Prove that \(\forall x\in\mathbb{R},\ |a\cos x+b\sin x|\le\sqrt{a^2+b^2}.\)

a. Prove that \(\forall x\in\mathbb{R}\), \(|a\cos x+b\sin x|\leq\sqrt{a^2+b^2}\).

b. Find the maximum and minimum of $$ f(x)=20\cos x+21\sin x+27. $$

Solution

a. Prove that \(\forall x\in\mathbb{R}\), \(|a\cos x+b\sin x|\leq\sqrt{a^2+b^2}\).

We have:

$$ a\cos x+b\sin x =\sqrt{a^2+b^2} \left(\frac{a}{\sqrt{a^2+b^2}}\cos x+ \frac{b}{\sqrt{a^2+b^2}}\sin x\right) $$

Let

$$ \sin\alpha=\frac{a}{\sqrt{a^2+b^2}} $$

where \(-1\leq\sin\alpha\leq1\). Since

$$ \sin^2\alpha+\cos^2\alpha=1, $$

we have

$$ \cos\alpha= \sqrt{1-\sin^2\alpha} =\sqrt{1-\frac{a^2}{a^2+b^2}} =\frac{b}{\sqrt{a^2+b^2}}. $$

Therefore,

$$ a\cos x+b\sin x =\sqrt{a^2+b^2} (\sin\alpha\cos x+\cos\alpha\sin x). $$

Using the identity

$$ \sin A\cos B+\sin B\cos A=\sin(A+B), $$

we get

$$ a\cos x+b\sin x =\sqrt{a^2+b^2}\sin(x+\alpha). $$

Hence,

$$ |a\cos x+b\sin x| =\sqrt{a^2+b^2}|\sin(x+\alpha)|. $$

Since

$$ -1\leq\sin(x+\alpha)\leq1, $$

we have

$$ |\sin(x+\alpha)|\leq1. $$

Therefore,

$$ |a\cos x+b\sin x| \leq\sqrt{a^2+b^2}. $$

The inequality is proved.

b. Find the maximum and minimum of

$$ f(x)=20\cos x+21\sin x+27. $$

We rewrite:

$$ \begin{aligned} f(x) &=27+20\cos x+21\sin x\\ &=27+29\left(\frac{20}{29}\cos x+\frac{21}{29}\sin x\right). \end{aligned} $$

Let

$$ \sin\alpha=\frac{20}{29},\qquad \cos\alpha=\frac{21}{29}. $$

Then,

$$ f(x)=27+29(\sin\alpha\cos x+\cos\alpha\sin x) $$

Using the identity above:

$$ f(x)=27+29\sin(x+\alpha). $$

Because

$$ -1\leq\sin(x+\alpha)\leq1, $$

we get

$$ -29\leq29\sin(x+\alpha)\leq29. $$

Adding \(27\):

$$ -2\leq f(x)\leq56. $$

Therefore,

$$ \boxed{\max f(x)=56} $$

$$ \boxed{\min f(x)=-2} $$

Solution by: Thin Sokkean