第11类RD Sharma解决方案–第9章多个和大约多个角度的三角比–练习9.3

证明：

问题1.罪² 72 ^o –罪² 60 ^o =(√5– 1)/ 8

解决方案：

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.

问题2. sin ² 24 ^o – sin ² 6 ^o =(√5– 1)/ 8

解决方案：

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.

问题3.罪² 42 ^o – cos ² 78 ^o =(√5+1)/ 8

解决方案：

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.

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

解决方案：

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.

问题5 $cos\frac{π}{15}cos\frac{2π}{15}cos\frac{4π}{15}cos\frac{7π}{15}=\frac{1}{16}$

解决方案：

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.

问题6。 $cos\frac{π}{15}cos\frac{2π}{15}cos\frac{3π}{15}cos\frac{4π}{15}cos\frac{5π}{15}cos\frac{6π}{15}cos\frac{7π}{15}=\frac{1}{128}$

解决方案：

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.

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

解决方案：

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.

问题8.罪6 ^O罪^42ö罪^66ö罪78 ^O = 1/16

解决方案：

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.

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

解决方案：

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.

问题10.罪孽36 ^o罪恶72 ^o罪恶108 ^o罪恶144 ^o = 5/16

解决方案：

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.