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Jhulian

  • one year ago

got stuck... 2sin(squared) theta =1

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  1. kawii2004
    • one year ago
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    okay

  2. Jhulian
    • one year ago
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    help me solve

  3. kawii2004
    • one year ago
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    umm wait

  4. anonymous
    • one year ago
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    Do you mean 2sin^2theta=1

  5. Jhulian
    • one year ago
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    yes

  6. anonymous
    • one year ago
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    \[2 \sin ^2\theta=1,or~1-\cos 2\theta=1\] \[\cos 2\theta=0=\cos (2n+1)\frac{ \pi }{ 2 }\] \[2 \theta=\left( 2n+1 \right)\frac{ \pi }{ 2 },\theta=\left( 2n+1 \right)\frac{ \pi }{ 4 }\]

  7. anonymous
    • one year ago
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    here n is an integer.

  8. Jhulian
    • one year ago
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    thanks

  9. anonymous
    • one year ago
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    2sin²(θ) = 1 sin²(θ) = 1/2 sin(θ) = ± 1/√2 sin(θ) = ± (√2)/2 First case: sin(θ) = + (√2)/2 θ = π/4 θ = π - (π/4) = 3π/4 Second case: sin(θ) = - (√2)/2 θ = 2π - (π/4) = 7π/4 θ = π + (π/4) = 5π/4 → Solution = { π/4 ; 3π/4 ; 5π/4 ; 7π/4 }

  10. Jhulian
    • one year ago
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    Thanks princess!

  11. anonymous
    • one year ago
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    yw

  12. anonymous
    • one year ago
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    You are quite welcome @Jhulian if you no longer have any questions about this subject please do close the question. Thank you very much.(:

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