Question

if  acosФ+bsinФ=m 
and asinФ-bcosФ=n
prove that, m²+n²=a²+b²

Answer 1

aCosФ + bSinФ=m ; asinФ-bcosФ=n

Taking L.H.S,
m²+n²
=a²Cos²Ф+b²Sin²Ф+2abCosФSinФ+b²Cos²Ф+a²Sin²Ф-2abCosФSinФ
Since 2abCosФSinФ is opposite in sign it gets cancelled,
=a²Cos²Ф+b²Sin²Ф+a²Sin²Ф+b²Cos²Ф
Pairing up the terms,
=a²Cos²Ф+a²Sin²Ф+b²Sin²Ф+b²Cos²Ф
Now taking a² and b² common,
=a²(Cos²Ф+Sin²Ф)+b²(Sin²Ф+Cos²Ф)
Since Cos²Ф+Sin²Ф=1
=a²+b²(Hence, Proved)
L.H.S=R.H.S
m²+n²=a²+b²

Answer 2

acosФ + bsinФ = m
 asinФ - bcosФ = n

m² + n² = a²cos²Ф + b²cos²Ф + 2absinФcosФ + a²sin²Ф + b²cos²Ф - 2abcosФsinФ
m² + n² = a²(cos²Ф + sin²Ф) + b²(cos²Ф + sin²Ф)
m² + n² = a² + b²     prove           (cos²Ф + sin²Ф = 1)