Question

maximum value of 3cosA+4sinA​

Answer 1

Answer:

4 sinA + 3 cosA

=4 sinA + 3 cosA * √(4^2 +3^2)        √(4^2 + 3^2)

Let √(4^2 + 3^2) = t

=(4 sinA + 3 cosA) * t

     t.             t

Let

          4.         =cosB

√(4^2 + 3^2)    

           5.         =sinB

√(4^2 + 3^2)

=(cosB.sinA + sinB.cosA)*t

=(sin(A+B))*√(4^2 + 3^2)

we know that max. value of sinβ=1

where β is any angle.

Therefore,

Max value of the function is

√(4^2 + 3^2)

=5

Step-by-step explanation:

Answer 2

Answer:

The answer of your question is 4