Class 11: Trigonometric Function of Sum and Differences

$\displaystyle \textbf{Question 1: }\text{Find the maximum and minimum values of each of the following}$
$\displaystyle \text{trigonometrical expressions:}$
$\displaystyle \text{(i) }12\sin x-5\cos x\qquad \text{(ii) }12\cos x+5\sin x+4$
$\displaystyle \text{(iii) }5\cos x+3\sin\left(\frac{\pi}{4}-x\right)+4\qquad \text{(iv) }\sin x-\cos x+1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that for all real }x,$
$\displaystyle -\sqrt{a^2+b^2}\leq a\sin x+b\cos x\leq\sqrt{a^2+b^2}.$

$\displaystyle \text{(i) Let }f(x)=12\sin x-5\cos x.$
$\displaystyle -\sqrt{12^2+(-5)^2}\leq f(x)\leq\sqrt{12^2+(-5)^2}$
$\displaystyle \Rightarrow -\sqrt{169}\leq f(x)\leq\sqrt{169}$
$\displaystyle \Rightarrow -13\leq f(x)\leq13$
$\displaystyle \therefore \text{The minimum and maximum values are }-13\text{ and }13\text{ respectively.}$

$\displaystyle \text{(ii) Let }f(x)=12\cos x+5\sin x.$
$\displaystyle -\sqrt{12^2+5^2}\leq f(x)\leq\sqrt{12^2+5^2}$
$\displaystyle \Rightarrow -13\leq f(x)\leq13$
$\displaystyle \Rightarrow -13+4\leq f(x)+4\leq13+4$
$\displaystyle \Rightarrow -9\leq12\cos x+5\sin x+4\leq17$
$\displaystyle \therefore \text{The minimum and maximum values are }-9\text{ and }17\text{ respectively.}$

$\displaystyle \text{(iii) Let }f(x)=5\cos x+3\sin\left(\frac{\pi}{4}-x\right)+4.$
$\displaystyle \sin\left(\frac{\pi}{4}-x\right)=\sin\frac{\pi}{4}\cos x-\cos\frac{\pi}{4}\sin x$
$\displaystyle =\frac{1}{\sqrt2}\cos x-\frac{1}{\sqrt2}\sin x$
$\displaystyle \therefore f(x)=5\cos x+\frac{3}{\sqrt2}\cos x-\frac{3}{\sqrt2}\sin x+4$
$\displaystyle =\left(5+\frac{3}{\sqrt2}\right)\cos x-\frac{3}{\sqrt2}\sin x+4$
$\displaystyle \text{Let }g(x)=\left(5+\frac{3}{\sqrt2}\right)\cos x-\frac{3}{\sqrt2}\sin x.$
$\displaystyle -R\leq g(x)\leq R,$
$\displaystyle \text{where }R=\sqrt{\left(5+\frac{3}{\sqrt2}\right)^2+\left(-\frac{3}{\sqrt2}\right)^2}$
$\displaystyle =\sqrt{25+15\sqrt2+\frac{9}{2}+\frac{9}{2}}$
$\displaystyle =\sqrt{34+15\sqrt2}$
$\displaystyle \therefore -\sqrt{34+15\sqrt2}\leq g(x)\leq\sqrt{34+15\sqrt2}$
$\displaystyle \Rightarrow 4-\sqrt{34+15\sqrt2}\leq f(x)\leq4+\sqrt{34+15\sqrt2}$
$\displaystyle \therefore \text{The minimum value is }4-\sqrt{34+15\sqrt2}.$
$\displaystyle \text{The maximum value is }4+\sqrt{34+15\sqrt2}.$

$\displaystyle \text{(iv) Let }f(x)=\sin x-\cos x.$
$\displaystyle -\sqrt{1^2+(-1)^2}\leq f(x)\leq\sqrt{1^2+(-1)^2}$
$\displaystyle \Rightarrow -\sqrt2\leq f(x)\leq\sqrt2$
$\displaystyle \Rightarrow 1-\sqrt2\leq f(x)+1\leq1+\sqrt2$
$\displaystyle \Rightarrow 1-\sqrt2\leq\sin x-\cos x+1\leq1+\sqrt2$
$\displaystyle \therefore \text{The minimum and maximum values are }1-\sqrt2\text{ and }1+\sqrt2\text{ respectively.}$
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$\displaystyle \textbf{Question 2: }\text{Reduce each of the following expressions to the sine and cosine of a}$
$\displaystyle \text{single expression:}$
$\displaystyle \text{(i) }\sqrt3\sin x-\cos x\qquad\text{(ii) }\cos x-\sin x$
$\displaystyle \text{(iii) }24\cos x+7\sin x$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Let }f(x)=\sqrt3\sin x-\cos x.$
$\displaystyle f(x)=2\left(\frac{\sqrt3}{2}\sin x-\frac{1}{2}\cos x\right)$
$\displaystyle =2\left(\sin x\cos\frac{\pi}{6}-\cos x\sin\frac{\pi}{6}\right)$
$\displaystyle \therefore f(x)=2\sin\left(x-\frac{\pi}{6}\right).$
$\displaystyle \text{Also,}$
$\displaystyle f(x)=-2\left(\frac{1}{2}\cos x-\frac{\sqrt3}{2}\sin x\right)$
$\displaystyle =-2\left(\cos x\cos\frac{\pi}{3}-\sin x\sin\frac{\pi}{3}\right)$
$\displaystyle \therefore f(x)=-2\cos\left(x+\frac{\pi}{3}\right).$
$\displaystyle \text{Hence, }\sqrt3\sin x-\cos x=2\sin\left(x-\frac{\pi}{6}\right)$
$\displaystyle =-2\cos\left(x+\frac{\pi}{3}\right).$

$\displaystyle \text{(ii) Let }f(x)=\cos x-\sin x.$
$\displaystyle f(x)=\sqrt2\left(\frac{1}{\sqrt2}\cos x-\frac{1}{\sqrt2}\sin x\right)$
$\displaystyle =\sqrt2\left(\sin\frac{\pi}{4}\cos x-\cos\frac{\pi}{4}\sin x\right)$
$\displaystyle \therefore f(x)=\sqrt2\sin\left(\frac{\pi}{4}-x\right).$
$\displaystyle \text{Also,}$
$\displaystyle f(x)=\sqrt2\left(\cos\frac{\pi}{4}\cos x-\sin\frac{\pi}{4}\sin x\right)$
$\displaystyle \therefore f(x)=\sqrt2\cos\left(x+\frac{\pi}{4}\right).$
$\displaystyle \text{Hence, }\cos x-\sin x=\sqrt2\sin\left(\frac{\pi}{4}-x\right)$
$\displaystyle =\sqrt2\cos\left(x+\frac{\pi}{4}\right).$

$\displaystyle \text{(iii) Let }f(x)=24\cos x+7\sin x.$
$\displaystyle f(x)=25\left(\frac{24}{25}\cos x+\frac{7}{25}\sin x\right).$
$\displaystyle \text{Let }\sin\alpha=\frac{24}{25}\text{ and }\cos\alpha=\frac{7}{25}.$
$\displaystyle \therefore \tan\alpha=\frac{24}{7}.$
$\displaystyle f(x)=25(\sin\alpha\cos x+\cos\alpha\sin x)$
$\displaystyle \therefore f(x)=25\sin(x+\alpha),\text{ where }\tan\alpha=\frac{24}{7}.$
$\displaystyle \text{Also, let }\cos\beta=\frac{24}{25}\text{ and }\sin\beta=\frac{7}{25}.$
$\displaystyle \therefore \tan\beta=\frac{7}{24}.$
$\displaystyle f(x)=25(\cos x\cos\beta+\sin x\sin\beta)$
$\displaystyle \therefore f(x)=25\cos(x-\beta),\text{ where }\tan\beta=\frac{7}{24}.$
$\displaystyle \text{Hence, }24\cos x+7\sin x=25\sin(x+\alpha),\text{ where }\tan\alpha=\frac{24}{7},$
$\displaystyle \text{or }24\cos x+7\sin x=25\cos(x-\beta),\text{ where }\tan\beta=\frac{7}{24}.$
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$\displaystyle \textbf{Question 3: }\text{Show that }\sin100^\circ-\sin10^\circ\text{ is positive.}$
$\displaystyle \text{Answer:}$
$\displaystyle \sin100^\circ-\sin10^\circ$
$\displaystyle =2\cos\frac{100^\circ+10^\circ}{2}\sin\frac{100^\circ-10^\circ}{2}$
$\displaystyle =2\cos55^\circ\sin45^\circ$
$\displaystyle =2\cos55^\circ\times\frac{1}{\sqrt2}$
$\displaystyle =\sqrt2\cos55^\circ$
$\displaystyle \text{Since }55^\circ\text{ lies in the first quadrant, }\cos55^\circ>0.$
$\displaystyle \text{Also, }\sqrt2>0.$
$\displaystyle \therefore \sin100^\circ-\sin10^\circ\text{ is positive.}$
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$\displaystyle \textbf{Question 4: }\text{Prove that }(2\sqrt3+3)\sin x+2\sqrt3\cos x\text{ lies between}$
$\displaystyle -(2\sqrt3+\sqrt{15})\text{ and }2\sqrt3+\sqrt{15}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }E=(2\sqrt3+3)\sin x+2\sqrt3\cos x.$
$\displaystyle \text{For an expression }a\sin x+b\cos x,$
$\displaystyle -\sqrt{a^2+b^2}\leq a\sin x+b\cos x\leq\sqrt{a^2+b^2}.$
$\displaystyle \text{Here, }a=2\sqrt3+3\text{ and }b=2\sqrt3.$
$\displaystyle \sqrt{a^2+b^2}=\sqrt{(2\sqrt3+3)^2+(2\sqrt3)^2}$
$\displaystyle =\sqrt{12+12\sqrt3+9+12}$
$\displaystyle =\sqrt{33+12\sqrt3}.$
$\displaystyle \therefore -\sqrt{33+12\sqrt3}\leq E\leq\sqrt{33+12\sqrt3}.$
$\displaystyle \text{Now, }(2\sqrt3+\sqrt{15})^2=12+15+4\sqrt{45}$
$\displaystyle =27+12\sqrt5.$
$\displaystyle (27+12\sqrt5)-(33+12\sqrt3)$
$\displaystyle =12(\sqrt5-\sqrt3)-6$
$\displaystyle =6\left[2(\sqrt5-\sqrt3)-1\right].$
$\displaystyle \text{Also, }\sqrt5-\sqrt3=\frac{2}{\sqrt5+\sqrt3}>\frac{1}{2},$
$\displaystyle \text{since }\sqrt5+\sqrt3<4.$
$\displaystyle \therefore 2(\sqrt5-\sqrt3)-1>0.$
$\displaystyle \Rightarrow 27+12\sqrt5>33+12\sqrt3$
$\displaystyle \Rightarrow 2\sqrt3+\sqrt{15}>\sqrt{33+12\sqrt3}.$
$\displaystyle \therefore -(2\sqrt3+\sqrt{15})<E<2\sqrt3+\sqrt{15}.$
$\displaystyle \text{Hence, }(2\sqrt3+3)\sin x+2\sqrt3\cos x\text{ lies between}$
$\displaystyle -(2\sqrt3+\sqrt{15})\text{ and }2\sqrt3+\sqrt{15}.$
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