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Class 11 RD Sharma Solutions – Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles – Exercise 9.3
  • Last Updated : 28 Apr, 2021

Prove that:

Question 1.  sin2 72o – sin2 60o = (√5 – 1)/8

Solution:

We have,

L.H.S. = sin2 72o – sin2 60o

= sin2 (90o–18o) – sin2 60o

= cos2 18o – sin2 60o



\left(\frac{\sqrt{10+2\sqrt{5}}}{4}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2

 \frac{10 + 2\sqrt{5}}{16} - \frac{3}{4}

\frac{10 + 2\sqrt{5} - 12}{16}

\frac{2\sqrt{5} - 2}{16}

 \frac{\sqrt{5}-1}{8}

= R.H.S.

Hence, proved.

Question 2. sin2 24o – sin2 6o = (√5 – 1)/8

Solution:



We have,

L.H.S. = sin2 24o – sin2 6o

= sin (24o + 6o) sin (24o – 6o)

= (sin 30o) (sin 18o)

= (1/2) × (√5 – 1)/4

= (√5 – 1)/8

= R.H.S.

Hence, proved.

Question 3. sin2 42o – cos2 78o = (√5 + 1)/8

Solution:

We have,

L.H.S. = sin2 42o – cos2 78o

= sin2 (90o–48o) – cos2 (90o–12o)

= cos2 48o – sin2 12o 

= cos (48o + 12o) cos (48o – 12o)

= cos 60o cos 36o

= (1/2) × (√5 + 1)/4

= (√5 + 1)/8

= R.H.S.

Hence, proved.

Question 4. cos 78o cos 42o cos 36o = 1/8

Solution:



We have,

L.H.S. = cos 78o cos 42o cos 36o

= (1/2) (2cos 78o cos 42o) (cos 36o

= 1/2 [cos (78o + 42o) + cos (78o – 42o)] (cos 36o) 

= 1/2 [(cos 120o + cos 36o)] (cos 36o)

= 1/2 (cos (180o – 60o) + cos 36o) (cos 36o)

= 1/2 (–cos 60o + cos 36o) (cos 36o)

\frac{1}{2}\left(\frac{-1}{2}+\frac{\sqrt{5}+1}{4}\right)\frac{\sqrt{5}+1}{4}

\frac{1}{2}(\frac{\sqrt{5} - 1}{4})(\frac{\sqrt{5} + 1}{4})

\frac{1}{2}×\frac{4}{16}

\frac{1}{8}

= R.H.S.

Hence proved.

Question 5. cos\frac{π}{15}cos\frac{2π}{15}cos\frac{4π}{15}cos\frac{7π}{15}=\frac{1}{16}

Solution:

We have,

L.H.S. = cos\frac{π}{15}cos\frac{2π}{15}cos\frac{4π}{15}cos\frac{7π}{15}

\frac{2sin\frac{π}{15}cos\frac{π}{15}cos\frac{2π}{15}cos\frac{4π}{15}cos\frac{7π}{15}}{2sin\frac{π}{15}}

\frac{2sin\frac{2π}{15}cos\frac{2π}{15}cos\frac{4π}{15}cos\frac{7π}{15}}{4sin\frac{π}{15}}

\frac{2sin\frac{4π}{15}cos\frac{4π}{15}cos\frac{7π}{15}}{8sin\frac{π}{15}}

\frac{2sin\frac{8π}{15}cos\frac{7π}{15}}{16sin\frac{π}{15}}

\frac{sin(\frac{8π}{15}+\frac{7π}{15})+sin(\frac{8π}{15}-\frac{7π}{15})}{16sin\frac{π}{15}}

\frac{sinπ+sin\frac{π}{15}}{16sin\frac{π}{15}}

\frac{sin\frac{π}{15}}{16sin\frac{π}{15}}

\frac{1}{16}

= R.H.S.

Hence proved.

Question 6. cos\frac{π}{15}cos\frac{2π}{15}cos\frac{3π}{15}cos\frac{4π}{15}cos\frac{5π}{15}cos\frac{6π}{15}cos\frac{7π}{15}=\frac{1}{128}

Solution:

We have,

L.H.S. = cos\frac{π}{15}cos\frac{2π}{15}cos\frac{3π}{15}cos\frac{4π}{15}cos\frac{5π}{15}cos\frac{6π}{15}cos\frac{7π}{15}

\left[cos\frac{π}{15}cos\frac{2π}{15}cos\frac{4π}{15}(-cos\frac{8π}{15})\right]\left(\frac{1}{2}cos\frac{3π}{15}cos\frac{6π}{15}\right)    

\left[\frac{-2sin\frac{π}{15}cos\frac{π}{15}cos\frac{2π}{15}cos\frac{4π}{15}cos\frac{8π}{15}}{2sin\frac{π}{15}}\right]\frac{2sin\frac{3π}{15}cos\frac{3π}{15}cos\frac{6π}{15}}{4sin\frac{3π}{15}}

\left[\frac{-2sin\frac{2π}{15}cos\frac{2π}{15}cos\frac{4π}{15}cos\frac{8π}{15}}{4sin\frac{π}{15}}\right]\frac{2sin\frac{6π}{15}cos\frac{6π}{15}}{8sin\frac{3π}{15}}

\left[\frac{-2sin\frac{4π}{15}cos\frac{4π}{15}cos\frac{8π}{15}}{8sin\frac{π}{15}}\right]\frac{sin\frac{12π}{15}}{8sin\frac{3π}{15}}

\left[\frac{-2sin\frac{8π}{15}cos\frac{8π}{15}}{16sin\frac{π}{15}}\right]\frac{sin(π-\frac{3π}{15})}{8sin\frac{3π}{15}}

\left[\frac{-sin\frac{16π}{15}}{16sin\frac{π}{15}}\right]\frac{sin\frac{3π}{15}}{8sin\frac{3π}{15}}

\left[\frac{-sin(π+\frac{π}{15})}{16sin\frac{π}{15}}\right](\frac{1}{8})

\left[\frac{-(-sin\frac{π}{15})}{16sin\frac{π}{15}}\right](\frac{1}{8})

(\frac{1}{16})(\frac{1}{8})

\frac{1}{128}

= R.H.S.

Hence proved.

Question 7. cos 6o cos 42o cos 66o cos 78o = 1/16

Solution:

We have,

L.H.S. = cos 6o cos 42o cos 66o cos 78o

= (1/4) (2cos 6o cos 66o) (2cos 42o cos 78o)

= (1/4) (cos 72o + cos 60o) (cos 120o + cos 36o)

= (1/4) (sin 18o + cos 60o) (cos 36o − cos 60o)

\frac{1}{4}(\frac{\sqrt{5}-1}{4}+\frac{1}{2})(\frac{\sqrt{5}+1}{4}-\frac{1}{2})

\frac{1}{4}(\frac{\sqrt{5}-1+2}{4})(\frac{\sqrt{5}+1-2}{4})

\frac{1}{64}(\sqrt{5}+1)(\sqrt{5}-1)

\frac{4}{64}

\frac{1}{16}

= R.H.S.

Hence proved.

Question 8. sin 6o sin 42o sin 66o sin 78o = 1/16

Solution:

We have,

L.H.S. = sin 6o sin 42o sin 66o sin 78o

=  (1/4) (2sin 6o sin 66o) (2sin 42o sin 78o)

= (1/4) (cos 60o − cos 72o) (cos 36o − cos 120o)

= (1/4) (cos 60o − sin 18o) (cos 36o + cos 60o)

\frac{1}{4}(\frac{1}{2}-\frac{\sqrt{5}-1}{4})(\frac{\sqrt{5}+1}{4}+\frac{1}{2})

\frac{1}{4}(\frac{2-\sqrt{5}+1}{4})(\frac{\sqrt{5}+1+2}{4})

\frac{1}{4}(\frac{3-\sqrt{5}}{4})(\frac{3+\sqrt{5}}{4})

\frac{4}{64}

 \frac{1}{16}

= R.H.S.

Hence proved.

Question 9. cos 36o cos 42o cos 60o cos 78o = 1/16

Solution:

We have,

L.H.S. = cos 36o cos 42o cos 60o cos 78o

= (1/2) cos 36o cos 60o (2cos 42o cos 78o)

= (1/2) cos 36o cos 60o (cos 120o + cos 36o)

= (1/2) cos 36o cos 60o (cos 36o − cos 60o)

\frac{1}{2}(\frac{\sqrt{5}+1}{4})\frac{1}{2}(\frac{\sqrt{5}+1}{4}-\frac{1}{2})

(\frac{\sqrt{5}+1}{16})(\frac{\sqrt{5}+1}{4}-\frac{1}{2})

(\frac{\sqrt{5}+1}{16})(\frac{\sqrt{5}+1-2}{4})

\frac{(\sqrt{5}+1)(\sqrt{5}-1)}{64}

\frac{4}{64}

\frac{1}{16}

= R.H.S.

Hence proved.

Question 10. sin 36o sin 72o sin 108o sin 144o = 5/16

Solution:

We have,

L.H.S. = sin 36o sin 72o sin 108o sin 144o 

= sin 36o sin 72o sin (180o−72o) sin (180o−36o

=  sin 36o sin 72o sin 72o sin 36o 

 = (1/4) (2sin 36o sin 72o)2

= (1/4) (2sin 36o cos 18o)2

\frac{1}{4}\left(2×\frac{\sqrt{10-2\sqrt{5}}}{4}×\frac{\sqrt{10+2\sqrt{5}}}{4}\right)^2

\frac{4}{4}\left(\frac{10-2\sqrt{5}}{16}×\frac{10+2\sqrt{5}}{16}\right)

\frac{80}{256}

\frac{5}{16}

= R.H.S.

Hence proved.

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