...... 0 ........... 1
- Five points (0, 1), (pi/2 ,0), (pi, -1), (1.5*pi, 0), (2*pi, 1)
- Since we know this is cosine curve
- Hence we can use these five points to sketch the cosine curve
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Q03. Sketch y = 2*cos(2*x) using five points method
Solution
- The period of sine finction is 2*x = 2*pi = pi
- Hence we can sketch the function from 0 to pi
- Five points : let t = 2*x
- t ...... 0 ...... pi/2 ..... pi ........ 1.5*pi ....... 2*pi
- x ...... 0 ...... pi/4 ..... pi/2 ...... 0.75*pi ...... 1*pi
- y ...... 2 ...... 0 ........ -2 ........ 0 ............ 2
- Five points (0, 2), (pi/4 ,0), (pi/2, -2), (0.75*pi, 0), (pi, 2)
- Since we know this is sine curve
- Hence we can use these five points to sketch the sine curve from x = 0 to pi
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Q4. Sketch y = Sqr(3)*cos(x) + sin(x) using five point method
Solution
- Since cos(30) = Sqr(3)/2 ans sin(x) = 1/2
- Hence y = 2*cos(x)*cos(60) + sin(x)*sin(30)
- Hence y = 2*cos(x - pi/6)
- The period of sine finction is 2*pi
- Hence we can sketch the function from 0 to pi
- Five points : let t = (x - pi/6)
- t ...... 0 ........ pi/2 ....... pi ......... 1.5*pi ...... 2*pi
- x ...... +pi/6..... 4*pi/6 ..... 7*pi/6 ..... 10*pi/6 ..... 13*pi/6
- y ...... 2 ........ 0 .......... -2 ......... 0 ........... 2
- Five points (0, 2), (4*pi/6 ,0), (7*pi/6, -2), (10*pi/6, 0), (13*pi/6, 2)
- Since we know this is sine curve
- Hence we can use these five points to sketch the sine curve from x = 0 to pi
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