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

Chapter 9 of RD Sharma’s Class 11 Mathematics textbook focuses on the trigonometric ratios of the multiple and submultiple angles. This chapter explores the concepts of the trigonometric functions evaluated at angles that are multiples or submultiples of the standard angles. It provides formulas and methods for computing these ratios which are essential for solving various trigonometric problems.

Trigonometric Ratios of Multiple and Submultiple Angles

The Trigonometric ratios for multiple and submultiple angles involve the angles that are integer multiples or fractions of the standard angles like 30°, 45°, and 60°. Understanding these ratios helps in simplifying complex trigonometric expressions and solving equations. Key formulas include those for the angles like 2θ, 3θ, and θ/2 and they rely on known values of standard angles to find the ratios for these multiples and submultiples. This knowledge is crucial for the advancement of trigonometric studies and applications.

Question 30(i). If 0 ≤ x≤ π and x lie 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.

Read More:

• Trigonometric Ratios
• Multiple Angle Formulas