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secondary 3 | A Maths
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do help with (b) thanks
Date Posted: 3 years ago
Views: 271
==> sin(5x)/cos(5x) = sin(2x)/cos(2x)
Cross multiplying, ==> sin(5x)cos(2x) = cos(3\5x)sin(2x)
==> sin(5x)cos(2x) - cos(5x)sin(2x) = 0
==>sin(5x - 2x) = 0 [Applying sin(A - B) = sinAcosB - cosAsinB]
==> sin(3x) = 0
==> sin(3x) = sin(nπ), where 'n' is an integer.
==> 3x = nπ
Thus 'x = nπ/3 radians'
Cross multiplying, ==> sin(5x)cos(2x) = cos(3\5x)sin(2x)
==> sin(5x)cos(2x) - cos(5x)sin(2x) = 0
==>sin(5x - 2x) = 0 [Applying sin(A - B) = sinAcosB - cosAsinB]
==> sin(3x) = 0
==> sin(3x) = sin(nπ), where 'n' is an integer.
==> 3x = nπ
Thus 'x = nπ/3 radians'
means x = n times 60 degree, where n = 1,2,3 etc--> in this case n can be 1 to 5
See 1 Answer
done
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clear
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Kindly verify the workings eh, thanks.
Date Posted:
3 years ago
thank you
thank you
You're welcome.
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