If cos theta+sin theta = 2^1/2cos theta the prove cos theta-sin theta = 2^1/2sin theta
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Hlw mate
--> 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 ^_^
Hope it helpful
--> 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 ^_^
Hope it helpful
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