solve sin²π/18+sin²π/9+sin²7π/18+sin²4π/9 =2
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We have sin² θ + cos² θ = 1 and sin² θ = cos² (π/2 - θ )
Using these formulae, we can proceed with this question.
Consider sin² π/18. It can be written as cos² (π/2 - π/18).
cos² (π/2 - π/18) = cos² (9π-π/18) = cos² (8π/18) = cos² (4π/9)
Consider sin² π/9. It can be written as cos² (π/2 - π/9).
cos² (π/2 - π/9) = cos² (9π-2π/18) = cos² (7π/18)
Thus the question takes the form
cos² (4π/9) + cos² (7π/18) + sin² (7π/18) + sin² (4π/9)
= { sin² (7π/18) + cos² (7π/18) } + { sin² (4π/9) + cos² (4π/9) }
As mentioned earlier, we have sin² θ + cos² θ = 1 and in this case, we have θ = 7π/18 and 4π/9 respectively.
Hence { sin² (7π/18) + cos² (7π/18) } + { sin² (4π/9) + cos² (4π/9) } = 1+1 = 2 which is the required answer.
Using these formulae, we can proceed with this question.
Consider sin² π/18. It can be written as cos² (π/2 - π/18).
cos² (π/2 - π/18) = cos² (9π-π/18) = cos² (8π/18) = cos² (4π/9)
Consider sin² π/9. It can be written as cos² (π/2 - π/9).
cos² (π/2 - π/9) = cos² (9π-2π/18) = cos² (7π/18)
Thus the question takes the form
cos² (4π/9) + cos² (7π/18) + sin² (7π/18) + sin² (4π/9)
= { sin² (7π/18) + cos² (7π/18) } + { sin² (4π/9) + cos² (4π/9) }
As mentioned earlier, we have sin² θ + cos² θ = 1 and in this case, we have θ = 7π/18 and 4π/9 respectively.
Hence { sin² (7π/18) + cos² (7π/18) } + { sin² (4π/9) + cos² (4π/9) } = 1+1 = 2 which is the required answer.
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