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Symbol Defintion
Example : Sqr(x) = square root of x
Q01. Sine Curve : y = sin(x)
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Q02. Sketch y = sin(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 ...... 0 ...... 1 ........ 0 ....... -1 .......... 0
- Five points (0, 0), (pi/2 ,1), (pi, 0), (1.5*pi, -1), (2*pi, 0)
- Since we know this is sine curve
- Hence we can use these five points to sketch the sine curve
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Q03. Sketch y = 2*sin(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 ...... 0 ...... 2 ........ 0 ......... -2 ........... 0
- Five points (0, 0), (pi/4 ,2), (pi/2, 0), (0.75*pi, -2), (pi, 0)
- 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 sin(60) = Sqr(3)/2 ans cos(x) = 1/2
- Hence y = 2*sin(60)*cos(x) + sin(x)*cos(60)
- Hence y = 2*sin(x + pi/3)
- The period of sine finction is 2*pi
- Hence we can sketch the function from 0 to pi
- Five points : let t = x + pi/3
- t ...... 0 ........ pi/2 ..... pi ......... 1.5*pi ...... 2*pi
- x ...... -pi/3..... pi/6 ..... 2*pi/3 ..... 7*pi/6 ...... 1*pi
- y ...... 0 ........ 2 ........ 0 .......... -2 .......... 0
- Five points (0, 0), (pi/6 ,2), (2*pi/3, 0), (7*pi/6, -2), (5*pi/3, 0)
- 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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