Question

If costita+sinteta=root2 costita prove costheta-sintheta=root 2 cos theta

Answer 1

Hi ,

Here I am using ' A ' instead of theta .

cos A + sinA = √2 cosA ( given )

=> ( cos A + sinA )² = ( √2 cosA )²

=> cos²A + sin²A + 2cosAsinA = 2cos²A

=> sin²A = 2cos²A - cos²A - 2cosAsinA

=> sin²A = cos²A - 2cosAsinA

=> sin²A + sin²A = sin²A + cos²A - 2cosAsinA

=> 2sin²A = ( cosA - sinA )²

=> √2 sinA = cosA - sinA

Therefore ,

cosA - sinA = √2sinA

Hence proved.

: )

Answer 2

-> cos θ + sin θ = √2 cos θ

--> sin θ = √2 cos θ - cos θ

=> sin θ = ( √2 - 1 ) cos θ

=> [ sin θ / ( √2 - 1 ) ] = cos θ

=> [ sin θ ( √2 + 1 ) / ( 2 - 1 ) ] = cos θ

0_0 --> We rationalized the denominator in the 2nd step ^_^ 

=> [ √2 sin θ + sin θ ] = cos θ

=> cos θ - sin θ = √2 sin θ