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junior college 1 | H2 Maths
2 Answers Below
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hi i need help with this, thank you!
Date Posted: 2 years ago
Views: 213
w = cosθ + i sinθ
w + 1/w
= cosθ + i sinθ + 1/(cosθ + i sinθ)
= cosθ + i sinθ + (cosθ - i sinθ)/[(cosθ - i sinθ)(cosθ + i sinθ)]
(Rationalising the denominator)
= cosθ + i sinθ + (cosθ - i sinθ)/(cos²θ - i² sin²θ)
(Recall that (a - b)(a + b) = a² - b²)
= cosθ + i sinθ + (cosθ - i sinθ)/(cos²θ + sin²θ)
(Since i² = -1)
= cosθ + i sinθ + (cosθ - i sinθ)/1
(Since cos²θ + sin²θ = 1)
= cosθ + i sinθ + cosθ - i sinθ
= 2 cosθ
There is no imaginary component, so the number is real.
w + 1/w
= cosθ + i sinθ + 1/(cosθ + i sinθ)
= cosθ + i sinθ + (cosθ - i sinθ)/[(cosθ - i sinθ)(cosθ + i sinθ)]
(Rationalising the denominator)
= cosθ + i sinθ + (cosθ - i sinθ)/(cos²θ - i² sin²θ)
(Recall that (a - b)(a + b) = a² - b²)
= cosθ + i sinθ + (cosθ - i sinθ)/(cos²θ + sin²θ)
(Since i² = -1)
= cosθ + i sinθ + (cosθ - i sinθ)/1
(Since cos²θ + sin²θ = 1)
= cosθ + i sinθ + cosθ - i sinθ
= 2 cosθ
There is no imaginary component, so the number is real.
thank you!
See 2 Answers
thank you!!
sorry can i ask for (i)(b) how do you get from e^itheta - 1/e^1theta + 1 to the next step! thank you ><
I will post an explanation
ohh i see, thank you so much!
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