Prove that:

Question 1: \sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x

Answer:

LHS = \sin 4x \cos x + \cos 4x \sin x

\big[ \because \cos 4x= \cos (2x+2x) = \cos 2x \cos 2x - \sin 2x \sin 2x \\ = \cos^2 2x - \sin^2 2x = 1- \sin^2 2x -  \sin^2 2x  =1 - 2 \sin^2 2x \big]

= 2 \sin 2x \cos 2x \cos x + (1 - 2 \sin^2 2x) \sin x

= 4 \sin x \cos x \cos 2x \cos x + \sin x - 2 \sin^2 2x \sin x

\big[ \because \cos 2x= \cos (x+x) = \cos x \cos x - \sin x \sin x \\ = \cos^2 x - \sin^2 x = 1- \sin^2 x -  \sin^2 x  =1 - 2 \sin^2 x \big]

= 4 \sin x \cos^2 x (1 - 2 \sin^2 x) + \sin x - 2 (2 \sin x \cos x)^2 \sin x

= 4 \sin x (1 - \sin^2 x) (1 - 2 \sin^2 x) + \sin x - 2 (4 \sin^2 x \cos^2 x) \sin x

= 4 \sin x (1 - 3 \sin^2 x + 2 \sin^4 x) + \sin x - 2 [4 \sin^3 x (1 - \sin^2 x)]

= 4 \sin x - 12 \sin^3 x + 8 \sin^5 x + \sin x - 8 \sin^3 x + 8 \sin^5 x

= 16 \sin^5 x - 20 \sin^3 x + 5 \sin x = RHS. Hence proved.

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Question 2: 4( \cos^3 10^o + \sin^3 20^o) = 3 ( \cos 10^o + \sin 20^o)

Answer:

\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta
4 \cos^3 \theta = \cos 3\theta + 3 \cos \theta
Taking \theta = 10^o
4 \cos^ 3 10^o = \cos 30^o + 3 \cos 10^o      … … … … … i)

 \sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta
4 \sin^3 \theta = 3 \sin \theta - \sin 3 \theta
Taking \theta = 20^o
4 \sin^3 20^o = 3 \sin 20^o - \sin 60^o    … … … … … ii)

Adding ( i ) and ( ii ) Equation.
4( \cos^3 10^o + \sin^3 20^o) = 3( \cos 10^o + \sin 20^o) + \cos 30^o - \sin 60^o
4( \cos^3 10^o + \sin^3 20^o) = 3( \cos 10^o + \sin 20^o) [as we know that \cos 30^o = \sin 60^o  ]

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Question 3: \cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x

Answer:

\sin 3x = \sin ( 2x + x)

= \sin 2x \cos x + \cos 2x \sin x

= 2 \sin^2 x \cos^2 x + ( \cos^2 x - \ sin^2 x) \sin x

= 2 \sin^2 x \cos^2 x + ( 1 - 2 \sin^2 x) \sin x

= 2 \sin x ( 1 - \sin^2 x) + ( 1- 2 \sin^2 x) \sin x

= 2 \sin x - 2 \sin^3 x + \sin x - 2 \sin^3 x

= 3 \sin x - 4 \sin^3 x

 \cos 3x = \cos ( 2x + x)

= \cos 2x \cos x - \sin 2 x \sin x

= ( \cos^2 x - \sin^2 x) \cos x - 2 \sin x \cos x \sin

= ( 2 \cos^2 x - 1) \cos x - 2 \cos x ( 1- \cos^2 x)

= 2 \cos^3 x - \cos x - 2 \cos x + 2 \cos^3 x

= 4 \cos^3 x - 3 \cos x

LHS = \cos^3 x \sin 3x + \sin^3 x \cos 3x

= \Big( \frac{\cos 3x + 3\cos x}{4} \Big) \sin 3x + \sin^3 x \Big( \frac{3 \sin x - \sin 3x}{4} \Big)

= \frac{1}{4} \Big[ \cos 3x \sin 3x + 3 \cos x \sin 3x + 3 \sin x \cos 3x - \sin 3x \cos x  \Big]

= \frac{3}{4} ( \cos x \sin 3x + \sin x \cos 3x )

= \frac{3}{4} \sin 4x

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Question 4:

\tan x \tan \Big( x + \frac{\pi}{3} \Big) + \tan x \tan \Big( x - \frac{\pi}{3} \Big) + \tan \Big( x + \frac{\pi}{3} \Big) \tan \Big( x - \frac{\pi}{3} \Big) = -3

Answer:

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

= \tan x \Big[ \frac{\tan x + \sqrt{3}}{1 - \tan x \sqrt{3}} \Big] + \tan x \Big[ \frac{\tan x - \sqrt{3}}{1 + \tan x \sqrt{3}} \Big] + \frac{\tan x - \sqrt{3}}{1 + \tan x \sqrt{3}} \times \frac{\tan x + \sqrt{3}}{1 - \tan x \sqrt{3}}

= \tan x \Big[ \frac{\tan x + \sqrt{3}}{1 - \tan x \sqrt{3}} + \frac{\tan x - \sqrt{3}}{1 + \tan x \sqrt{3}} \Big]  + \frac{\tan x - \sqrt{3}}{1 + \tan x \sqrt{3}} \times \frac{\tan x + \sqrt{3}}{1 - \tan x \sqrt{3}}

= \tan x \Big[ \frac{(\tan x + \sqrt{3}) (1 + \tan x \sqrt{3}) +(\tan x - \sqrt{3}) (1 - \tan x \sqrt{3}) }{(1 - \tan x \sqrt{3})(1 + \tan x \sqrt{3} ) }  \Big] +  \frac{\tan x - \sqrt{3}}{1 + \tan x \sqrt{3}} \times \frac{\tan x + \sqrt{3}}{1 - \tan x \sqrt{3}}

= \tan x \Big[  \frac{ \tan x +\sqrt{3} + \sqrt{3} \tan^2 x + 3 \tan x + \tan x -\sqrt{3} - \sqrt{3} \tan^2 x + 3 \tan x }{ 1 - 3 \tan^2 x  } \Big] + \frac{\tan^2 x - 3 }{1 - 3 \tan^2 x }

= \frac{8 \tan x}{1 - 3 \tan^2 x} + \frac{\tan^2 x - 3}{1 - 3 \tan^2 x}

= \frac{9\tan^2 x - 3}{1 - 3 \tan^2 x}

= -3 \Big( \frac{1 - 3 \tan^2 x}{1 - 3 \tan^2 x} \Big)

= - 3 = RHS. Hence proved.

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Question 5: \tan x + \tan \Big( \frac{\pi}{3} + x \Big) - \tan \Big( \frac{\pi}{3} - x \Big) = 3 \tan 3x

Answer:

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

= \tan x + \frac{\tan 60^o + \tan x }{1 - \tan 60^o \tan x } + \frac{\tan 60^o - \tan x }{1 + \tan 60^o \tan x }

= \tan x + \frac{\sqrt{3} + \tan x }{1 - \sqrt{3} \tan x } + \frac{\sqrt{3} - \tan x }{1 + \sqrt{3} \tan x }

= \tan x + \frac{(\sqrt{3} + \tan x)(1 + \sqrt{3} \tan x) + (\sqrt{3} - \tan x)(1 - \sqrt{3} \tan x)}{(1 - \sqrt{3} \tan x)(1 + \sqrt{3} \tan x)}

= \tan x + \frac{\tan x + \sqrt{3} + 3 \tan x + \sqrt{3} \tan^2 x - \sqrt{3} + \tan x + 3 \tan x - \sqrt{3} \tan^2 x}{1 - 3 \tan^2 x}

= \tan x + \frac{8 \tan x}{1 - 3 \tan^2 x}

= \frac{\tan x - 3 \tan^3 x + 8 \tan x }{1 - 3 \tan^2 x}

= \frac{9\tan x - 3 \tan^3 x  }{1 - 3 \tan^2 x}

= 3 \times \frac{3\tan x -  \tan^3 x  }{1 - 3 \tan^2 x}

= 3 \tan 3x = RHS. Hence proved.

Since

\tan 3 x = \tan ( 2x + x)

= \frac{\tan 2x + \tan x }{1 - \tan 2x \tan x }

= \frac{\big( \frac{2\tan x}{1-\tan^2 x} \big) + \tan x }{1 - \big( \frac{2\tan x}{1-\tan^2 x} \big) \tan x }

= \frac{2 \tan x + \tan x (1-\tan^2 x) }{1-\tan^2 x - 2 \tan^2 x}

= \frac{2 \tan x + \tan x - \tan^3 x }{1-3\tan^2 x }

= \frac{3 \tan x  - \tan^3 x }{1-3\tan^2 x }

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Question 6: \cot x + \cot \Big( \frac{\pi}{3} + x \Big) - \cot \Big( \frac{\pi}{3} - x \Big) = 3 \cot 3x

Answer:

LHS = \cot x + \cot ( 60^o + x) - \cot ( 60^o - x)

= \frac{1}{\tan x } + \frac{1}{\tan (60^o + x)} - \frac{1}{\tan (60^o - x)}

= \frac{1}{\tan x } + \frac{1 - \tan 60^o \tan x }{\tan 60^o + \tan x} - \frac{1 + \tan 60^o \tan x }{\tan 60^o - \tan x}

= \frac{1}{\tan x } + \frac{1 - \sqrt{3} \tan x }{\sqrt{3} + \tan x} + \frac{1 + \sqrt{3} \tan x }{\sqrt{3} - \tan x}

= \frac{1}{\tan x } + \frac{(1 - \sqrt{3} \tan x)(\sqrt{3} - \tan x) - (1 + \sqrt{3} \tan x)(\sqrt{3} + \tan x) }{(\sqrt{3} + \tan x) (\sqrt{3} - \tan x)}

= \frac{1}{\tan x } + \frac{\sqrt{3} - 3 \tan x - \tan x + \sqrt{3} \tan^2 x - \sqrt{3} - 3 \tan x - \tan x - \sqrt{3} \tan^2 x}{3 - \tan^2 x}

= \frac{1}{\tan x } - \frac{8 \tan x}{3 - \tan^2 x}

= \frac{3 - \tan^2 x - 8 \tan^2 x }{\tan x (3 - \tan^2 x)}

= \frac{3 - 9\tan^2 x }{3\tan x - \tan^3 x}

= 3 \times \frac{1 - 3\tan^2 x }{3\tan x - \tan^3 x}

= \frac{3}{\tan 3x}

= 3 \cot 3x = RHS. Hence proved.

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Question 7: \cot x + \cot \Big( \frac{\pi}{3} + x \Big) + \cot \Big( \frac{2\pi}{3} + x \Big) = 3 \cot 3x

Answer:

LHS = \cot x + \cot ( 60^o + x) + \cot ( 120^o + x)

= \cot x + \cot ( 60^o + x) - \cot [ 180 - ( 120^o + x) ]

= \cot x + \cot ( 60^o + x) - \cot ( 60^o - x)

= \frac{1}{\tan x } + \frac{1}{\tan (60^o + x)} - \frac{1}{\tan (60^o - x)}

= \frac{1}{\tan x } + \frac{1 - \tan 60^o \tan x }{\tan 60^o + \tan x} - \frac{1 + \tan 60^o \tan x }{\tan 60^o - \tan x}

= \frac{1}{\tan x } + \frac{1 - \sqrt{3} \tan x }{\sqrt{3} + \tan x} + \frac{1 + \sqrt{3} \tan x }{\sqrt{3} - \tan x}

= \frac{1}{\tan x } + \frac{(1 - \sqrt{3} \tan x)(\sqrt{3} - \tan x) - (1 + \sqrt{3} \tan x)(\sqrt{3} + \tan x) }{(\sqrt{3} + \tan x) (\sqrt{3} - \tan x)}

= \frac{1}{\tan x } + \frac{\sqrt{3} - 3 \tan x - \tan x + \sqrt{3} \tan^2 x - \sqrt{3} - 3 \tan x - \tan x - \sqrt{3} \tan^2 x}{3 - \tan^2 x}

= \frac{1}{\tan x } - \frac{8 \tan x}{3 - \tan^2 x}

= \frac{3 - \tan^2 x - 8 \tan^2 x }{\tan x (3 - \tan^2 x)}

= \frac{3 - 9\tan^2 x }{3\tan x - \tan^3 x}

= 3 \times \frac{1 - 3\tan^2 x }{3\tan x - \tan^3 x}

= \frac{3}{\tan 3x}

= 3 \cot 3x = RHS. Hence proved.

Question 8: \sin 5x = 5 \cos^4 x \sin x - 10 \cos^2 x \sin^3 x + \sin^5 x

Answer:

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

\sin 3x = \sin ( 2x + x)

= \sin 2x \cos x + \cos 2x \sin x

= 2 \sin x \cos^2 x + ( \cos^2 x - \sin^2 x) \sin x

= 2 \sin x \cos^2 x + \sin x \cos^2 x - \sin^3 x

= 3 \sin x \cos^2x - \sin^3 x

\cos 2 x = \cos x \cos x - \sin x \sin x = \cos^2 x - \sin^2 x

\sin 3x \cos 2x = (3 \sin x \cos^2x - \sin^3 x ) (\cos^2 x - \sin^2 x)

= 3 \sin x \cos^4 x - 3 \sin^3 x \cos^2 x - \sin^3 x \cos^2 x + \sin^5 x

= 3 \sin x \cos^4 x - 4 \sin^3 x \cos^2 x + \sin^5 x

\cos 3x = \cos ( 2x + x)

= \cos 2x \cos x - \sin 2x \sin x

= ( \cos^2 x - \sin^2 x ) \cos x - 2 \sin^2 x \cos x

= \cos^3 x - \sin^2 x \cos x - 2 \sin^2 x \cos x

= \cos^3 x - 3 \sin^2 x \cos x

\cos 3x \sin 2x = ( \cos^3 x - 3 \sin^2 x \cos x ) (2 \sin x \cos x)

= 2 \cos^4 x \sin x - 6 \sin^3 x \cos^2 x

\therefore \sin 5x = 3 \sin x \cos^4 x - 4 \sin^3 x \cos^2 x + \sin^5 x + 2\cos^4 x \sin x - 6 \sin^3 x \cos^2 x

= 5 \cos^4 x \sin x - 10 \cos^2 x \sin^3 x + \sin^5 x  = RHS. Hence proved.

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Question 9: \sin^3 x + \sin^3 \Big( \frac{2\pi}{3} + x \Big) + \sin^3 \Big( \frac{4\pi}{3} + x \Big) = - \frac{3}{4} \sin 3x

Answer:

LHS = \sin^3 x + \sin^3 (120^o+x) + \sin^3 (240^o+x)

=  \sin^3 x + [\sin (120^o+x)]^3 + [\sin (240^o+x) ]^3

=  \sin^3 x + [\sin 120^o \cos x + \cos 120^o \sin x]^3 + [\sin 240^o \cos x + \cos 240^o \sin x ]^3

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

=  \sin^3 x + \frac{1}{8} (\sqrt{3} \cos x - \sin x)^3 - \frac{1}{8} (\sqrt{3} \cos x + \sin x)^3

=  \sin^3 x + \frac{1}{8} \Big[ (\sqrt{3} \cos x - \sin x)^3 - (\sqrt{3} \cos x + \sin x)^3 \Big]

\because a^3 - b^3 = ( a-b)(a^2 + ab + b^2)

=  \sin^3 x + \frac{1}{8} \Big[ (\sqrt{3} \cos x - \sin x - \sqrt{3} \cos x - \sin x ) \big(  (\sqrt{3} \cos x - \sin x)^2 +(\sqrt{3} \cos x - \sin x)(\sqrt{3} \cos x + \sin x) + (\sqrt{3} \cos x + \sin x)^2  \big)   \Big]

=  \sin^3 x + \frac{1}{8} \Big[ (-2 \sin x) (3 \cos^2 x + \sin^2 x - 2 \sqrt{3} \cos x \sin x + 3 \cos^2 x - \sqrt{3} \sin x \cos x  + \sqrt{3} \sin x \cos x - \sin^2 x + 3 \cos^2 x + \sin^2 x + 2 \sqrt{3} \cos x \sin x ) \Big]

=  \sin^3 x + \frac{1}{8} \Big[ (-2 \sin x) ( 9 \cos^2 x + \sin^2 x)  \Big]

= \sin^3 x - \frac{1}{4} \sin x ( 9 \cos^2 x + \sin^2 x)

= \frac{1}{4} \Big[ 4 \sin^3 x - 9 \sin x \cos^2 x - \sin^2 x \Big]

= \frac{1}{4} ( 3 \sin^2 x - 9 \sin x \cos^2 x)

= - \frac{3}{4} \sin 3x = RHS. Hence proved.

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Question 10: \Big| \sin x \sin \Big( \frac{\pi}{3} - x \Big) \sin \Big( \frac{\pi}{3} + x \Big) \Big| \leq \frac{1}{4} for all values of x

Answer:

LHS = \Big| \sin x \sin ( 60- x) \sin ( 60 + x) \Big|

= \Big| \sin x ( \sin 60 \cos x - \cos 60 \sin x) ( \sin 60 \cos x + \cos 60 \sin x) \Big|

= \Big| \sin x ( \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x) ( \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x) \Big|

= \Big| \sin x ( \frac{3}{4} \cos^2 x - \frac{1}{4} \sin^2 x ) \Big|

= \Big| \frac{1}{4} \sin x ( 3 \cos^2 x - \sin^2 x ) \Big|

= \Big| \frac{1}{4} \sin x ( 3  - 4 \sin^2 x ) \Big|

= \Big| \frac{1}{4} ( 3 \sin x  - 4 \sin^3 x ) \Big|

= \Big| \frac{1}{4} \sin 3x \Big| \leq \frac{1}{4}                  Since \Big| \sin 3x \Big| \leq 1

Hence proven.

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Question 11: \Big| \cos x \cos \Big( \frac{\pi}{3} - x \Big) \cos \Big( \frac{\pi}{3} + x \Big) \Big| \leq \frac{1}{4} for all values of x

Answer:

LHS = \Big| \cos x \cos ( 60- x) \cos ( 60 + x) \Big|

= \Big| \cos x ( \cos 60 \cos x + \sin 60 \sin x) ( \cos 60 \cos x - \sin 60 \sin x) \Big|

= \Big| \cos x ( \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x) ( \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x) \Big|

= \Big| \cos x ( \frac{1}{4} \cos^2 x - \frac{3}{4} \sin^2 x ) \Big|

= \Big| \frac{1}{4} \cos x (  \cos^2 x - 3 \sin^2 x ) \Big|

= \Big| \frac{1}{4} \cos x (  4 \cos^2 x - 3 ) \Big|

= \Big| \frac{1}{4} (  4 \cos^3 x - 3 \cos x ) \Big|

= \Big| \frac{1}{4} \cos 3x \Big| \leq \frac{1}{4}                  Since \Big| \cos 3x \Big| \leq 1

Hence proven.