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Real or imaginary part of a fraction in sin theta or cos theta is given: find theta

This type of question, asked 5 times: 2021 Paper 12, 2021 Paper 17, 27 Jul 2022, 8 Apr 2023, 8 Apr 2024. All 5 questions below, oldest first.

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  1. Question 1

    2021 Paper 12

    45
    If the real part of the complex number (1−cos⁡θ+2isinθ)−1(1 - \cos\theta+ 2isin\theta)^{- 1} is 15\frac{1}{5} for θ∈(0,π)\theta\in (0,\pi), then the value of the integral ∫0θsin⁡xdx\int_{0}^{\theta} \sin xdx is equal to :

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  2. Question 2

    2021 Paper 17

    83
    If the real part of the complex number z=3+2icosθ1−3icosθ,θ∈(0,π2)z =\frac{3 + 2icos\theta}{1 - 3icos\theta},\theta \in\left( 0,\frac{\pi}{2} \right) is zero, then the value of sin⁡23θ+cos⁡2θ\sin^{2}3\theta +\cos^{2}\theta is equal to
  3. Question 3

    27 Jul 2022

    152
    Let SS be the set of all (α,β),π<α,β<2π(\alpha,\beta),\pi < \alpha,\beta < 2\pi, for which the complex number 1−isinα1+2isinα\frac{1 - isin\alpha}{1 + 2isin\alpha} is purely imaginary and 1+icosβ1−2icosβ\frac{1 + icos\beta}{1 - 2icos\beta} is purely real. Let Zαβ=sin⁡2α+icos2β,(α,β)∈SZ_{\alpha\beta}= \sin2\alpha + icos2\beta,(\alpha,\beta) \in S. Then ∑(α,β)∈S(iZαβ+1iZˉαβ)\sum_{(\alpha,\beta) \in S} \left( iZ_{\alpha\beta}+\frac{1}{i{\bar{Z}}_{\alpha\beta}} \right) is equal to:

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  4. Question 4

    8 Apr 2023

    154
    Let A=θ∈(0,2π):1+2isinθ1−isinθA = \theta \in (0,2\pi):\frac{1 + 2isin\theta}{1 - isin\theta} is purely imaginary. Then the sum of the elements in AA is

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  5. Question 5

    8 Apr 2024

    156
    The sum of all possible values of θ∈[−π,2π]\theta \in \lbrack - \pi,2\pi\rbrack, for which 1+icosθ1−2icosθ\frac{1 + icos\theta}{1 - 2icos\theta} is purely imaginary, is equal to

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