# Find the maximum and minimum values of each of the following trigonometrical expressions:

Question:

Find the maximum and minimum values of each of the following trigonometrical expressions:

(i) 12 sin x − 5 cos x

(ii) 12 cos x + 5 sin x + 4

(iii) $5 \cos x+3 \sin \left(\frac{\pi}{6}-x\right)+4$

(iv) sin x − cos x + 1

Solution:

(i) $\operatorname{Le} t f(x)=12 \sin x-5 \cos x$

We know that

$-\sqrt{12^{2}+(-5)^{2}} \leq 12 \sin x-5 \cos x \leq \sqrt{12^{2}+(-5)^{2}}$

$-\sqrt{144+25} \leq 12 \sin x-5 \cos x \leq \sqrt{144+25}$

$-13 \leq 12 \sin x-5 \cos x \leq 13$

Hence the maximum and minumun values of $f(x)$ are 13 and $-13$, respectively.

(ii) Let $f(x)=12 \cos x+5 \sin x+4$

We know that

$-\sqrt{12^{2}+5^{2}} \leq 12 \cos x+5 \sin x \leq \sqrt{12^{2}+5^{2}} \quad$ for all $x$

$\Rightarrow-\sqrt{169} \leq 12 \cos x+5 \sin x \leq \sqrt{169}$

$\Rightarrow-13 \leq 12 \cos x+5 \sin x \leq 13$

$\Rightarrow-9 \leq 12 \cos x+5 \sin x+4 \leq 17$

Hence, the maximum and minimum vaues of $f(x)$ are 17 and $-9$, respectively.

(iii) Let $f(x)=5 \cos x+3 \sin \left(\frac{\pi}{6}-x\right)+4$

Now $f(x)=5 \cos x+3\left(\sin 30^{\circ} \cos x-\cos 30^{\circ} \sin x\right)+4$

$=5 \cos x+\frac{3}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x+4$

$=\frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x+4$

We know that

$-\sqrt{\left(\frac{13}{2}\right)^{2}+\left(-\frac{3 \sqrt{3}}{2}\right)^{2}} \leq \frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x \leq \sqrt{\left(\frac{13}{2}\right)^{2}+\left(-\frac{3 \sqrt{3}}{2}\right)^{2}} \quad$ for all $x$

Therefore,

$-\sqrt{\frac{169+27}{4}} \leq \frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x \leq \sqrt{\frac{169+27}{4}}$

$\Rightarrow-\frac{14}{2}+4 \leq \frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x+4 \leq \frac{14}{2}+4$

$\Rightarrow-3 \leq \frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x+4 \leq 11$

Hence, maximum and minimun values of $f(x)$ are 11 and $-3$, respectively .

(iv) Let $f(x)=\sin x-\cos x+1$

We know that

$-\sqrt{1^{2}+(-1)^{2}} \leq \sin x-\cos x \leq \sqrt{1^{2}+(-1)^{2}} \quad$ for all $x$

$\Rightarrow-\sqrt{2} \leq \sin x-\cos x \leq \sqrt{2}$

$\Rightarrow-\sqrt{2}+1 \leq \sin x-\cos x+1 \leq \sqrt{2}+1$

Hence maximum and minimum values of $f(x)$ are $1+\sqrt{2}$ and $1-\sqrt{2}$, respectively.