Question

y=3sintheta + 4costheta find the maximum value of y​

Answer 1

Answer:

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Answer 2

Answer:

Given:

3\sin \theta+4\ cos \theta3sinθ+4 cosθ

To find:

Maximum value of above expression

Solution:

As we know the general form :

a\sin \theta+b\cos \thetaasinθ+bcosθ

Here, a=3,b=4a=3,b=4

So, the maximum value would be :

\begin{gathered}\sqrt{a^2+b^2}\\\\=\sqrt{3^2+4^2}\\\\=\sqrt{9+16}\\\\=\sqrt{25}\\\\=5\end{gathered}

a

2

+b

2

=

3

2

+4

2

=

9+16

=

25

=5

Hence, the maximum value is 5.