Question

If a cos + b sin = 4and a sin -b cos = 3, then find the value of a2 + b2​

Answer 1

Answer :

a² + b² = 25

Solution :

• Given : a·cos∅ + b·sin∅ = 4

a·sin∅ – b·cos∅ = 3

• To find : a² + b² = ?

We have ,

a·cos∅ + b·sin∅ = 4

Now ,

Squaring both the sides , we get ;

=> (a·cos∅ + b·sin∅)² = 4²

=> (a·cos∅)² + (b·sin∅)² + 2·(a·cos∅)·(b·sin∅) = 16

=> a²·cos²∅ + b²·sin²∅ + 2ab·sin∅·cos∅ = 16 --------(1)

Also ,

We have ,

a·sin∅ – b·cos∅ = 3

Now ,

=> (a·sin∅ – b·cos∅)² = 3²

=> (a·sin∅)² + (b·cos∅)² – 2·(a·sin∅)·(b·cos∅) = 9

=> a²·sin²∅ + b²·cos²∅ – 2ab·sin∅·cos∅ = 9 ------(2)

Now ,

Adding eq-(1) and (2) , we get ;

=> a²·cos²∅ + b²·sin²∅ + 2ab·sin∅·cos∅ + a²·sin²∅ + b²·cos²∅ – 2ab·sin∅·cos∅ = 16 + 9

=> a²·cos²∅ + b²·sin²∅ + a²·sin²∅ + b²·cos²∅ = 25

=> a²(sin²∅ + cos²∅) + b²(sin²∅ + cos²∅) = 25

=> (sin²∅ + cos²∅)·(a² + b²) = 25

=> 1·(a² + b²) = 25

=> a² + b² = 25

Hence ,

a² + b² = 25