If sin θ − cos θ = 0 and 0°<θ<90 find theta
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Answered by
4
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Given:
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Or,
Answered by
7
Heya user ,
Here is your answer !!
Given ,
sin θ − cos θ = 0
=> sin θ = cos θ
=> tan θ = 1
=> tan θ = tan 45 degree
=> θ = 45 degree .... [ Ans ]
# ALTERNATIVE PROCESS #
Given ,
sin θ − cos θ = 0 .... ( i )
=> ( sin θ − cos θ )^2 = 0
=> sin^2 θ - 2sin θ cos θ + cos^2 θ = 0
=> 1 = 2sin θ cos θ .
Now ,
( sin θ + cos θ )^2
= sin^2 θ + cos^2 θ + 2 sin θ cos θ
= 1 + 1
= 2 .
Therefore , sin θ + cos θ = √2 ... ( ii )
Solving ( i ) and ( ii ) , we get ,
sin θ + cos θ = √2
sin θ − cos θ = 0
=> 2 sin θ = √2
=> sin θ = √2 / 2
=> sin θ = 1 / √2
=> sin θ = sin 45 degree
=> θ = 45 degree ....... [ Ans ]
Hope it helps you !!
Here is your answer !!
Given ,
sin θ − cos θ = 0
=> sin θ = cos θ
=> tan θ = 1
=> tan θ = tan 45 degree
=> θ = 45 degree .... [ Ans ]
# ALTERNATIVE PROCESS #
Given ,
sin θ − cos θ = 0 .... ( i )
=> ( sin θ − cos θ )^2 = 0
=> sin^2 θ - 2sin θ cos θ + cos^2 θ = 0
=> 1 = 2sin θ cos θ .
Now ,
( sin θ + cos θ )^2
= sin^2 θ + cos^2 θ + 2 sin θ cos θ
= 1 + 1
= 2 .
Therefore , sin θ + cos θ = √2 ... ( ii )
Solving ( i ) and ( ii ) , we get ,
sin θ + cos θ = √2
sin θ − cos θ = 0
=> 2 sin θ = √2
=> sin θ = √2 / 2
=> sin θ = 1 / √2
=> sin θ = sin 45 degree
=> θ = 45 degree ....... [ Ans ]
Hope it helps you !!
EmadAhamed:
Great answer! ^_^
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