140. Find the maximum value of the function

$$
y(x)=3 \sin x+4 \cos x
$$
Solution. $y(x)=5\left(\frac{3}{5} \sin x+\frac{4}{5} \cos x\right)$. Since $\left(\frac{3}{5}\right)^{2}+$ $+\left(\frac{4}{5}\right)^{2}=1$, there exists an angle $x_{0}$ such that $\frac{3}{5}=\cos x_{0}, \frac{4}{5}=\sin x_{0}$. Then $y(x)=5 \sin \left(x+x_{0}\right)$. It is obvious that the maximum value of $y(x)$ is 5.

Answer: 5.

$\square$ 141. Prove the identity $\cos 25^{\circ}+\cos 47^{\circ}=\cos 11^{\circ}+\cos 61^{\circ}+\sin 7^{\circ}$.

Solution. Method I. Multiply both sides of the identity by $2 \cos 18^{\circ}$. In the left part, we will have:

$2 \cos 18^{\circ}\left(\cos 25^{\circ}+\cos 47^{\circ}\right)=\cos 43^{\circ}+\cos 7^{\circ}+\cos 65^{\circ}+\cos 29^{\circ}$.

In the right part, we will get:

$$
\begin{aligned}
& 2 \cos 18^{\circ}\left(\cos 11^{\circ}+\cos 61^{\circ}+\sin 7^{\circ}\right)= \\
&= \cos 29^{\circ}+\cos 7^{\circ}+\cos 79^{\circ}+\cos 43^{\circ}+\sin 25^{\circ}-\sin 11^{\circ}
\end{aligned}
$$

Since $\cos 79^{\circ}=\sin 11^{\circ}$, and $\sin 25^{\circ}=\cos 65^{\circ}$, the identity is proven.

Method II. Apply the formula for the sum of cosines of two angles to the left part of the identity and to the sum $\cos 61^{\circ}+\cos 83^{\circ}\left(\sin 7^{\circ}=\right.$ $=\cos 83^{\circ}$ ) in the right part of the identity. We will have:

$$
2 \cos 36^{\circ} \cos 11^{\circ}=\cos 11^{\circ}+2 \cos 72^{\circ} \cos 11^{\circ}
$$

Move the product $2 \cos 72^{\circ} \cos 11^{\circ}$ to the left part and divide both sides of the obtained identity by $\cos 11^{\circ} \neq 0$. We will get the identity

$$
\begin{aligned}
& 2 \cos 36^{\circ}-2 \cos 72^{\circ}=1 \\
& 2 \cdot 2 \sin 54^{\circ} \sin 18^{\circ}=1
\end{aligned}
$$

To prove the obtained identity (1), multiply and divide its left part by $\cos 18^{\circ}$ (to use the formula for the sine of a double argument). We will have:

$$
\begin{gathered}
2 \cdot 2 \sin 54^{\circ} \sin 18^{\circ}=\frac{2 \cdot 2 \sin 54^{\circ} \sin 18^{\circ} \cos 18^{\circ}}{\cos 18^{\circ}}= \\
=\frac{2 \sin 54^{\circ} \sin 36^{\circ}}{\cos 18^{\circ}}=\frac{2 \cos 36^{\circ} \sin 36^{\circ}}{\cos 18^{\circ}}= \\
=\frac{\sin 72^{\circ}}{\cos 18^{\circ}}=1
\end{gathered}
$$

The identity (1), and thus the given identity, is proven.

Remark. It is advisable to solve this problem simultaneously with problems 1217 and 1223 from [3].