Question 1: 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 \Big( \frac{\pi}{4} - x \Big) + 4     iv) \sin x - \cos x + 1

Answer:

i) Let f(x) = 12 \sin x - 5 \cos x

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

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

\Rightarrow - 13 \leq f(x) \leq  13

Hence minimum and maximum value of 12 \sin x - 5 \cos x are -13 and 13 respectively.

ii) Given 12 \cos x + 5 \sin x + 4

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

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

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

\Rightarrow - 13 \leq f(x) \leq  13

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

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

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

 

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

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

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

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

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

We know

- \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 }

- \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} }

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

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

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

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

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

iv) Given \sin x - \cos x + 1

Let f(x) = \sin x - \cos x

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

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

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

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

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

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

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

Answer:

i) f(x) = \sqrt{3} \sin x  - \cos x

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

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]

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

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

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

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

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

Also 

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

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

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

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

ii) f(x) = \cos x - \sin x

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

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]

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

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

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

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

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

Similarly,

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

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

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

iii) f(x) = 24 \cos x +7 \sin x

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

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]

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

Let \sin \alpha  = \frac{24}{25} and \cos \alpha \frac{7}{25}

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

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

Similarly,

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

Let \cos \alpha  = \frac{24}{25} and \sin \alpha \frac{7}{25}

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

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

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Question 3: Show that \sin 100^o - \sin 10^o is positive

Answer:

Given \sin 100^o - \sin 10^o

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

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

= \sqrt{2} \sin (100^o - 45^o)

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

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

Answer:

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

Assume a = (2\sqrt{3} + 3) and b = 2 \sqrt{3}

\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 \sqrt{33+ 12\sqrt{3}} we get

\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]

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

\Rightarrow \sqrt{33+ 12\sqrt{3}} \ \ ( \sin \phi \sin x + \cos \phi \cos x) = \sqrt{33+ 12\sqrt{3}} \ \ \cos ( x - \phi)

We know that the max. and minimum value of any cosine is + 1 and -1

We know, \sqrt{33+ 12\sqrt{3}} = \sqrt{15+12+6+12\sqrt{3}}

We know 12 \sqrt{3} + 6  < 12 \sqrt{5} because value of \sqrt{5} -\sqrt{3} is more that 0.5

So we replace  12 \sqrt{3} + 6 with 12 \sqrt{5} . The above inequality still holds true.

So range of above expression can be \sqrt{15+12+12\sqrt{5}} = 2\sqrt{3}+ \sqrt{15}

-(2\sqrt{3}+ \sqrt{15}) < \sqrt{33+ 12\sqrt{3}} \ \ \cos ( x - \phi)    <   (2\sqrt{3}+ \sqrt{15})