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9.2

证明：

问题1罪5θ=θ5sin – 20罪^3θ+ 16罪^5θ

解决方案：

We have,

L.H.S. = sin 5θ

= sin (3θ + 2θ) 

= sin 3θ cos 2θ + cos 3θ sin 2θ

= (3sin θ – 4sin³ θ) (1 – 2sin² θ) + (4cos³ θ – 3cos θ) (2sin θ cos θ)

= 3sin θ – 6sin³ θ – 4sin³ θ + 8sin⁵ θ + 8sin θ cos⁴ θ – 6sin θ cos² θ

= 3sin θ – 10sin³ θ + 8sin⁵ θ + 8sin θ (1 – sin² θ)² – 6sin θ (1–sin² θ)

= 3sin θ – 10sin³ θ + 8sin⁵ θ + 8sin θ (1 + sin⁴ θ – 2sin²θ) – 6sin θ + 6sin³ θ

= 3sin θ – 4sin³ θ + 8sin⁵ θ + 8sin θ + 8sin⁵ θ – 16sin³ θ – 6sin θ

= 5sin θ – 20 sin³ θ + 16 sin⁵ θ

= R.H.S.

Hence, proved.

问题2. 4(cos ³ 10 ^o + sin ³ 20 ^o )= 3(cos 10 ^o + sin 20 ^o )

解决方案：

Now we know, 

sin θ = cos (90–θ)

For θ = 60^o, we have,

=> sin 60^o = cos 30^o

=> sin (3×20^o) = cos (3×10^o)

=> 3sin 20^o – 4sin³ 20^o = 4cos³ 10^o – 3cos 10^o

=> 4cos³ 10^o + 4sin³ 20^o = 3cos 10^o + 3sin 20^o

=> 4 (cos³ 10^o + sin³ 20^o) = 3 (cos 10^o + sin 20^o)

Hence, proved.

问题3 COS ^3θ罪3θ+罪^3θ余弦3θ=(3sin4θ)/ 4

解决方案：

We have,

L.H.S. = cos³ θ sin 3θ + sin³ θ cos 3θ

= sin 3θ (cos 3θ + 3cos θ)/4  + cos 3θ (3sin θ – sin 3θ)/4 

= (sin 3θ cos 3θ + 3sin 3θ cos θ + 3cos 3θ sin θ – cos 3θ sin 3θ)/4

= [3(sin 3θ cos θ + cos 3θ sin θ) ]/4

= [3sin (3θ+θ)]/4

= (3sin 4θ)/4

= R.H.S.

Hence, proved.

问题4.正弦5A = 5 cos ⁴ A正弦A – 10 cos ² A正弦³ A +正弦⁵ A

解决方案：

We have, 

L.H.S. = sin 5A

= sin (3A + 2A)

= sin 3A cos 2A + cos 3A sin 2A

= (3sin A – 4sin³ A) (2cos² A – 1) + (4cos³ A – 3cos A) (2sin A cos A)

= 6sin A cos² A – 3sin A – 8sin³ A cos² A + 4sin³ A + 8sin A cos⁴ A – 6sin A cos² A

= – 3sin A – 8sin³ A cos² A + 4sin³ A + 8sin A cos⁴ A

= – 3sin A – 10 sin³ A cos² A + 2sin³ A cos² A + 4sin³ A + 5sin A cos⁴ A + 3sin A cos⁴ A

= 5sin A cos⁴ A – 10 sin³ A cos² A – 3sin A (1 – cos⁴ A) + 2sin³ A (2 + cos² A)

= 5sin A cos⁴ A – 10 sin³ A cos² A – 3sin A (1 – cos² A) (1 + cos² A) + 2sin³ A (2 + cos² A)

= 5sin A cos⁴ A – 10 sin³ A cos² A – 3sin³ A (1 + cos² A) + 2sin³ A (2 + cos² A)

= 5sin A cos⁴ A – 10 sin³ A cos² A – sin³ A (3 + 3cos² A – 4 – 2cos² A)

= 5sin A cos⁴ A – 10 sin³ A cos² A – sin³ A (cos² A – 1)

= 5sin A cos⁴ A – 10 sin³ A cos² A + sin⁵ A

= R.H.S.

Hence, proved.

问题5.棕褐色棕褐色(A + 60 ^o )+棕褐色棕褐色(A – 60 ^o )+棕褐色(A + 60 ^o )棕褐色(A – 60 ^o )= –3

解决方案：

We have,

L.H.S. = tan A tan (A + 60^o) + tan A tan (A – 60^o) + tan (A + 60^o) tan (A – 60^o)

= $tanA\left[\frac{tanA+tan60^0}{1-tanAtan60^0}\right]+tanA\left[\frac{tanA-tan60^0}{1+tanAtan60^0}\right]+\left[\frac{tanA+tan60^0}{1-tanAtan60^0}\right]\left[\frac{tanA-tan60^0}{1+tanAtan60^0}\right]$

= $tanA\left[\frac{(tanA+tan60^0)(1+tanAtan60^0)}{1-tan^2Atan^260^0}\right]+tanA\left[\frac{(tanA-tan60^0)(1-tanAtan60^0)}{1-tan^2Atan^260^0}\right]+\left[\frac{tan^2A-tan^260^0}{1-tan^2tan^260^0}\right]$

= $\left[\frac{tanA(4tanA-\sqrt{3}-\sqrt{3}tan^2A+4tanA+\sqrt{3}+\sqrt{3}tan^2A)+tan^2A-3}{1-3tan^2A}\right]$

= $\frac{9tan^2A-3}{1-3tan^2A}$

= $\left[\frac{-3(1-3tan^2A)}{1-3tan^2A}\right]$

= –3

= R.H.S.

Hence, proved.

问题6.棕褐色+棕褐色(60 ^o + A)–棕褐色(60 ^o – A)= 3棕褐色3A

解决方案：

We have,

L.H.S. = tan A + tan (60^o + A) – tan (60^o – A)

= $tanA+\frac{tan60^o+tanA}{1-tan60^otanA}-\frac{tan60^o-tanA}{1+tan60^otanA}$

= $tanA+\frac{\sqrt{3}+tanA}{1-\sqrt{3}tanA}-\frac{\sqrt{3}-tanA}{1+\sqrt{3}tanA}$

= $tanA+\frac{8tanA}{1-3tan^2A}$

= $\frac{9tanA-3tan^3A}{1-3tan^2A}$

= $\frac{3(3tanA-tan^3A)}{1-3tan^2A}$

= 3 tan 3A

= R.H.S.

Hence, proved.

问题7.婴儿床A +婴儿床(60 ^o + A)–婴儿床(60 ^o – A)= 3婴儿床3A

解决方案：

We have,

L.H.S. = cot A + cot (60^o + A) – cot (60^o – A)

= $\frac{1}{tanA}+\frac{1-tan60^otanA}{tan60^o+tanA}-\frac{1+tan60^otanA}{tan60^o-tanA}$

= $\frac{1}{tanA}+\frac{1-\sqrt{3}tanA}{\sqrt{3}+tanA}-\frac{1+\sqrt{3}tanA}{\sqrt{3}-tanA}$

= $\frac{3-tan^2A-8tan^2A}{3tanA-tan^3A}$

= $\frac{3(1-3tan^2A)}{3tanA-tan^3A}$

= $\frac{3}{tan3A}$

= 3 cot 3A

= R.H.S.

Hence, proved.

问题8.婴儿床A +婴儿床(60 ^o + A)+婴儿床(120 ^o + A)= 3婴儿床3A

解决方案：

We have,

L.H.S. = cot A + cot (60^o + A) + cot (120^o + A)

= cot A + cot (60^o + A) – cot (180 – (120^o + A))

= cot A + cot (60^o + A) – cot (60^o – A)

= $\frac{1}{tanA}+\frac{1-tan60^otanA}{tan60^o+tanA}-\frac{1+tan60^otanA}{tan60^o-tanA}$

= $\frac{1}{tanA}+\frac{1-\sqrt{3}tanA}{\sqrt{3}+tanA}-\frac{1+\sqrt{3}tanA}{\sqrt{3}-tanA}$

= $\frac{3-tan^2A-8tan^2A}{3tanA-tan^3A}$

= $\frac{3(1-3tan^2A)}{3tanA-tan^3A}$

= $\frac{3}{tan3A}$

= 3 cot 3A

= R.H.S.

Hence, proved.

问题9。 $sin^3A+sin^3(\frac{2π}{3}+A)+sin^3(\frac{4π}{3}+A)=\frac{-3}{4}sin3A$

解决方案：

We have,

L.H.S. = $sin^3A+sin^3(\frac{2π}{3}+A)+sin^3(\frac{4π}{3}+A)$ 

= $(\frac{3sinA-sin3A}{4})+\left[\frac{3sin(\frac{2π}{3}+A)-sin3(\frac{2π}{3}+A)}{4}\right]+\left[\frac{3sin(\frac{4π}{3}+A)-sin3(\frac{4π}{3}+A)}{4}\right]$

= $(\frac{3sinA-sin3A}{4})+\left[\frac{3sin(π-(\frac{π}{3}-A))-sin(2π+3A)}{4}\right]+\left[\frac{3sin(π+(\frac{π}{3}+A))-sin(4π+3A)}{4}\right]$

= $\frac{1}{4}\left[3sinA-sin3A+3sin(\frac{π}{3}-A)-sin3A-{3sin(\frac{π}{3}+A)-sin3A}\right]$

= $\frac{1}{4}\left[3sinA-3sin3A+3(sin(\frac{π}{3}-A)-sin(\frac{π}{3}+A))\right]$

= $\frac{1}{4}\left[3sinA-3sin3A+3(2cos\frac{\frac{π}{3}-A+\frac{π}{3}+A}{2}sin\frac{\frac{π}{3}-A-\frac{π}{3}-A}{2})\right]$

= $\frac{1}{4}\left[3sinA-3sin3A+6cos\frac{π}{3}sin(-A)\right]$ 

= $\frac{1}{4}\left[3sinA-3sin3A-3sinA\right]$

= $\frac{-3}{4}sin3A$

= R.H.S.

Hence, proved.

问题10。| sinθsin (60–θ)sin(60 +θ)|对于所有θ值，≤1/4。

解决方案：

We have,

= |sin θ sin (60–θ) sin (60+θ)|

= |sin θ (sin² 60 – sin² θ)|

= |sin θ (3/4 – sin² θ)|

= |sin θ/4 (3 – 4sin² θ)|

= |1/4 (3sin θ – 4sin³ θ)|

= |1/4 (sin 3θ)| ≤ 1/4

Hence, proved.

问题11。| cosθcos (60–θ)cos(60 +θ)|对于所有θ值，≤1/4 。

解决方案：

We have,

= |cos θ cos (60–θ) cos (60+θ)|

= |cos θ (cos² 60 – sin² θ)|

= |cos θ (1/4 – sin² θ)|

= |cos θ/4 (1 – 4sin² θ)|

= |cos θ/4 (1 – 4 (1 – cos² θ))|

= |cos θ/4 (–3 + 4cos² θ)|

= |1/4 (4cos³ θ – 3cos θ)|

= |1/4 (cos 3θ)| ≤ 1/4

Hence, proved.