Class 11 RD Sharma Solutions - Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles - Exercise 9.3

Prove that:

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

Solution:

We have,

L.H.S. = sin² 72^o – sin² 60^o

= sin² (90^o–18^o) – sin² 60^o

= cos² 18^o – sin² 60^o

= \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 24^o – sin² 6^o = (√5 – 1)/8

Solution:

We have,

L.H.S. = sin² 24^o – sin² 6^o

= sin (24^o + 6^o) sin (24^o – 6^o)

= (sin 30^o) (sin 18^o)

= (1/2) × (√5 – 1)/4

= (√5 – 1)/8

= R.H.S.

Hence, proved.

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

Solution:

We have,

L.H.S. = sin² 42^o – cos² 78^o

= sin² (90^o–48^o) – cos² (90^o–12^o)

= cos² 48^o – sin² 12^o 

= cos (48^o + 12^o) cos (48^o – 12^o)

= cos 60^o cos 36^o

= (1/2) × (√5 + 1)/4

= (√5 + 1)/8

= R.H.S.

Hence, proved.

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

Solution:

We have,

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

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

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

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

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

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

= \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 6^o cos 42^o cos 66^o cos 78^o = 1/16

Solution:

We have,

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

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

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

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

= \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 6^o sin 42^o sin 66^o sin 78^o = 1/16

Solution:

We have,

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

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

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

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

= \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 36^o cos 42^o cos 60^o cos 78^o = 1/16

Solution:

We have,

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

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

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

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

= \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 36^o sin 72^o sin 108^o sin 144^o = 5/16

Solution:

We have,

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

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

=  sin 36^o sin 72^o sin 72^o sin 36^o 

 = (1/4) (2sin 36^o sin 72^o)²

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