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Class 11 RD Sharma Solutions – Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles – Exercise 9.1 | Set 3

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  • Last Updated : 08 May, 2021
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Question 30(i). If 0 ≤ x≤ π and x lies in the 2nd quadrant such that sinx = 1/4, Find the values of cos(x/2), sin(x/2), and tan(x/2).

Solution:

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 .

Question 30(ii). If cosx = 4/5 and x is acute, find tan2x.

Solution:

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

Question 30(iii). If sinx = 4/5 and 0 < x < π/2, then find the value of sin4x.

Solution:

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)

Question 31. If tanx = b/a, then find the value of \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}

Solution:

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}}

Question 32. If tanA = 1/7 and tanB = 1/3, show that cos2A = sin4B

Solution:

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

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

Solution:

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.

Question 34. Proved that, cos(2π/15)cos(4π/15)cos(8π/15)cos(16π/15) = 1/16

Solution:

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.

Question 35. Proved that, cos(π/5)cos(2π/5)cos(4π/5)cos(8π/5) = -1/16

Solution:

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.

Question 36. Proved that, cos(π/65)cos(2π/65)cos(4π/65)cos(8π/65)cos(16π/65)cos(32π/65) = 1/64

Solution:

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

Question 37. If 2tanα = 3tanβ, prove that tan(α – β) = sin2β / (5 – cos2β)

Solution:

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.

Question 38(i). If sinα + sinβ = a and cosα + cosβ = b, prove that sin(α + β) = 2ab/(a2 + b2)

Solution:

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

Question 38(ii). If sinα + sinβ = a and cosα + cosβ = b, prove that cos(α – β) = (a2 + b2 – 2)/2

Solution:

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.

Question 39. If 2tan(α/2) = tan(β/2), prove that cosα = \frac{3+5cosβ}{5+3cosβ}

Solution:

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.

Question 40. If cosx = \frac{cosα+cosβ}{1+cosαcosβ}  , prove that tan(x/2) = ± tan(α/2)tan(β/2).

Solution:

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.

Question 41. If sec(x + α) + sec(x – α) = 2secx, prove that cosx = ± √2 cos(α/2).

Solution:

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

Question 42. If cosα + cosβ = 1/3 and sinα + sinβ = 1/4, prove that cos(α – β)/2 = ±5/24.

Solution:

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.

Question 43. If sinα = 4/5 and cosβ = 5/13, prove that cos{(α – β)/2} = 8/√65.

Solution:

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.

Question 44. If acos2θ + bsin2θ = c has α and β as its roots prove that,

(i) tanα + tanβ = 2b/(a + c)

(ii) tanα tanβ = (c – a)/(c + a)

(iii) tan(α + β) = b/a

Solution:

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.

Question 45. If cosα + cosβ = 0 = sinα + sinβ, then prove that cos2α + cos2β = -2cos(α + β).

Solution:

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.


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