Question

If acosФ+bsinФ=8, asinФ+bcosФ=4, find a²+b²

Answer 1

acosФ+bsinФ=8_______(1)
asinФ - bcosФ=4________(2)
squaring equations (1) and (2) then, add
=> (a²cos²Φ + b²sin²Φ+ 2abcosΦ.sinΦ)+( b²cos²Φ + a²sin²Φ - 2abcosΦ.sinΦ) = 8² + 4²
=> a²(sin ²Φ + cos²Φ)+b²(sin²Φ+cos²Φ) = 64 + 16
a²+b² = 80

Answer 2

acosø + bsinø = 8
asinø + bcosø = 4

So this equation squaring add

= ( a² cos² ø + b²sin²ø +2abCosø.Sinø)+(b²cos² ø + a²sin²ø 2abcosø.sinø) 8²+4²

We taken common

a² (sin² ø + cos²ø) + b² (sin² ø - cos²ø) 64 + 16

So

a² + b² = 80