11类RD Sharma解决方案–第9章复数角和约数角的三角比–练习9.1 |套装3

问题30(i)。如果0≤x≤π并且x位于第二象限中，使得sinx = 1/4，则求出cos(x / 2)，sin(x / 2)和tan(x / 2)的值。

解决方案：

Given that,

 sinx = 1/4

As we know that, sinx = √(1 – cos²x)

So, 

⇒ (1/4)² = (1 – cos²x)

⇒ (1/16) – 1 = – cos²x

cosx = ± √15/4

It is given that x is in 2nd quadrant, so cosx is negative.

cosx = – √15/4

Now,

As we know that, cosx = 2 cos²(x/2) – 1

So, 

⇒ – √15/4 = 2cos²(x/2) – 1 

⇒ cos²(x/2) = – √15/8 + 1/2

cos(x/2) = ± (4-√15)/8

It is given that, x is in 2nd quadrant, so cos(x/2) is positive.

cos(x/2) = (4 – √15)/8

Again,

cosx = cos²(x/2) – sin²(x/2)

⇒ – √15/4 = {(4 – √15)/8}² – sin²(x/2)

⇒ sin²(x/2) = (4 + √15)/8

⇒ sin(x/2) = ± √{(4 + √15)/8} = √{(4 + √15)/8}

Now,

tan(x/2) = sin(x/2) / cos(x/2)

= $\frac{\sqrt{\frac{4+√15}{8}}}{\sqrt{\frac{4-√15}{8}}}$

= $\frac{\sqrt{4+√15}}{\sqrt{4-√15}}$

= $\sqrt{\frac{(4+√15)(4+√15)}{(4-√15)(4+√15)}}$

= $\frac{4+√15}{4^2-(√15)^2}$

= $\frac{4+√15}{16-15}$

= 4 + √15

Hence, the value of cos(x/2) = (4 – √15)/8, sin(x/2) = √{(4 + √15)/8}, and tan(x/2) = 4 + √15 .

问题30(ii)。如果cosx = 4/5且x为锐角，则找到tan2x。

解决方案：

Given that,  

cosx = 4/5

As we know that, sinx = √(1 – cos²x)

So, 

= √(1 – (4/5)²)

= √(1 – 16/25)

= √{(25 – 16)/25}

= √(9/25)

= 3/5

Since, tanx = sinx/cosx, so

= (3/5) / (4/5)

= 3/4

As we know that,

tan2x = 2tanx / (1 – tan²x)

= 2(3/4) / {1 – (3/4)²}

= 2(3/4) / (1 – 9/16)

= (3/2) / (7/16)

= 24/7

Hence, the value of tan2x is 24/7

问题30(iii)。如果sinx = 4/5且0

解决方案：

Given that, 

sinx = 4/5

As we know that, sinx = √(1 – cos²x)

So, 

⇒ (4/5)² = 1 – cos²x

⇒ 16/25 – 1 = -cos²x

⇒ 9/25 = cos²x

⇒ cosx = ±3/5

It is given that, x is ln the 1st quadrant 

So, cosx = 3/5

Now,

sin4x = 2 sin2x cos2x

= 2 (2 sinx cosx)(1 – 2sin²x)

= 2(2 × 4/5 × 3/5)(1 – 2(4/5)²)

= 2(24/25)(1-32/25)

= 2(24/25)((25-32)/25)

= 2(24/25)(-7/25)

= -336/625

Hence, the value of sin4x is (- 336/625)

问题31.如果tanx = b / a，则求出值$\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$

解决方案：

We have to find the value of $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$

So, 

= $\sqrt{\frac{1+\frac{b}{a}}{1-\frac{b}{a}}}+\sqrt{\frac{1-\frac{b}{a}}{1+\frac{b}{a}}}$

It is given that tanx = b/a, so

= $\sqrt{\frac{1+tanx}{1-tanx}}+\sqrt{\frac{1-tanx}{1+tanx}}$

= $\sqrt{\frac{1+\frac{sinx}{cosx}}{1-\frac{sinx}{cosx}}}+\sqrt{\frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}}}$

= $\sqrt{\frac{cosx+sinx}{cosx-sinx}}+\sqrt{\frac{cosx-sinx}{cosx+sinx}}$

= $\frac{cosx+sinx+cosx-sinx}{\sqrt{(cosx-sinx)(cosx+sinx)}}$

= $\frac{2cosx}{\sqrt{cos^2x-sin^2x}}$

= $\frac{2cosx}{\sqrt{cos2x}}$

Hence, the value of $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$ is $\frac{2cosx}{\sqrt{cos2x}}$

问题32.如果tanA = 1/7和tanB = 1/3，则表明cos2A = sin4B

解决方案：

Given that, tanA = 1/7 and tanB = 1/3

Show: cos2A = sin4B

As we know that, tan2B = 2tanB / (1 – tan²B)

= (2 × 1/3)(1 – 1/9) = 3/4

So, cos2A = (1 – tan²A)/(1 + tan²A)  

= {1-(1/7)²}/{1+(1/7)²}                     

= 48/50                                             

= 24/25

And sin4B = 2tan2B / (1 + tan²2B)

= {2 × 3/4}{1 + (3/4)²}

= 24/25

Hence, cos2A = sin4B

问题33. cos7°cos14°cos28°cos56°= sin68°/ 16cos83°

解决方案：

Lets solve LHS 

= cos7° cos14° cos28° cos56°

On dividing and multiplying by 2sin7°, we get

= $\frac{1}{2sin7°}$ × 2sin7° × cos7° × cos14° × cos28° × cos56°

= $\frac{2sin14°}{2×2sin7°}$ × cos28° × cos56°

= $\frac{2sin56°}{2×8sin7°}$ × cos56°

= $\frac{sin112°}{16sin7°}$

= $\frac{sin(180°-68°)}{16sin(90°-83°)}$

= $\frac{sin68°}{16cos83°}$

LHS = RHS

Hence proved.

问题34.证明，cos(2π/ 15)cos(4π/ 15)cos(8π/ 15)cos(16π/ 15)= 1/16

解决方案：

Let’s solve LHS 

= cos(2π/15)cos(4π/15)cos(8π/15)cos(16π/15)

On dividing and multiplying by 2sin(2π/15), we get

= $\frac{1}{2sin(\frac{2π}{15})}×2sin(\frac{2π}{15})×cos(\frac{2π}{15})×cos(\frac{4π}{15})×cos(\frac{8π}{15})×cos(\frac{16π}{15})$

= $\frac{1}{2×4sin(\frac{2π}{15})}[2sin(\frac{8π}{15})×cos(\frac{8π}{15})]×cos(\frac{16π}{15})$

= $\frac{1}{2×8sin(\frac{2π}{15})}[2sin(\frac{16π}{15})×cos(\frac{16π}{15})]$

= $\frac{1}{16sin(\frac{2π}{15})}sin(\frac{32π}{15})$

= $-\frac{1}{16sin(\frac{2π}{15})}sin(2π-\frac{32π}{15})$

= $-\frac{1}{16sin(\frac{2π}{15})}sin(-\frac{2π}{15})$

= 1/16 

LHS = RHS

Hence proved.

问题35.证明，cos(π/ 5)cos(2π/ 5)cos(4π/ 5)cos(8π/ 5)= -1/16

解决方案：

Lets solve LHS 

= cos(π/5)cos(2π/5)cos(4π/5)cos(8π/5)

On dividing and multiplying by 2sin(2π/5), we get

= $\frac{1}{2sin(\frac{π}{5})}$ × 2sin(π/5)cos(π/5)cos(2π/5)cos(4π/5)cos(8π/5)

= $\frac{1}{2sin(\frac{π}{5})}$(sin(2π/5)cos(2π/5)cos(4π/5)cos(8π/5))

= $\frac{1}{4sin(\frac{π}{5})}$[2sin(2π/5)cos(2π/5)cos(4π/5)cos(8π/5)]

= $\frac{1}{4sin(\frac{π}{5})}$[sin(4π/5)cos(4π/5)cos(8π/5)]

= $\frac{1}{8sin(\frac{π}{5})}$[2sin(4π/5)cos(4π/5)cos(8π/5)]

=$\frac{1}{8sin(\frac{π}{5})}$ [sin(8π/5)cos(8π/5)]  

=$\frac{1}{16sin(\frac{π}{5})}$[2sin(8π/5)cos(8π/5)]

= $\frac{sin(\frac{16π}{5})}{16sin(\frac{π}{5})}$

=$\frac{sin(3π+\frac{π}{5})}{16sin(\frac{π}{5})}$

= $-\frac{sin(\frac{π}{5})}{16sin(\frac{π}{5})}$

= -1/16 

LHS = RHS

Hence proved.

问题36.证明，cos(π/ 65)cos(2π/ 65)cos(4π/ 65)cos(8π/ 65)cos(16π/ 65)cos(32π/ 65)= 1/64

解决方案：

Lets solve LHS 

= cos(π/65)cos(2π/65)cos(4π/65)cos(8π/65)cos(16π/65)cos(32π/65)

Now on dividing and multiplying by 2sin(π/65), we get

= $\frac{1}{2sin(\frac{π}{65})}$ × 2sin(π/65)cos(π/65)cos(2π/65)cos(4π/65)cos(8π/65)cos(16π/65)cos(32π/65)

= $\frac{2×sin2(\frac{π}{65})}{2×2sin(\frac{π}{65})}$ × [cos(2π/65) × cos(4π/65) × cos(8π/65) × cos(16π/65) × cos(32π/65)]

= $\frac{2×sin4(\frac{π}{65})}{2×4sin(\frac{π}{65})}$ × cos(4π/65) × cos(8π/65) × cos(16π/65) × cos(32π/65)

= $\frac{2×sin8(\frac{π}{65})}{2×8sin(\frac{π}{65})}$ × cos(8π/65) × cos(16π/65) × cos(32π/65)

= $\frac{2×sin16(\frac{π}{65})}{2×16sin(\frac{π}{65})}$ × cos(16π/65) × cos(32π/65)

= $\frac{2×sin32(\frac{π}{65})}{2×32sin(\frac{π}{65})}$ × cos(32π/65)

= $\frac{sin64(\frac{π}{65})}{64sin(\frac{π}{65})}$

= $\frac{sin(π-\frac{π}{65})}{64sin(\frac{π}{65})}$

= $\frac{sin(\frac{π}{65})}{64sin(\frac{π}{65})}$

= 1/64 

LHS = RHS

Hence proved

问题37.如果2tanα=3tanβ，则证明tan(α–β)=sin2β/(5 –cos2β)

解决方案：

Given that,

2tanα = 3tanβ

Prove: tan(α – β) = sin2β / (5 – cos2β)

Proof:

Lets solve LHS 

= $\frac{tanα-tanβ}{1+tanαtanβ}$

= $\frac{3/2tanβ-tanβ}{1+3/2tan^2β}$

= $\frac{1/2tanβ}{1+3/2tan^2β}$

= $\frac{tanβ}{2+3tan^2β}$

= $\frac{\frac{sinβ}{cosβ}}{2+3\frac{sin^2β}{cos^2β}}$

= $\frac{\frac{sinβ}{cosβ}cos^2β}{2cos^2β+3sin^2β}$

= $\frac{sinβcosβ}{2cos^2β+2sin^2β+sin^2β}$

= $\frac{1}{2}\frac{2sinβcosβ}{2(cos^2β+sin^2β)+sin^2β}$ 

= $1/2\frac{sin2β}{(2+sin^2β)}$

= $\frac{sin2β}{4+2sin^2β}$

= $\frac{sin2β}{4+2(1-cos^2β)}$

= $\frac{sin2β}{6-2cos^2β}$

= $\frac{sin2β}{6-(1+cos2β)}$

= $\frac{sin2β}{5-cos2β}$

LHS = RHS

Hence proved.

问题38(i)。如果sinα+sinβ= a且cosα+cosβ= b，则证明sin(α+β)= 2ab /(a ² + b ² )

解决方案：

Given that, 

sinα + sinβ = a and cosα + cosβ = b

Prove: sin(α + β) = 2ab/(a² + b²)

Proof:

As we know that, $sinC+sinD=2sin\frac{C+D}{2}cos\frac{C-D}{2}$

So  $2sin\frac{α+β}{2}cos\frac{α-β}{2}=a$     ……(i)

Now, using the identity

$cosα+cosβ=2cos\frac{α+β}{2}cos\frac{α-β}{2}=b$   …..(ii)

Now on dividing  eq(i) and (ii), we get

tan(α + β)/2 = a/b

As we know that,

sin2x = 2tanx/(1 + tan²x)

$sin(α+β)=\frac{2tan[\frac{(α+β)}{2}]}{1+tan^2[\frac{(α+β)}{2}]}$

= $\frac{2\frac{a}{b}}{1+\frac{a^2}{b^2}}$

= 2ab/(a² + b²)

LHS = RHS

Hence proved

问题38(ii)。如果sinα+sinβ= a且cosα+cosβ= b，则证明cos(α–β)=(a ² + b ² – 2)/ 2

解决方案：

Given that, 

sinα + sinβ = a ……(i) 

cosα + cosβ = b …….(ii)

Now on squaring eq(i) and (ii) and then adding them, we get

sin²α + sin²β + 2sinαsinβ + cos²α + cos²β + 2cosαcosβ = a² + b²

⇒ 1 + 1 + 2(sinαsinβ + cosαcosβ) = a² + b²   

⇒ 2(sinαsinβ + cosαcosβ) = a² + b² – 2   

⇒ 2cos(α – β) = a² + b² – 2  

⇒ cos(α – β) = (a² + b² – 2)/2 

Hence proved.

问题39.如果2tan(α/ 2)= tan(β/ 2)，则证明cosα= $\frac{3+5cosβ}{5+3cosβ}$

解决方案：

Given that, 

2tan(α/2) = tan(β/2)

Prove: cosα = $\frac{3+5cosβ}{5+3cosβ}$

Proof:

Let us solve RHS 

= $\frac{3+5cosβ}{5+3cosβ}$

= $\frac{3+5(\frac{1-tan^2β}{1+tan^2β})}{5+3(\frac{1-tan^2β}{1+tan^2β})}$

= $\frac{3+3tan^2(\frac{β}{2})+5-5tan^2(\frac{β}{2})}{5+5tan^2(\frac{β}{2})+3-3tan^2(\frac{β}{2})}$

= $\frac{8-2tan^2(\frac{β}{2})}{8+2tan^2(\frac{β}{2})}$

= $\frac{8-8tan^2(\frac{α}{2})}{8+8tan^2(\frac{α}{2})}$

= $\frac{8(1-tan^2(\frac{α}{2}))}{8(1+tan^2(\frac{α}{2}))}$

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

= cosα 

RHS = LHS

Hence proved.

问题40.如果cosx = $\frac{cosα+cosβ}{1+cosαcosβ}$证明tan(x / 2)=±tan(α/ 2)tan(β/ 2)。

解决方案：

Given that,

$cosx=\frac{cosα+cosβ}{1+cosαcosβ}$    …..(i)

⇒ $\frac{1-tan^2(\frac{x}{2})}{1+tan^2(\frac{x}{2})}=\frac{cosα+cosβ}{1+cosαcosβ}$

Now, by componendo and dividendo, we get

⇒ $\frac{(1-tan^2(\frac{x}{2}))+(1+tan^2(\frac{x}{2}))}{(1-tan^2(\frac{x}{2}))-(1+tan^2(\frac{x}{2}))}=\frac{1+cosαcosβ+cosα+cosβ}{-(1+cosαcosβ-cosα-cosβ)}$

⇒ $\frac{2}{2tan^2(\frac{x}{2})}=\frac{(1+cosα)(1+cosβ)}{(1-cosα)(1-cosβ)}$

⇒ $tan^2(\frac{x}{2})=\frac{(1-cosα)(1-cosβ)}{(1+cosα)(1+cosβ)}$

⇒ $tan^2(\frac{x}{2})=\frac{2sin^2(\frac{α}{2})2sin^2(\frac{β}{2})}{2cos^2(\frac{α}{2})(2cos^2(\frac{β}{2})}$

⇒ tan²(x/2) = tan²(α/2)tan²(β/2)

⇒ tan(x/2) = ±tan(α/2)tan(β/2)

Hence Proved.

问题41.如果sec(x +α)+ sec(x –α)= 2secx，则证明cosx =±√2cos(α/ 2)。

解决方案：

Given that, 

sec(x + α) + sec(x – α) = 2secx

So, 

⇒ $\frac{1}{cosxcosα-sinxsinα}+\frac{1}{cosxcosα+sinxsinα}=\frac{2}{cosx}$

⇒ $\frac{2cosxcosα}{cos^2xcos^2α-sin^2xsin^2α}=\frac{2}{cosx}$

⇒ $\frac{cosxcosα}{cos^2xcos^2α-(1-cos^2x)sin^2α}=\frac{1}{cosx}$

⇒ cos²xcosα = cos²x(cos²α + sin²α) – sin²α

⇒ cos²x(1 – cosα) = sin²α

⇒ $cos^2x=\frac{sin^2α}{2sin^2(\frac{α}{2})}$

= $\frac{4sin^2(\frac{α}{2}).cos^2(\frac{α}{2})}{2sin^2(\frac{α}{2})}$

⇒ cosx = ± √2 cos(α/2)

Hence Proved

问题42.如果cosα+cosβ= 1/3且sinα+sinβ= 1/4，则证明cos(α–β)/ 2 =±5/24。

解决方案：

Given that, 

cosα + cosβ = 1/3  

sinα + sinβ = 1/4, we get

Prove:  cos(α – β)/2 = ±5/24

Proof:

(cos²α + cos²β + cosαcosβ) + (sin²α + sin²β + 2sinαsinβ) = 1/9 + 1/16

1 + 1 + 2(cosαcosβ + sinαsinβ) = 25/144

2 + 2cos(α – β) = -263/288  …..(i)

Now,

 $cos^2(\frac{α-β}{2})=\frac{1+cos(α-β)}{2}$

= $\frac{1-\frac{263}{288}}{2}$  [From (i)]

= 25/576

= ± 5/24

Hence proved.

问题43.如果sinα= 4/5且cosβ= 5/13，则证明cos {(α–β)/ 2} = 8 /√65。

解决方案：

Given that,

sinα = 4/5 and cosβ = 5/13

As we know that.

cosα = √(1 – sin²α)

So, 

= √{1 – (4/5)²}

= 3/5

Also, sinβ = √(1 – cos²β)

= √{1 – (5/13)²}

= 12/13

Now,

cos(α – β) = cosα cosβ + sinα sinβ

= (3/5)(5/13)(4/5)(12/13)

= 63/65

Thus,

cos{(α – β)/2} = $\sqrt{\frac{1+cos(α-β)}{2}}$

= $\sqrt{\frac{1+63/65}{2}}$

= 8/√65

Hence Proved.

问题44.如果acos2θ+bsin2θ= c以α和β为根，则证明，

(i)tanα+tanβ= 2b /(a + c)

(ii)tanαtanβ=(c – a)/(c + a)

(iii)tan(α+β)= b / a

解决方案：

As we know that

$cos2θ=\frac{1-tan^2θ}{1+tan^2θ}$

$sin2θ=\frac{2tanθ}{1+tan^2θ}$

Now substitute these values in the given equation, we get

a(1 – tan²θ) + b(2tanθ) = c(1 + tan²θ)

(c + a)tan²θ + 2btanθ + c – a = 0 

(i) As α and β are roots

So, sum of the roots:

tanα + tanβ = 2b / (c + a)

(ii) As α and β are roots

So, product of roots:

tanα tanβ = (c – a) / (c + a)

 (iii) tan(α + β)= $\frac{tanα+tanβ}{1-tanαtanβ}$

= $\frac{2b}{c+a-c+a}$

= b/a

Hence proved.

问题45.如果cosα+cosβ= 0 =sinα+sinβ，则证明cos2α+cos2β= -2cos(α+β)。

解决方案：

Given that,

 cosα + cosβ = 0 = sinα + sinβ

Prove: cos2α + cos2β = -2cos(α + β)

Proof:

cosα + cosβ = 0

On squaring on both sides, we get

cos²α + cos²β + 2 cosα cosβ = 0     ….(i)

Similarly

sinα + sinβ = 0

On squaring on both sides, we get

sin²α + sin²β + 2 sinα sinβ = 0       …..(ii)

Now, subtract eq (ii) from (i), we get

⇒ (cos²α + cos²β + 2 cosα cosβ) – (sin²α + sin²β + 2 sinα sinβ) = 0

⇒ cos²α – sin²α + cos²β – sin²β + 2(cosα cosβ – sinα sinβ) = 0

⇒ cos2α + cos2β + 2cos(α + β) = 0

⇒ cos2α + cos2β = -2cos(α + β)

Hence proved.