Question 1: Express each of the following as a sum or difference of sine and cosines:

i) 2 \sin 3x \cos x    ii) 2 \cos 3x \sin 2x     iii) 2 \sin 4x \sin 3x     iv) 2 \cos 7x \cos 3x

Answer:

i) 2 \sin 3x \cos x = \sin (3x + x) + \sin (3x - x) = \sin 4x + \sin 2x

ii) 2 \cos 3x \sin 2x = \sin ( 3x + 2x) - \sin (3x - 2x) = \sin 5x - \sin x

iii) 2 \sin 4x \sin 3x = \cos ( 4x - 3x) - \cos (4x + 3x) = \cos x - \cos 7x

iv) 2 \cos 7x \cos 3x = \cos ( 7x + 3x) + \cos (7x - 3x) = \cos 10 x + \cos 4x

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Question 2: Prove that: 

i) 2 \sin \frac{5\pi}{12} \sin \frac{\pi}{12} = \frac{1}{2}     ii) 2 \cos \frac{5\pi}{12} \cos \frac{\pi}{12} = \frac{1}{2}     iii) 2 \sin \frac{5\pi}{12} \cos \frac{\pi}{12} = \frac{\sqrt{3}+2}{2}

Answer:

i) 2 \sin \frac{5\pi}{12} \sin \frac{\pi}{12}

= \cos \Big(  \frac{5\pi}{12} - \frac{\pi}{12} \Big) - \cos \Big(  \frac{5\pi}{12} + \frac{\pi}{12} \Big)

= \cos \frac{4\pi}{12} - \cos \frac{6\pi}{12}

= \cos \frac{\pi}{3} - \cos \frac{\pi}{2}

= \frac{1}{2} - 0

= \frac{1}{2} = RHS. Hence proved.

ii) 2 \cos \frac{5\pi}{12} \cos \frac{\pi}{12}

= \cos \Big(  \frac{5\pi}{12} + \frac{\pi}{12} \Big) + \cos \Big(  \frac{5\pi}{12} - \frac{\pi}{12} \Big)

= \cos \frac{\pi}{2} + \cos \frac{\pi}{3}

= 0 + \frac{1}{2}

= \frac{1}{2} = RHS. Hence proved.

iii) 2 \sin \frac{5\pi}{12} \cos \frac{\pi}{12}

= \sin \Big(  \frac{5\pi}{12} + \frac{\pi}{12} \Big) + \sin \Big(  \frac{5\pi}{12} - \frac{\pi}{12} \Big)

= \sin \frac{\pi}{2} + \sin \frac{\pi}{3}

= 1 + \frac{\sqrt{3}}{2}

= \frac{2 + \sqrt{3}}{2} = RHS. Hence proved.

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Question 3: Show that:

i) \sin 50^o \cos 85^o = \frac{1 - \sqrt{2} \sin 35^o}{2\sqrt{2}}     ii) \sin 25^o \cos 115^o = \frac{1}{2} ( \sin 40^o - 1)

Answer:

i) LHS = \sin 50^o \cos 85^o

= \frac{1}{2} \Big[ 2 \sin 50^o \cos 85^o   \Big]

 = \frac{1}{2} \Big[  \sin (50^o + 85^o) +  \sin (50^o - 85^o)   \Big]

 = \frac{1}{2} \Big[  \sin 135^o +  \sin ( - 35^o)   \Big]

= \frac{1}{2} \Big[  \sin (90^o + 45^o) -  \sin  35^o   \Big]

= \frac{1}{2} \Big[  \cos  45^o -  \sin  35^o   \Big]

= \frac{1}{2} \Big[ \frac{1}{\sqrt{2}} -  \sin  35^o   \Big]

= \frac{1 - \sqrt{2} \sin  35^o}{2 \sqrt{2}} = RHS. Hence Proved.

ii) LHS = \sin 25^o \cos 115^o

= \frac{1}{2} \Big[ 2 \sin 25^o \cos 115^o   \Big]

 = \frac{1}{2} \Big[  \sin (25^o + 115^o) +  \sin (25^o - 115^o)   \Big]

 = \frac{1}{2} \Big[  \sin 140^o +  \sin ( - 90^o)   \Big]

= \frac{1}{2} \Big[  \sin (90^o + 50^o) -  \sin  90^o   \Big]

= \frac{1}{2} \Big[  \cos  50^o -  1   \Big]

= \frac{1}{2} \Big[  \cos  (90^o - 40^o) -  1   \Big]

= \frac{1}{2} \Big[  \sin  40^o -  1   \Big]   RHS. Hence Proved.

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Question 4: Prove that 4 \cos x \cos \Big( \frac{\pi}{3} +x \Big) \cos \Big( \frac{\pi}{3} -x \Big) = \cos 3x

Answer:

LHS = 4 \cos x \cos \Big( \frac{\pi}{3} + x \Big)  \cos \Big( \frac{\pi}{3} - x \Big)

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

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

= 2 \cos x  \Big[ \cos \frac{2\pi}{3} + \cos 2x   \Big]

= 2 \cos x  \Big[ \cos \Big( \frac{\pi}{2} + \frac{\pi}{6} \Big)  + \cos 2x   \Big]

= 2 \cos x  \Big[ - \sin \frac{\pi}{6} + \cos 2x   \Big]

= 2 \cos x  \Big[ - \frac{1}{2} + \cos 2x   \Big]

= - \cos x + 2 \cos x \cos 2x

= - \cos x + [ \cos (2x + x) + \cos ( 2x - x) ]

= - \cos x + \cos 3x + \cos x

= \cos 3x = RHS

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Question 5: Prove that:

i) \cos 10^o \cos 30^o \cos 50^o \cos 70^o = \frac{3}{16}    ii) \cos 40^o \cos 80^o \cos 160^o  = - \frac{1}{8}

iii) \sin 20^o \sin 40^o \sin 80^o  = \frac{\sqrt{3}}{8}    iv) \cos 20^o \cos 40^o \cos 80^o  = \frac{1}{8}

v) \tan 20^o \tan 40^o \tan 60^o \tan 80^o = 3     vi) \tan 20^o \tan 30^o \tan 40^o \tan 80^o = 1  

vii) \sin 10^o \sin 50^o \sin 60^o \sin 70^o = \frac{\sqrt{3}}{16}    viii) \sin 20^o \sin 40^o \sin 60^o \sin 80^o = \frac{3}{16}

Answer:

i) LHS = \cos 10^o \cos 30^o \cos 50^o \cos 70^o

= \frac{\sqrt{3}}{2} \Big[  (\cos 10^o  \cos 50^o) \cos 70^o \Big]

= \frac{\sqrt{3}}{4} \Big[  (2 \cos 50^o  \cos 10^o) \cos 70^o \Big]

= \frac{\sqrt{3}}{4} \Big[  \Big( \cos (50^o+10^o) + \cos (50^o-10^o)  \Big) \cos 70^o \Big]

= \frac{\sqrt{3}}{4} \Big[  ( \cos 60^o + \cos 40^o )  \cos 70^o \Big]

= \frac{\sqrt{3}}{4} \Big[  ( \frac{1}{2} + \cos 40^o )  \cos 70^o \Big]

= \frac{\sqrt{3}}{8} \cos 70^o + \frac{\sqrt{3}}{4} \cos 40^o   \cos 70^o

= \frac{\sqrt{3}}{8} \cos 70^o + \frac{\sqrt{3}}{8} (2\cos 40^o   \cos 70^o)

= \frac{\sqrt{3}}{8} \cos 70^o + \frac{\sqrt{3}}{8} ( \cos (70^o+40^o) + \cos (70^o-40^o) )

= \frac{\sqrt{3}}{8} \cos 70^o + \frac{\sqrt{3}}{8} \cos 110 + \frac{\sqrt{3}}{8} \cos 30^o

= \frac{\sqrt{3}}{8} \Big[ \cos 70^o +  \cos 110 +  \cos 30 \Big]

= \frac{\sqrt{3}}{8} \Big[ \cos 70^o +  \cos (180 - 70) +  \cos 30 \Big]

= \frac{\sqrt{3}}{8} \Big[ \cos 70^o -  \cos 70^o +  \frac{\sqrt{3}}{2} \Big]

= \frac{\sqrt{3}}{8} \Big[  \frac{\sqrt{3}}{2} \Big]

= \frac{3}{16} = RHS. Hence proved.

ii) LHS = \cos 40^o \cos 80^o \cos 160^o

= \frac{1}{2} \Big[  2 \cos 40^o \cos 80^o \cos 160^o \Big]

= \frac{1}{2} \Big[  \cos 80^o ( 2 \cos 160^o \cos 40^o) \Big]

= \frac{1}{2} \Big[  \cos 80^o [ \cos ( 160^o+40^o) + \cos ( 160^o-40^o) ] \Big]

= \frac{1}{2} \Big[  \cos 80^o ( \cos 200^o + \cos 120^o  ) \Big]

= \frac{1}{2} \Big[  \cos 80^o [ \cos ( 180^o+20^o) + \cos ( 180^o-60^o) ] \Big]

= \frac{1}{2} \Big[  \cos 80^o ( \cos 20^o + \cos 60^o  ) \Big]

= - \frac{1}{2} \cos 80^o \cos 20^o + \frac{1}{2} \cos 80^o \cos 60^o

= - \frac{1}{4} [  2 \cos 80^o \cos 20^o + 2 \cos 80^o \cos 60^o ]

= - \frac{1}{4} [  \cos (80^o+20^o) + \cos (80^o-20^o)+ 2 \cos 80^o \times \frac{1}{2} ]

= - \frac{1}{4} [  \cos 100^o + \cos 60^o+ \cos 80^o ]

= - \frac{1}{4} [  \cos (180^o - 80^o) + \cos 60^o+ \cos 80^o ]

= - \frac{1}{4} [  -\cos 80^o + \cos 60^o+ \cos 80^o ]

= - \frac{1}{4} [   \cos 60^o ]

= - \frac{1}{8} = RHS. Hence proved.

iii) LHS = \sin 20^o \sin 40^o \sin 80^o 

= \frac{1}{2} \Big[  2 \sin 20^o \sin 40^o \Big] \sin 80^o 

= \frac{1}{2} \Big[  \cos (40^o-20^o) - \cos (40^o+20^o)  \Big] \sin 80^o  

= \frac{1}{2} \Big[  \cos 20^o  - \cos 60^o  \Big] \sin 80^o 

= \frac{1}{2} \Big[  \cos 20^o  - \frac{1}{2} \Big] \sin 80^o 

= \frac{1}{2} \Big[  \cos 20^o \sin 80^o \Big]  - \frac{1}{4}   \sin 80^o  

= \frac{1}{4} \Big[  2 \cos 20^o \sin 80^o \Big]  - \frac{1}{4}   \sin 80^o  

= \frac{1}{4} \Big[  \sin (80^o+20^o) + \sin (80^o-20^o) \Big]  - \frac{1}{4}   \sin 80^o  

= \frac{1}{4} \Big[  \sin 100^o + \sin  60^o \Big]  - \frac{1}{4}   \sin 80^o  

= \frac{1}{4} \Big[  \sin (180^o - 80^o) + \sin  60^o \Big]  - \frac{1}{4}   \sin 80^o  

= \frac{1}{4} \sin 80^o + \frac{\sqrt{3}}{8} - \frac{1}{4} \sin 80^o

= \frac{\sqrt{3}}{8} = RHS. Hence proved.

iv) LHS = \cos 20^o \cos 40^o \cos 80^o 

= \frac{1}{2} \Big[ 2 \cos 20^o \cos 40^o  \Big] \cos 80^o 

= \frac{1}{2} \Big[ \cos ( 40^o+ 20^o) + \cos ( 40^o - 20^o)  \Big] \cos 80^o 

= \frac{1}{2} \Big[ \cos 60^o + \cos 20^o  \Big] \cos 80^o 

= \frac{1}{2} \Big[ \frac{1}{2} + \cos 20^o  \Big] \cos 80^o 

= \frac{1}{4} \Big[ \cos 80^o + 2\cos 20^o \cos 80^o  \Big]  

= \frac{1}{4} \Big[ \cos 80^o + \cos ( 80^o + 20^o) + \cos ( 80^o - 20^o)  \Big]  

= \frac{1}{4} \Big[ \cos 80^o + \cos 100^o + \cos 60^o  \Big]  

= \frac{1}{4} \Big[ \cos 80^o + \cos (180^o - 80^o) + \cos 60^o  \Big]  

= \frac{1}{4} \Big[ \cos 80^o - \cos 80^o + \frac{1}{2} \Big]  

= \frac{1}{8} =  RHS. Hence Proved.

v) LHS = \tan 20^o \tan 40^o \tan 60^o \tan 80^o

= \sqrt{3} [ \tan 20^o \tan 40^o \tan 80^o ]

= \sqrt{3} \Big[  \frac{\sin 20^o \sin 40^o \sin 80^o }{\cos 20^o \cos 40^o \cos 80^o }   \Big] 

= \sqrt{3} \Big[  \frac{ [ \cos ( 40^o-20^o) - \cos ( 40^o + 20^o) ] \sin 80^o }{ [ \cos ( 20^o+40^o) + \cos ( 40^o - 20^o) ] \cos 80^o }   \Big] 

= \sqrt{3} \Big[  \frac{ [ \cos 20^o - \cos 60^o ] \sin 80^o }{ [ \cos 60^o + \cos 20^o ] \cos 80^o }   \Big] 

= \sqrt{3} \Big[  \frac{ [ 2 \cos 20^o - 1 ] \sin 80^o }{ [ 1 + 2 \cos 20^o ] \cos 80^o }   \Big] 

= \sqrt{3} \Big[  \frac{  2 \cos 20^o \sin 80^o - \sin 80^o  }{  \cos 80^o + 2 \cos 20^o \cos 80^o   }   \Big] 

= \sqrt{3} \Big[  \frac{  \sin (80+20) + \sin (80^o-20^o) - \sin 80^o  }{  \cos 80^o + \cos (80^o+20^o) + \cos (80^o-20^o)   }   \Big] 

= \sqrt{3} \Big[  \frac{  \sin 100^o + \sin 60^o - \sin 80^o  }{  \cos 80^o + \cos 100^o + \cos 60^o   }   \Big] 

= \sqrt{3} \Big[  \frac{  \sin (180^o-80^o) + \sin 60^o - \sin 80^o  }{  \cos 80^o + \cos (180^o-80^o) + \cos 60^o   }   \Big] 

= \sqrt{3} \Big[  \frac{  \sin 80^o + \sin 60^o - \sin 80^o  }{  \cos 80^o - \cos 80^o + \cos 60^o   }   \Big]

= \sqrt{3} \tan 60^o

= \sqrt{3} \times \sqrt{3} = 3 = RHS. Hence proved.

vi) LHS = \tan 20^o \tan 30^o \tan 40^o \tan 80^o

= \frac{1}{\sqrt{3}} [ \tan 20^o \tan 40^o \tan 80^o ]

= \frac{1}{\sqrt{3}} \Big[  \frac{\sin 20^o \sin 40^o \sin 80^o }{\cos 20^o \cos 40^o \cos 80^o } \Big] 

= \frac{1}{\sqrt{3}} \Big[  \frac{ (2 \sin 20^o \sin 40^o)  \sin 80^o }{ (2 \cos 20^o \cos 40^o) \cos 80^o }   \Big] 

= \frac{1}{\sqrt{3}} \Big[  \frac{ [ \cos ( 40^o-20^o) - \cos ( 40^o + 20^o) ] \sin 80^o }{ [ \cos ( 20^o+40^o) + \cos ( 40^o - 20^o) ] \cos 80^o }   \Big] 

= \frac{1}{\sqrt{3}} \Big[  \frac{ [ \cos 20^o - \cos 60^o ] \sin 80^o }{ [ \cos 60^o + \cos 20^o ] \cos 80^o }   \Big] 

= \frac{1}{\sqrt{3}} \Big[  \frac{ [ 2 \cos 20^o - 1 ] \sin 80^o }{ [ 1 + 2 \cos 20^o ] \cos 80^o }   \Big] 

= \frac{1}{\sqrt{3}} \Big[  \frac{  2 \cos 20^o \sin 80^o - \sin 80^o  }{  \cos 80^o + 2 \cos 20^o \cos 80^o   }   \Big] 

= \frac{1}{\sqrt{3}} \Big[  \frac{  \sin (80+20) + \sin (80^o-20^o) - \sin 80^o  }{  \cos 80^o + \cos (80^o+20^o) + \cos (80^o-20^o)   }   \Big] 

= \frac{1}{\sqrt{3}} \Big[  \frac{  \sin 100^o + \sin 60^o - \sin 80^o  }{  \cos 80^o + \cos 100^o + \cos 60^o   }   \Big] 

= \frac{1}{\sqrt{3}} \Big[  \frac{  \sin (180^o-80^o) + \sin 60^o - \sin 80^o  }{  \cos 80^o + \cos (180^o-80^o) + \cos 60^o   }   \Big] 

= \frac{1}{\sqrt{3}} \Big[  \frac{  \sin 80^o + \sin 60^o - \sin 80^o  }{  \cos 80^o - \cos 80^o + \cos 60^o   }   \Big]

= \frac{1}{\sqrt{3}} \tan 60^o

= \frac{1}{\sqrt{3}} \times \sqrt{3} = 1 = RHS. Hence proved.

vii) LHS = \sin 10^o \sin 50^o \sin 60^o \sin 70^o

= \frac{\sqrt{3}}{2} \Big[  \sin 10^o \sin 50^o  \sin 70^o   \Big]

= \frac{\sqrt{3}}{2} \Big[  \sin (90^o - 80^o) \sin (90^o - 40^o)   \sin (90^o - 20^o)   \Big]

= \frac{\sqrt{3}}{2} \Big[  \cos 80^o \cos 40^o  \cos 20^o   \Big]

= \frac{\sqrt{3}}{4} \Big[  \cos 80^o (2 \cos 40^o  \cos 20^o)   \Big]

= \frac{\sqrt{3}}{4} \Big[   \cos (40^o + 20^o) +   \cos (40^o - 20^o)   \Big] \cos 80^o

= \frac{\sqrt{3}}{4} \Big[   \cos 60^o+   \cos 20^o    \Big] \cos 80^o

= \frac{\sqrt{3}}{8} \Big[   \cos 80^o+  2 \cos 20^o \cos 80^o    \Big]

= \frac{\sqrt{3}}{8} \Big[   \cos 80^o+  \cos ( 80^o + 20^o) + \cos (80^o - 20^o)    \Big]

= \frac{\sqrt{3}}{8} \Big[   \cos 80^o+  \cos 100^o + \cos 60^o    \Big]

= \frac{\sqrt{3}}{8} \Big[   \cos 80^o+  \cos (180 - 80^o) + \cos 60^o    \Big]

= \frac{\sqrt{3}}{8} \Big[   \cos 80^o - \cos 80^o + \cos 60^o    \Big]

= \frac{\sqrt{3}}{8} \cos 60^o

= \frac{\sqrt{3}}{16} = RHS. Hence proved.

viii) LHS = \sin 20^o \sin 40^o \sin 60^o \sin 80^o

= \frac{\sqrt{3}}{2} \Big[  \sin 20^o \sin 40^o  \sin 80^o   \Big]

= \frac{\sqrt{3}}{4} \Big[  (2 \sin 20^o \sin 40^o)  \sin 80^o   \Big]

= \frac{\sqrt{3}}{4} \Big[   \cos (40^o - 20^o) -   \cos (40^o + 20^o)   \Big] \sin 80^o

= \frac{\sqrt{3}}{4} \Big[   - \cos 60^o+   \cos 20^o    \Big] \sin 80^o

= \frac{\sqrt{3}}{4} \Big[   - \frac{1}{2}+   \cos 20^o    \Big] \sin 80^o

= \frac{\sqrt{3}}{4} \Big[   - \frac{1}{2} \sin 80^o +   \cos 20^o  \sin 80^o   \Big] 

= \frac{\sqrt{3}}{8} \Big[   -  \sin 80^o +   2 \cos 20^o  \sin 80^o   \Big] 

= \frac{\sqrt{3}}{8} \Big[   -  \sin 80^o +   \sin (80^o + 20^o) + \sin ( 80^o - 20^o)   \Big] 

= \frac{\sqrt{3}}{8} \Big[   -  \sin 80^o +   \sin 100^o  + \sin  60^o    \Big] 

= \frac{\sqrt{3}}{8} \Big[   -  \sin 80^o +   \sin (180^o - 80^o)  + \sin  60^o    \Big] 

= \frac{\sqrt{3}}{8} \Big[   -  \sin 80^o +   \sin  80^o  + \sin  60^o    \Big] 

= \frac{\sqrt{3}}{8} \sin 60^o

= \frac{\sqrt{3}}{16} = RHS. Hence proved.

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Question 6: Show that:

i) \sin A \sin (B-C) + \sin B \sin (C-A) + \sin C \sin (A-B) = 0

ii) \sin (B-C) \cos (A - D) + \sin (C-A) \cos (B-D) + \sin (A-B) \cos (C-D) = 0

Answer:

i) LHS = \sin A \sin (B-C) + \sin B \sin (C-A) + \sin C \sin (A-B) 

= \frac{1}{2}  \Big[  2\sin A \sin (B-C) + 2\sin B \sin (C-A) +  2\sin C \sin (A-B)  \Big] 

= \frac{1}{2}  \Big[  \cos ( A - B + C) - \cos ( A + B - C)  + \cos ( B - C + A )  - \cos ( B + C - A )  +  \cos ( C - A + B ) - \cos ( C + A - B)  \Big]  

= \frac{1}{2}  \Big[  \cos ( A - B + C) - \cos ( A - B + C)  - \cos ( A + B - C ) +  \cos ( A + B - C )  - \cos ( B + C - A ) + \cos ( B + C - A )  \Big]  

= \frac{1}{2}  \Big[  0  \Big] = 0 =  RHS. Hence proved.

ii) LHS = \sin (B-C) \cos (A - D) + \sin (C-A) \cos (B-D) + \sin (A-B) \cos (C-D)

= \frac{1}{2} \Big[ 2\sin (B-C) \cos (A - D) + 2\sin (C-A) \cos (B-D) + 2\sin (A-B) \cos (C-D)  \Big]

= \frac{1}{2} \Big[ \sin ( B - C + A - D) + \sin ( B - C - A + D) + \sin ( C - A + B - D) + \sin ( C - A - B + D)  + \sin ( A - B + C - D) + \sin ( A - B - C + D) \Big]

= \frac{1}{2} \Big[ \sin ( A + B -C-D) + \sin ( B + D - C - A) + \sin ( -( A + D - B - C) ) + \sin (- ( A + B - C - D) )  + \sin ( -( B + D - A - C) ) + \sin ( A + D - B - C) \Big]

= \frac{1}{2} \Big[ \sin ( A + B -C-D) + \sin ( B + D - C - A) - \sin ( A + D - B - C)  - \sin  ( A + B - C - D)   - \sin ( B + D - A - C)  + \sin ( A + D - B - C) \Big]

= \frac{1}{2}  \Big[  0  \Big] = 0 =  RHS. Hence proved.

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Question 7: Prove that: \tan x \tan \Big( \frac{\pi}{3} - x \Big) \tan \Big( \frac{\pi}{3} + x \Big) = \tan 3x

Answer:

LHS = \tan x \tan (60-x) \tan (60+x)

= \frac{2 \sin x \sin ( 60-x) \sin ( 60+x)}{2 \cos x \cos ( 60-x) \cos ( 60+x)}

= \frac{\sin x }{\cos x} \Big[ \frac{2 \sin ( 60-x) \sin ( 60+x)}{2 \cos ( 60-x) \cos ( 60+x)}   \Big]

= \frac{\sin x }{\cos x} \Big[ \frac{\cos ( 60-x- 60 - x) - \cos ( 60 - x + 60 + x) }{\cos ( 60 - x + 60 + x) + \cos ( 60 - x - 60 - x)}   \Big]

= \frac{\sin x }{\cos x} \Big[ \frac{\cos ( -2x) - \cos 120^o }{\cos 120^o + \cos ( -2x)}   \Big]

= \frac{\sin x }{\cos x} \Big[ \frac{\cos 2x - \cos 120^o }{\cos 120^o + \cos 2x }   \Big]

= \frac{\sin x }{\cos x} \Big[ \frac{\cos 2x - \cos (90^o + 30^o) }{\cos (90^o + 30^o) + \cos 2x }   \Big]

= \frac{\sin x }{\cos x} \Big[ \frac{\cos 2x + \sin 30^o }{- \sin 30^o + \cos 2x }   \Big]

= \frac{\sin x }{\cos x} \Big[ \frac{2\cos 2x + 1 }{2 \cos 2x - 1 }   \Big]

= \frac{2 \sin x \cos 2x + \sin x}{2 \cos x \cos 2x - \cos x}

= \frac{\sin ( x + 2x) + \sin ( x - 2x) + \sin x}{\cos ( x + 2x) + \cos ( x - 2x) - \cos x}

= \frac{\sin 3x - \sin x + \sin x}{\cos 3x + \cos x - \cos x}

= \frac{\sin 3x}{\cos 3x}

= \tan 3x = RHS. Hence proved.

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Question 8: If \alpha + \beta = \frac{\pi}{2} , show that the maximum value of \cos \alpha \cos \beta is \frac{1}{2}

Answer:

Let x =  \cos \alpha \cos \beta

\Rightarrow x = \frac{1}{2} \Big[ 2 \cos \alpha \cos \beta \Big]

\Rightarrow x = \frac{1}{2} \Big[ \cos (\alpha + \beta) + \cos (\alpha - \beta)  \Big]

\Rightarrow x = \frac{1}{2} \Big[ \cos 90^o + \cos (\alpha - \beta)  \Big]

\Rightarrow x = \frac{1}{2} \Big[ 0+ \cos (\alpha - \beta)  \Big]

\Rightarrow x = \frac{1}{2}   \cos (\alpha - \beta)

Now we know-1 \leq \cos (\alpha - \beta) \leq 1

\Rightarrow - \frac{1}{2} \leq \frac{1}{2} \cos (\alpha - \beta) \leq \frac{1}{2}

\Rightarrow - \frac{1}{2} \leq x \leq \frac{1}{2}

Therefore the max value of x is \frac{1}{2}