Class 11: Trigonometric Function of Sum and Differences – Exercise 7.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the maximum and minimum values of each of the following trigonometrical expressions:

$\displaystyle \text{i) } 12 \sin x - 5 \cos x \hspace{1.0cm} \text{ii) } 12 \cos x + 5 \sin x + 4$

$\displaystyle \text{iii) } 5 \cos x + 3 \sin \Big( \frac{\pi}{4} - x \Big) + 4 \hspace{1.0cm} \text{iv) } \sin x - \cos x + 1$

Answer:

$\displaystyle \text{i) } \text{Let } f(x) = 12 \sin x - 5 \cos x$

$\displaystyle \text{We know that } -\sqrt{(12)^2 + (-5)^2} \leq f(x) \leq \sqrt{(12)^2 + (-5)^2}$

$\displaystyle \Rightarrow -\sqrt{169} \leq f(x) \leq \sqrt{169}$

$\displaystyle \Rightarrow - 13 \leq f(x) \leq 13$

$\displaystyle \text{Hence minimum and maximum value of } 12 \sin x - 5 \cos x \text{ are } -13 13 \text{ respectively. }$

$\displaystyle \text{ii) } \text{Given } 12 \cos x + 5 \sin x + 4$

$\displaystyle \text{Let } f(x) = 12 \cos x + 5 \sin x$

$\displaystyle \text{We know that } -\sqrt{(12)^2 + (5)^2} \leq f(x) \leq \sqrt{(12)^2 + (5)^2}$

$\displaystyle \Rightarrow -\sqrt{169} \leq f(x) \leq \sqrt{169}$

$\displaystyle \Rightarrow - 13 \leq f(x) \leq 13$

$\displaystyle \Rightarrow - 13+4 \leq f(x)+4 \leq 13+4$

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

$\displaystyle \text{Hence minimum and maximum value of } 12 \cos x + 5 \sin x + 4 \text{ are } -9 17 \text{ respectively. }$

$\displaystyle \text{iii) } \text{Let } f(x) = 5 \cos x + 3 \sin \Big( \frac{\pi}{4} - x \Big) + 4$

$\displaystyle \Rightarrow f(x) = 5 \cos x + 3 \Big[ \sin \frac{\pi}{6} \cos x - \cos \frac{\pi}{6} \sin x \Big] + 4$

$\displaystyle \Rightarrow f(x) = 5 \cos x + 3 \Big[ \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x \Big] + 4$

$\displaystyle \Rightarrow f(x) = (5 + \frac{3}{2} ) \cos x - \frac{3\sqrt{3}}{2} \sin x + 4$

$\displaystyle \Rightarrow f(x) = \frac{13}{2} \cos x - \frac{3\sqrt{3}}{2} \sin x + 4$

We know

$\displaystyle - \sqrt{ \Big( \frac{13}{2} \Big)^2 + \Big( \frac{-3\sqrt{3}}{2}\Big)^2 } \leq \frac{13}{2} \cos x - \frac{3\sqrt{3}}{2} \sin x \leq \sqrt{ \Big( \frac{13}{2} \Big)^2 + \Big( \frac{-3\sqrt{3}}{2} \Big)^2 }$

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

$\displaystyle - \frac{14}{2} \leq \frac{13}{2} \cos x - \frac{3\sqrt{3}}{2} \sin x \leq \frac{14}{2}$

$\displaystyle - 7 \leq \frac{13}{2} \cos x - \frac{3\sqrt{3}}{2} \sin x \leq 7$

$\displaystyle - 7 + 4 \leq \frac{13}{2} \cos x - \frac{3\sqrt{3}}{2} + 4 \sin x \leq 7 + 4$

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

$\displaystyle \text{Hence minimum and maximum value of } 5 \cos x + 3 \sin \Big( \frac{\pi}{4} - x \Big) + 4 \text{ are } -3 11 \text{ respectively. }$

$\displaystyle \text{iv) } \text{Given } \sin x - \cos x + 1$

$\displaystyle \text{Let } f(x) = \sin x - \cos x$

$\displaystyle \text{We know that } -\sqrt{(-1)^2 + (1)^2} \leq f(x) \leq \sqrt{(-1)^2 + (1)^2}$

$\displaystyle \Rightarrow -\sqrt{2} \leq f(x) \leq \sqrt{2}$

$\displaystyle \Rightarrow - \sqrt{2} \leq f(x) \leq \sqrt{2}$

$\displaystyle \Rightarrow - \sqrt{2} + 1 \leq f(x)+ 1 \leq \sqrt{2} + 1$

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

$\displaystyle \text{Hence minimum and maximum value of } \sin x - \cos x + 1 \text{ are } 1 - \sqrt{2} \sqrt{2} + 1 \text{ respectively. }$

$\displaystyle \\$

Question 2: Reduce each of the following expressions to the sine and cosine of a single expression: $\displaystyle \text{i) } \sqrt{3} \sin x - \cos x \hspace{1.0cm} \text{ii) } \cos x - \sin x \hspace{1.0cm} \text{iii) } 24 \cos x + 7 \sin x$

Answer:

$\displaystyle \text{i) } f(x) = \sqrt{3} \sin x - \cos x$

Multiplying and dividing by $\displaystyle \sqrt{(\sqrt{3})^2 + (-1)^2}$ we get

$\displaystyle f(x) = \sqrt{(\sqrt{3})^2 + (-1)^2} \Big[ \frac{\sqrt{3} \sin x}{\sqrt{(\sqrt{3})^2 + (-1)^2}} - \frac{\cos x}{\sqrt{(\sqrt{3})^2 + (-1)^2}} \Big]$

$\displaystyle \Rightarrow f(x) = \sqrt{(3+1)} \Big[ \frac{\sqrt{3} \sin x}{\sqrt{(3+1)}} - \frac{\cos x}{\sqrt{(3+1)}} \Big]$

$\displaystyle \Rightarrow f(x) = 2 \Big[ \frac{\sqrt{3} \sin x}{2} - \frac{\cos x}{2} \Big]$

$\displaystyle \Rightarrow f(x) = 2 \Big[ \frac{\sqrt{3}}{2} \sin x - \frac{1}{2} \cos x \Big]$

$\displaystyle \Rightarrow f(x) = 2 \Big[ \sin x \cos \frac{\pi}{6} - \cos x \cos \frac{\pi}{6} \Big]$

$\displaystyle \Rightarrow f(x) = 2 \sin (x - \frac{\pi}{6} )$

Also

$\displaystyle f(x) = 2 \Big[ \frac{\sqrt{3} \sin x}{2} - \frac{\cos x}{2} \Big]$

$\displaystyle \Rightarrow f(x) = - 2 \Big[ \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x \Big]$

$\displaystyle \Rightarrow f(x) = - 2 \Big[ \cos \frac{\pi}{3} \cos x - \cos \frac{\pi}{3} \sin x \Big]$

$\displaystyle \Rightarrow f(x) = -2 \cos (x + \frac{\pi}{3} )$

$\displaystyle \text{ii) } f(x) = \cos x - \sin x$

Multiplying and dividing by $\displaystyle \sqrt{(1)^2 + (1)^2}$ we get

$\displaystyle f(x) = \sqrt{(1)^2 + (1)^2} \Big[ \frac{ \cos x}{\sqrt{(1)^2 + (1)^2}} - \frac{\sin x}{\sqrt{(1)^2 + (1)^2}} \Big]$

$\displaystyle \Rightarrow f(x) = \sqrt{(1+1)} \Big[ \frac{ \cos x}{\sqrt{(1+1)}} - \frac{\sin x}{\sqrt{(1+1)}} \Big]$

$\displaystyle \Rightarrow f(x) = \sqrt{2} \Big[ \frac{ \cos x}{\sqrt{2}} - \frac{\sin x}{\sqrt{2}} \Big]$

$\displaystyle \Rightarrow f(x) = \sqrt{2} \Big[ \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x \Big]$

$\displaystyle \Rightarrow f(x) = \sqrt{2} \Big[ \sin \frac{\pi}{4} \cos x - \cos \frac{\pi}{4} \sin x \Big]$

$\displaystyle \Rightarrow f(x) = \sqrt{2} \sin ( \frac{\pi}{4} - x )$

Similarly,

$\displaystyle \Rightarrow f(x) = \sqrt{2} \Big[ \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x \Big]$

$\displaystyle \Rightarrow f(x) = \sqrt{2} \Big[ \cos \frac{\pi}{4} \cos x - \sin \frac{\pi}{4} \sin x \Big]$

$\displaystyle \Rightarrow f(x) = \sqrt{2} \cos ( \frac{\pi}{4} + x )$

$\displaystyle \text{iii) } f(x) = 24 \cos x +7 \sin x$

Multiplying and dividing by $\displaystyle \sqrt{(24)^2 + (7)^2}$ we get

$\displaystyle f(x) = \sqrt{(24)^2 + (7)^2} \Big[ \frac{ \cos x}{\sqrt{(24)^2 + (7)^2}} - \frac{\sin x}{\sqrt{(24)^2 + (7)^2}} \Big]$

$\displaystyle \Rightarrow f(x) = 25 \Big[ \frac{ 24}{25} \cos x + \frac{7 }{25} \sin x \Big]$

$\displaystyle \text{Let } \sin \alpha = \frac{24}{25} \cos \alpha \frac{7}{25}$

$\displaystyle \therefore f(x) = 25 [\sin \alpha \cos x + \cos \alpha \sin x ]$

$\displaystyle f(x) = 25 \sin ( x + \alpha) \text{ where } \tan \alpha = \frac{\sin \alpha }{\cos \alpha} = \frac{24}{7}$

Similarly,

$\displaystyle \Rightarrow f(x) = 25 \Big[ \frac{ 24}{25} \cos x + \frac{7 }{25} \sin x \Big]$

$\displaystyle \text{Let } \cos \alpha = \frac{24}{25} \sin \alpha \frac{7}{25}$

$\displaystyle \therefore f(x) = 25 [\cos \alpha \cos x + \sin \alpha \sin x ]$

$\displaystyle f(x) = 25 \sin ( x - \alpha) \text{ where } \tan \alpha = \frac{\sin \alpha }{\cos \alpha} = \frac{7}{24}$

$\displaystyle \\$

Question 3: Show that $\displaystyle \sin 100^{\circ} - \sin 10^{\circ}$ is positive

Answer:

$\displaystyle \text{Given } \sin 100^{\circ} - \sin 10^{\circ}$

$\displaystyle = \sqrt{2} \Big[ \frac{1}{\sqrt{2}} \sin 100^{\circ} - \frac{1}{\sqrt{2}} \sin 10^{\circ} \Big]$

$\displaystyle = \sqrt{2} \Big[ \sin 45^{\circ} \sin 100^{\circ} - \cos 45^{\circ} \sin 10^{\circ} \Big]$

$\displaystyle = \sqrt{2} \sin (100^{\circ} - 45^{\circ})$

$\displaystyle = \sqrt{2} \sin 55$ which is a positive number as $\displaystyle \sin 55^{\circ}$ is in first quadrant.

$\displaystyle \\$

Question 4: Prove that $\displaystyle (2\sqrt{3} + 3) \sin x + 2 \sqrt{3} \cos x$ lies between $\displaystyle -(2\sqrt{3}+ \sqrt{15}) (2\sqrt{3}+ \sqrt{15})$

Answer:

$\displaystyle \text{Given } (2\sqrt{3} + 3) \sin x + 2 \sqrt{3} \cos x$

$\displaystyle \text{Assume } a = (2\sqrt{3} + 3) b = 2 \sqrt{3}$

$\displaystyle \sqrt{ a^2 +b^2} = \sqrt{ (2\sqrt{3} + 3)^2 + (2 \sqrt{3})^2 }= \sqrt{12+ 9 + 12\sqrt{3} + 12 }= \sqrt{33+ 12\sqrt{3}}$

Dividing and multiplying by $\displaystyle \sqrt{33+ 12\sqrt{3}}$ we get

$\displaystyle \sqrt{33+ 12\sqrt{3}} \Big[ \frac{(2\sqrt{3} + 3) \sin x}{\sqrt{33+ 12\sqrt{3}} } + \frac{2 \sqrt{3} \cos x}{\sqrt{33+ 12\sqrt{3}} } \Big]$

$\displaystyle \text{Assume } \tan \phi = \frac{a}{b} \text{ we get } \sin \phi = \frac{a}{\sqrt{a^2 + b^2}}$ , $\displaystyle \cos \phi = \frac{b}{\sqrt{a^2 + b^2}}$