Jhulian
  • Jhulian
got stuck... 2sin(squared) theta =1
Trigonometry
schrodinger
  • schrodinger
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kawii2004
  • kawii2004
okay
Jhulian
  • Jhulian
help me solve
kawii2004
  • kawii2004
umm wait

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anonymous
  • anonymous
Do you mean 2sin^2theta=1
Jhulian
  • Jhulian
yes
anonymous
  • anonymous
\[2 \sin ^2\theta=1,or~1-\cos 2\theta=1\] \[\cos 2\theta=0=\cos (2n+1)\frac{ \pi }{ 2 }\] \[2 \theta=\left( 2n+1 \right)\frac{ \pi }{ 2 },\theta=\left( 2n+1 \right)\frac{ \pi }{ 4 }\]
anonymous
  • anonymous
here n is an integer.
Jhulian
  • Jhulian
thanks
anonymous
  • anonymous
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 }
Jhulian
  • Jhulian
Thanks princess!
anonymous
  • anonymous
yw
anonymous
  • anonymous
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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