Question

If x=3 sinФ and y=4 cosФ, find the value of √ 16x^{2}+9y^{2} [/tex]

Answer 1

the answer is 12
{16*(3sinϴ)^2+9*(4COSϴ)^2}^1/2
={16*9SINϴ^2 +9*16COSϴ^2}^1/2
={16*9(SINϴ^2 +COSϴ^2)}^1/2
={16*9*1}^1/2
=4*3
=12