Mad max: Sine of Cosine

First we multiply and divide given expression by $\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } }$ to get

$\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \times \left( \frac { 3\sin { \theta } }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } +\frac { 4\cos { \theta } }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } \right)$

Consider above triangle.So we have $\sin { \angle B } =\frac { 3 }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } }$ and $\cos { \angle B } =\frac { 4 }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } }$ .

Therefore

$\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \times \left( \frac { 3\sin { \theta } }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } +\frac { 4\cos { \theta } }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } \right) =\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \left( \sin { \angle B } \sin { \theta } +\cos { \angle B } \cos { \theta } \right) =\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \left( \cos { \theta -\angle B } \right)$

i.e $3\sin { \theta } +4\cos { \theta =\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \left( \cos { \theta -\angle B } \right) }$

Now we know

$-1\le \left( \cos { \theta -\angle B } \right) \le 1\Rightarrow -\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \le \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \left( \cos { \theta -\angle B } \right) \le \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } }$

Therefore maximum value of $\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \left( \cos { \theta -\angle B } \right)$ or $3\sin { \theta } +4\cos { \theta }$ is $\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } =5$ .

The max value $a\sin { \theta } +b\cos { \theta }$ is $\sqrt { { a }^{ 2 }+{ b }^{ 2 } }$