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Question-1

Find the radian measure corresponding to the following degree measures:
   a) 15o
   
b) -370 30
   
c) 2400
   
d) 5300    

    
    

Solution:
a) 1o = radian
15o == radian

b) 1o =radian
-37o30’ ==

c) 1o = radian
240o = =

d) 1o =radian

530o ==

Question-2

Find the degree measure corresponding to the following radian measures:
    a)
    b) –4
    c)
    d)

Solution:
a)1 radian =
=× == = 42.95 = 42o57

b)1 radian =
-4radian = × -4 == -32.72 × 7 = -229o5’24’’

c)1 radian =

== 3000

d) 1 radian =
= 210o

Question-3

A wheel makes 360 revolutions in one minute. Through how many radians does it turn is one second?

Solution:
One complete revolution = 2π360 revolutions = 360 × 2π= 720π radians / minute = 720π /60 radians/sec = 12π radians/sec.   

Question-4

Find the degree measure of the angle subtended at the centre of a circle of diameter 200cm by an arc of length 22cm.

Solution:
Radius (r) = (200/2)cm = 100cm
Length of an arc (l) = 22cm
θ = l/r = 22/100radians = =12o36’

Question-5

In a circle of diameter 40cm. The length of a chord is 20cm. Find the length of minor arc corresponding to the chord.

Solution:

Question-6

If, in two circles, arcs of the same length subtend angles of 60o and 75o at the centre, find the ratio of their radii.

Solution:

Let the radii be r1 and r2. Let the angles subtend by the arcs in two circles be θ 1 and θ 2.
l = θ r where θ is the angle, l the length of an arc and r the radius of the circle.
θ 1 = 60o = 60 × =radian
θ 2 = 75o = 75 × =radian
r1 = r2
r1 / r2 = =
Therefore the required ratio is 5:4.

Question-7

Find the angle in radian through which a pendulum swings if its length is 75cm and the tip described an arcs of length
    (i) 10cm
   (ii) 15cm
  (iii) 21cm

Solution:
(i) Length of the pendulum (r) = 75cm
Length of an arc (l) = 10cm
θ = l/r = 10/75radians = 2/15radians

(ii) Length of the pendulum (r) = 75cm
Length of an arc (l) = 15cm
θ = l/r = 15/75 radians = 1/5radians

(iii) Length of the pendulum (r) = 75cm
Length of an arc (l) = 21cm
θ = l/r = 21/75 radians = 7/25radians

Question-8

Find the values of the other five trigonometric functions in the following problem: cosθ = -1/2, is quadrant III

Solution:
cosθ and secθ are negative; tan θ and cot θ are positive.
cosθ = -
sin2θ = 1 - =

sin θ = -

cosecθ =-=-

Question-9

Find the values of the other five trigonometric functions in the following problem: sin is quadrant II

Solution:
sinθ and cosecθ positive; tanθ , cosθ , cotθ and secθ all are negative.
sin2θ + cos2θ = 1
cos2θ = 1 - sin2θ = 1 - ==
cosθ = -

tanθ = = -
cotθ = = -


sec θ =

Question-10

Find the values of the other five trigonometric functions in the following problem: tanθ = is quadrant III

Solution:
sinθ , cosθ , cosecθ and secθ are negative in III rd quadrant tanθ and cotθ are positive in IIIrd quadrant.
1+tan2θ = sec2θ
1+
1+

secθ =
cosθ =
sin2θ = 1-

sinθ =
cosecθ =

Question-11

Find the values of the other five trigonometric functions in the following problem: sec θ = 13/5, θ lies in fourth quadrant.

Solution:
sin θ , tan θ , cosec θ and cot θ are negative in IV th quadrant sec θ and cos θ are positive in IVth quadrant.
sec θ = 13/5
cos θ = 1/sec θ = 5/13

sin2 θ = 1-
sin θ =
cosec θ =    
tan θ = =  
cot θ = =

Question-12

Find the value of the following trigonometric function: sin 765o

Solution:
sin 765o = sin= sin= sin=

Question-13

Find the value of the following trigonometric function: cosec(-1410o)

Solution:
cosec(-1410o) = -cosec 1410o = -cosec (8π – 30o) = cosec 30o = 2

Question-14

Find the value of the following trigonometric function: tan

Solution:
tan = tan = tan=

Question-15

Find the value of the following trigonometric function: cot

Solution:
cot= -cot= -cot= cot= 1

Question-16

Prove that: sin2+ cos2- tan2=

Solution:
L.H.S = sin2+ cos2- tan2
       
= + - 1
       =– 1
       =
= R.H.S

Question-17

Prove that:  2sin2+ coseccos2 = 0

Solution:
L.H.S = 2sin2+ coseccos2
       
= 2× + cosec×
       
=- cosec×
        =- 2×
        = 0 = R.H.S

Question-18

Prove that: 3cos2+ sec + 5tan2 =

Solution:
L.H.S = 3cos2+ sec+ 5tan2
       
= 3× - sec + 5tan2
       
= - 2 + 5×
       = - 2 + 15
       = 13 +
      
=
      
= R.H.S

Question-19

Prove that:  cot2 + cosec + 3tan2 = 6

Solution:
L.H.S = cot2 + cosec + 3tan2
      
= cot2 + cosec + 3tan2
      
= cot2 + cosec + 3tan2
      
=+ 2 + 3
       = 3 + 2 + 1
       = 6
       = R.H.S

Question-20

Prove that:  2sin2+2cos2+2sec2 = 10

Solution:
L.H.S = 2sin2+2cos2+2sec2
       
= 2sin2+2cos2+2sec2
      
= 2sin2 + 2cos2+ 2sec2
      
= 2× + 2× + 2× 4
       = 1 + 1+ 8
       = 10
       = R.H.S

Question-21

Show that:
        cos 70o cos 10o + sin 70osin10o = ½

Solution:
We know that
cos (θ - φ ) =
cosθ cosφ + sinθ sinφ∴ L.H.S = cos 70o cos 10o + sin 70osin10o
            = cos (70o - 10o)
            = cos60o
            = 1/2

Question-22

Show that:
          cos 130o cos 40o + sin 130o sin40o = 0

Solution:
We know that
cos (θ - φ ) =cosθ cosφ + sinθ sinφ∴ L.H.S = cos 130o cos 40o + sin 130o sin40o
           
= cos(130o - 40o)
            = cos90o
           
= 0

Question-23

Show that:
          sin(40o + θ )cos (10o + θ ) – cos (40o + θ )sin(10o + θ ) = ½

Solution:
We know that
sin (θ - φ ) = sinθ cosφ - cosθ sinφL.H.S = sin(40o + θ )cos (10o + θ ) – cos (40o + θ )sin(10o + θ )
            = sin30o
            =1/2
            = R.H.S

Question-24

Prove that:
          coscos - sin sin= sin(θ + φ)

Solution:
We know that
cos (θ + φ ) = cosθ cosφ - sinθ sinφL.H.S = coscos - sin sin
            = cos
            = cos
            = sin(θ + φ )
            = R.H.S

Question-25

Prove that:
        

Solution:
L.H.S =
         
          =
          =
          = R.H.S

Question-26

Prove that :
       

Solution:
L.H.S =
       
=
      
= cot2θ
        = R.H.S

Question-27

cosθ + sin (270o + θ ) – sin (270o - θ ) + cos (180o + θ ) = 0.

Solution:
L.H.S = cosθ + sin (270o + θ ) – sin (270o - θ ) + cos (180o + θ )
        = cosθ - cosθ + cos θ - cos θ
= 0= R.H.S   

Question-28

coscos(2π + θ )=1.

Solution:
L.H.S = coscos(2π + θ )
       = (- sinθ )(-cos θ )(tanθ + cotθ )
       = sinθ cos θ (tanθ + cotθ )
       = sin2θ + cos2θ
= 1 = R.H.S       

Question-29

sin (n + 1)x sin (n + 2)x + cos(n + 1)xcos(n + 2)x = cosx.

Solution:
L.H.S = sin (n + 1)x sin (n + 2)x + cos(n + 1)xcos(n + 2)x
         = cos[(n + 1)x - (n + 2)x]
         = cos (-x)
         = cosx
         = RH.S

Question-30

Find the value of:
   (i) cos 210o
   
(ii) sin 225o
  
(iii) tan 330o
  
(iv) cot (-315o)

Solution:
(i)cos 210o = cos (180o + 30o) = - cos 30o = -

(ii) sin 225o = sin (270o + 45o) = cos 45o =
(iii) tan 330o = tan (360o - 30o) = - tan 30o =

(iv) cot (-315o) = cot (315o) = cot (360o – 45o) = cot 45o = 1

Question-31

Find the value of tan(α + β ), given that
   cot α = ½, α and sec β = -5/3, β

Solution:
cot α = ½, α
and sec β = -5/3, β
tanβ = -4/3
tan(α + β ) =

               =
              =
              = 2/11

Question-32

Prove the following identity:
   sin(150o + x ) + sin(150o –x) = cosx

Solution:
We know that
2sinθ cosφ = sin(θ + φ ) + sin(θ - φ )
L.H.S = sin(150o + x ) + sin(150o –x)

         = 2sin150o cosx
         = 2sin(180o – 30o) cosx
         = 2sin30o cosx
         = 2(1/2)cosx
         = cosx
         = R.H.S

Question-33

Prove the following identity:
          cos - cos= - sinx

Solution:
We know that
-2sinθ sinφ = cos(θ + φ ) - cos(θ - φ )
L.H.S = -2sinsinx = -2 sincosx = - 2 sinsinx= - 2× sinx= - sinx = R.H.S

Question-34

Prove the following identity:
          cos + cos=cosx

Solution:
We know that
2cosθ cosφ = cos(θ + φ ) + cos(θ - φ )
L.H.S = 2coscosx = 2× cosx =cosx = R.H.S

Question-35

sin2x + 2sin4x + sin6x = 4cos2x sin4x

Solution:
L.H.S = sin2x + 2sin4x + sin6x
         = 2sincos + 2sin4x
         = 2sin4xcos2x + 2sin4x
         = 2sin4x(cos2x + 1)
         = 4sin4xcos2x
         = R.H.S

Question-36

sin2 6x – sin2 4x = sin2x sin10x

Solution:
L.H.S = sin2 6x – sin2 4x
[Using the formulae 2sin2x = 1 - cos2x]

         =
         = -(cos12x – cos8x)
         = -[cos(10x + 2x) – cos(10x - 2x)]
[Using the formulae -2sinθ sinφ = cos(θ + φ ) - cos(θ - φ )]

         = × 2 sin10x sin2x
         = sin10x sin2x

Question-37

cos22x – cos26x = sin4x sin8x

Solution:
L.H.S = cos22x – cos26x
[Using the formulae 2cos2x = 1+ cos2x]

         =
         = -(cos12x – cos4x)
         = -[cos(8x + 4x) – cos(8x - 4x)]
[Using the formulae -2sinθ sinφ = cos(θ + φ ) - cos(θ - φ )]

         =× 2 sin8x sin4x
         = sin4x sin8x

Question-38

cos7x + cos5x + cos3x + cosx = 4 cosx cos2x cos4x

Solution:
L.H.S = cos7x + cosx + cos5x + cos3x
         = cos(4x + 3x) + cos(4x – 3x) + cos(4x + x) + cos(4x – x)
[Using the formula 2cosθ cosφ = cos(θ + φ ) + cos(θ - φ )]

         = 2cos4xcos3x + 2cos4xcosx
         = 2cos4x(cos3x + cosx)
         = 2cos4x(cos3x + cosx)
[Using the formula
cosx + cosy = 2 coscos]
         = 4cos4xcoscos]
         = 4cos4xcos2xcosx
         = R.H.S

Question-39

cot4x(sin5x + sin3x) = cotx(sin5x – sin3x)

Solution:
L.H.S = cot4x(sin5x + sin3x)
[Using the formula sin5x + sin3x = 2 sincos]
         = 2cot4x sincos
        = 2cot4x sincos
         = 2cot4x sin4x cosx
         = 2cos4x cosx
R.H.S = cotx(sin5x – sin3x)
[Using the formula sin5x - sin3x = 2 cossin]

         = 2cotx cossin
         = 2cotx cossin
         = 2cos4x cosx
L.H.S = R.H.S

Question-40

tan3xtan2xtanx = tan3x – tan2x – tanx.

Solution:
tan3x = tan(2x + x)
          = (tan2x + tanx)/(1 - tan2xtanx)
tan3x(1 - tan2xtanx) = tan2x + tanx
tan3x - tan3xtan2xtanx = tan2x + tanx
tan3xtan2xtanx = tan3x – tan2x – tanx.

Question-41

Prove that =

Solution:
L.H.S =
        =
        =
        =
        =
        = R.H.S

Question-42

Prove that = tan4x

Solution:
L.H.S =
        =
        =
        =
        = tan4x
        = R.H.S

Question-43

Prove that = tan

Solution:
L.H.S =
        =
        =
        = tan
        = R.H.S

Question-44

Prove that (sinA + sin3A)/(cosA + cos3A) = tan2A

Solution:
L.H.S = (sinA + sin3A)/(cosA + cos3A) = 2sin2AcosA/2cos2AcosA = tan2A = R.H.S                                      

Hence proved

Question-45

Prove that = tan

Solution:
L.H.S =
      
=
      
=
      
= tan
       = R.H.S

Question-46

Prove that (tan5θ+ tan3θ)/(tan5θ - tan3θ)= 4cos2θcos4θ

Solution:
L.H.S       =

               =  

[sin(A+B) = sinAcosB + cosAsinB and sin (A-B) = sinAcosB - cosAsinB]

               

               (sin2θ = 2sinθ cosθ)

                 =     4 cos2θ cos4θ

                 =     R.H.S

 Hence proved

Question-47

Prove that =2sinx

Solution:
L.H.S =
       =
       =
[Using the identity cos2x = cos2x – sin2x and sinx – siny = 2cos sin]

       =
       = 2sinx
       = R.H.S

Question-48

Prove that (3sinA - sin3A)2/3 + (3cosA + cos3A)2/3= 42/3

Solution:
L.H.S =(3sinA - sin3A)2/3 + (3cosA + cos3A)2/3
        =[3sinA - (3sinA - 4sin3A)]2/3 + [3cosA + (4cos3A - 3 cosA)]2/3
        =(4sin3A)]2/3 + [4cos3A]2/3
       
= 42/3(sin2A + cos2A)2/3
       
= 42/3 12/3
        = 42/3
        = R.H.S

Question-49

Prove that = tan6x

Solution:
L.H.S =
       =
       =
       =
       = tan6x
       = R.H.S

Question-50

Prove that cos4x   =  1 - 8sin2xcos2x.

Solution:
L.H.S = cos4x = 1-2sin22x = 1-2 (2sinx cosx)2 = 1-8 sin2x cos2x = R.H.S

Hence proved.

Question-51

Prove that tan4θ   =    

Solution:
R.H.S      =     

               =   

               =    
              
             

               =     tan4θ
   
               =  L.H.S

 Hence proved.

Question-52

Prove that (sin3A + sin A) sinA + (cos 3A - cosA) cosA = 0.

Solution:
L.H.S.      =    sin3AsinA +  sin2A +  cos3AcosA -  cos2A

               =    sin3AsinA +  cos3AcosA +  sin2A -  cos2A


[cosA cosB +  sinA sinB   =  cos(A - B)]

               =     cos(3A - A)  -  cos2A

[cos2A -  sin2A =  cos 2A]

               =  cos2A -  cos2A

               =  0

               =  R.H.S

Hence proved.

Question-53

Prove that (cosα+cosβ)2  + (sinα+sinβ)2  =  4cos2 .

Solution:
L.H.S   = cos2α + cos2β + 2cosαcosβ + sin2α + sin2β + 2sinαsinβ 

           = 1+1+2(cosαcosβ + sinαsinβ)

          = 2+2 cos (αβ)

         = 2(1 + cos (αβ))

        = 2[2cos2(α β)/2]

       = 4cos2(α β)/2
       = R.H.S

Hence proved.

Question-54

Prove that   .

Solution:
L.H.S   =   

         

           =   cot3x

           =   R.H.S

Hence proved

Question-55

Prove that sin3A + sin2A - sinA = 4 sinA cosA/2cos3A/2

Solution:
L.H.S   = sin3A - sinA + sin2A

           = 2cos2A sinA + 2sinA cosA

           = 2sinA (cos2A + cosA)

           =
           = 4 sinA

           = R.H.S

Hence proved

Question-56

Prove that cos6A = 32cos6A - 48cos4A + 18cos2A-1

Solution:
cos6A      = 4 cos32A - 3 cos2A

                   = 4(2cos2A-1)3 - 3(2cos2A-1)

                   = 4[(2cos2A)3 - 3×(2cos2A)2×1 + 3 × 2 cos2A×1 - 13]-6cos2A + 3

                   = 32cos6A - 48cos4A + 18cos2A -1

                   = R.H.S

Hence proved

Question-57

Find sine, cosine and tangent of x/2 if 0 x where tanx= -4/3, x in quadrant II.

Solution:


          -4+4tan2-6 tan = 0

         4tan2-6 tan-4=0

 Put    tan=z

         4z2-6z-4=0

        z =

       

    
tanis in the I quadrant.

...  tanis +ve

 Hence, z = 2

 tan = 2

 sin =

 and cos =

Question-58

Find sine, cosine and tangent of x/2 if 0 x where cosx =-1/3, x in quadrant III.

Solution:
If x is in the III quadrant, lies in the II Quadrant. Hence sine is positive whereas cos θ, tan θ is negative.



sin θ, cos θis - ve in III quadrant

Question-59

Find sine, cosine and tangent of x/2 if 0 x £ 2p where sinx = 1/4, x in quadrant II.

Solution:
x lies in II quadrant
... lies in I quadrant
where sine, cosine and tan are positive.
sin =

2 sin cos =

2 sin =

Put sin() = y


Squaring both sides

4y2 (1-y2) =

64y2 - 64y4 = 1

64y4 - 64y2 + 1 = 0

y2 =

... sin =

and cos =  

tan =

Question-60

If A + B + C = π , then prove the following identity sin2A – sin2B + sin2C = 4cosAsinBcosC

Solution:
L.H.S = sin2A – sin2B + sin2C
        = 2cos(A + B)sin(A - B) + 2sinCcosC
        = 2cos(π - C)sin(A - B) + 2sinCcosC
        = -2cosCsin(A - B) + 2sinCcosC
        = 2cosC[-sin(A - B) + sin{π - (A + B)}]
        = 2cosC[-sin(A - B) + sin(A + B)]
        = 2cosC(2cosAsinB)
        = 4cosAsinBcosC

Question-61

If A + B + C = π , then prove the following identity:
cos2A + cos2B + cos2C = -1 - 4cosAcosBcosC

Solution:
L.H.S = cos2A + cos2B + cos2C
        = 2cos(A + B)cos(A - B) + cos2C
        = 2cos(π - C)cos(A - B) + cos2C
        = - 2cosCcos(A - B) + 2cos2C - 1
        = -1 - 2cosC[cos(A - B) - cosC]
        = -1 - 2cosC[cos(A - B) – cos{π - (A + B)}]

        = -1 - 2cosC[cos(A - B) + cos(A + B)]
        = -1 - 2cosC[2cosAcosB]
        = -1 - 4cosAcosBcosC
        = R.H.S

Question-62

If A + B + C = π , then prove the following identity:
sin2A + sin2B + sin2C = 4sinAsinBsinC

Solution:
L.H.S = sin2A + sin2B + sin2C
        = 2sin(A + B)cos(A - B) + 2sinCcosC
        = 2sin(π - C)cos(A - B) + 2sinCcosC
        = 2sinCcos(A - B) + 2sinCcosC
        = 2sinC[cos(A - B) + cos{π - (A + B)}]
        = 2sinC[cos(A - B) - cos(A + B)]
        = 2sinC[2sinAsinB]
        = 4 sinAsinBsinC
        = R.H.S

Question-63

If A + B + C = π , then prove the following identity:
          sinA + sinB + sinC = 4coscossin

Solution:
L.H.S = sinA + sinB + sinC
        = 2sincos + sinC
        = 2sincos + 2sincos
        = 2coscos + 2sincos
        = 2cos
        = 2cos
        = 2cos
        = 2cos
        = 4coscossin
        = R.H.S

Question-64

If A + B + C = π , then prove the following identity:
     cosA + cosB - cosC = 4coscossin - 1

Solution:
L.H.S = cosA + cosB – cosC
        = 2coscos + 2sin2 - 1
        = 2coscos + 2sin2 - 1
        = 2sincos + 2sin2 - 1
        = 2sin [cos + sin] – 1
        = 2sin [cos + cos] - 1
        = 2sin [2coscos] - 1
        = 4coscossin - 1

Question-65

If A + B + C = π , then prove the following identity:
         = 8sinsinsin

Solution:
sin2A + sin2B + sin2C = 4sinAsinBsinC
sinA + sinB + sinC = 4coscossin
L.H.S =

        =
        =
        =
        = 8sinsinsin
        = R.H.S

Question-66

If A + B + C = π , then prove the following identity:
      cos2A + cos2B - cos2C = 1 - 2sinAsinBcosC

Solution:
L.H.S = cos2A + cos2B - cos2C
        = 1 - sin2A + cos2B - cos2C
        = 1 + (– sin2A + cos2B) - cos2C
        = 1 + cos(A+B)cos(A-B) - cos2C
        = 1 + cos(π - C)cos(A-B) - cos2C
        = 1 - cosC[cos(A-B) + cos(π - A - B)]
        = 1 - cosC[cos(A-B) - cos(A + B)]
        = 1 - cosC[2sinAsinB]
        = 1 - 2sinAsinBcosC
        = R.H.S

Question-67

If A + B + C = π , then prove the following identity
          sin2 + sin2 + sin2 = 1 - 2sinsinsin

Solution:
L.H.S = 1 - cos2 + sin2 + sin2
        = 1 – [cos2 - sin2] + sin2
        = 1 – coscos + sin2
        = 1 – coscos + sin2
        = 1 – sincos + sin2
        = 1 – sin[cos - sin]
        = 1 – sin[cos- cos]
        = 1 – sin[2sinsin]
        = 1 – 2sinsinsin
        = R.H.S

Question-68

If A + B + C = π , then prove the following identity:
          tanA + tanB + tanC = tanAtanBtanC

Solution:
tan(A + B) = tan(π− C)
(tanA + tanB)/[1-tanAtanB]  = -
tanC
(tanA + tanB)  = -
tanC[1-tanAtanB]
tanA + tanB = -
tanC + tanAtanBtanC
tanA + tanB + tanC = tanAtanBtanC

Question-69

If A + B + C = π , then prove the following identity
    cotBcotC + cotCcotA + cotAcotB = 1

Solution:
tanA + tanB + tanC = tanAtanBtanC
Dividing both sides by tanAtanBtanC we get,
= 1
cotBcotC + cotCcotA + cotAcotB = 1
Hence proved

Question-70

If A + B + C = π , then prove the following identity
   cot+ cot + cot = cotcotcot

Solution:
tan= tan
tan= cot
cotcot = 1
cot = 1
cot
cot+ cot + cot = cotcotcot

Question-71

If A + B + C = π , then prove the following identity:
   tan2A + tan2B + tan2C = tan2Atan2Btan2C

Solution:
tan(2A + 2B) = tan(2π− 2C)
(
tan2A + tan2B)/[1-tan2Atan2B]  = -tan2C
(tan2A + tan2B)  = -
tan2C[1-tan2Atan2B]
tan2A + tan2B = -
tan2C + tan2Atan2Btan2C
tan2A + tan2B + tan2C = tan2Atan2Btan2C

Question-72

Sketch the following graph:
           y = 3sin2x

Solution:

x

-π/2

-π/4

0

π/4

π/2

3π/4

π

3sin2x

0

-3

0

3

0

-3

0

Question-73

Sketch the following graph:
          y = 2tanx

Solution:

x

-π/2

-π/4

0

 π/4

π/2

π

3π/2

2tanx

-2

0

2

0

 

Question-74

Sketch the following graph:
          y = sin

Solution:

x

-2π

-π

0

π

2π

3π

4π

sin

0

-1

0

1

0

-1

0

 

Question-75

Sketch the following graph:
   y = sinx, y = cosx, 0 x 2π

Solution:

x

0

π/2

π

3π/2

2π

sinx

0

1

0

-1

0

 

x

0

π/2

π

3π/2

cosx

1

0

-1

0

 

Question-76

Sketch the following graphs:
    y = cosx, y = cos2x, 0 x 2π

Solution:

x

0

π/2

3π/2

2π

cosx

1

0

0

1

 

x

0

π /4

π /2

3π /4

π

3π /2

cos2x

1

0

-1

0

1

-1

 

Question-77

Find the following:
     (i) cos 20o10'
     (ii) sin 48o
    
(iii) tan 54o 30'
     (iv) cot 33o40'

Solution:
(i) cos 20o10' = 0.9387
(ii) sin 48o = 0.7431
(iii) tan 54o 30' = 1.402
(iv) cot 33o40' = 1.501

Question-78

Find the angle θ, 0 θ < 90o
     
(i)  sin θ = 0.5373
    (ii) cos θ= 0.6087
    (iii) tan θ= 34.37
    (iv) cot θ= 3.018

Solution:
(i) θ = 32o30'
(ii) θ = 89o30'
iii) θ = 88o20'
(iv) θ = 18o20'

Question-79

Find the following:
     (i) sin 34o 22'
     (ii) cos64o 34'
     (iii) tan 42o 6'
     (iv) cot 47o26'

Solution:
(i) sin 34o 22' = 0.5645
(ii) cos64o 34' = 0.4295
(iii) tan 42o 6' = 0.9037
(iv) cot 47o26'= 0.9185

Question-80

Prove that:
If A + B + C = π, then
sin(B + C - A) + sin(C + A - B) – sin(A + B - C) = 4cosA cosB cosC.

Solution:
L.H.S = sin2A + sin2B – sin2C
        = 2sinAcosA + 2sin(B - C)cos(B + C)
        = 2sinAcosA - 2sin(B - C)cosA
        = 2cosA[sinA - sin(B - C)]
        = 2cosA[sin(B + C) - sin(B - C)]
        = 2cosA[2cosBsinC]
        = 4cosAcosBsinC
        = R.H.S

Question-81

Prove that:
  If A + B + C = π , then
  cosAcos2Acos4Acos8A =

Solution:
L.H.S = cosAcos2Acos4Acos8A
        =
        =
        =
        =
        =
        =
        = R.H.S

Question-82

Prove that:
  = tanθ

Solution:
L.H.S =
        =
        =
        = tanθ        = R.H.S

Question-83

Prove that:
  2tan2x = -

Solution:
R.H.S = -
        =
        =
         =
        = 2tan2x = L.H.S

Question-84

Prove that:
  sin10o sin30o sin50o sin70o = 1/16

Solution:
L.H.S = sin10o sin30o sin50o sin70o
        = cos80o cos60o cos40o cos20o
        = cos80o cos60o cos40o
        = cos80o cos60o      
        = cos80o
        =
        =
        =
        = R.H.S

Question-85

Prove that:
  (cosx + cosy)2 + (sinx + siny)2 = 4cos2

Solution:
L.H.S = (cosx + cosy)2 + (sinx + siny)2
       
= cos2x + cos2y + 2cosxcosy + sin2x + sin2y + 2sinxsiny
        = 2 + 2cosxcosy + 2sinxsiny
[Using the formula cos (θ + φ ) = cosθ cosφ + sinθ sinφ ]
        = 2[1 + cos(x-y)]                                                        
        = 4cos2[(x-y)/2]

        = R.H.S

Question-86

Prove that:
  (cosx - cosy)2 + (sinx + siny)2 = 4sin2

Solution:
L.H.S = (cosx - cosy)2 + (sinx + siny)2
        = cos2x + cos2y - 2cosxcosy + sin2x + sin2y + 2sinxsiny
        = 2 - 2cosxcosy + 2sinxsiny
        = 2 – 2(cosxcosy - sinxsiny)
[Using the formula cos (θ - φ ) = cosθ cosφ - sinθ sinφ ]
        = 2[1 - cos(x+y)]                                                  
        = 4sin2                                                         
        = R.H.S

Question-87

Prove that :
          sinθ + sin3θ + sin5θ + sin7θ = 4cosθ cos2θ sin4θ.

Solution:

L.H.S = sinθ + sin3θ + sin5θ + sin7θ          = 2sin2θ cosθ + 2sin6θ cosθ          = 2cosθ [sin2θ + sin6θ ]
       = 2cosθ 2sin4θ cos2θ
         = 4cosθ cos2θ sin4θ         = R.H.S

Question-88

Prove that:
 

Solution:
L.H.S =
       =
       =
      =
      =
      = R.H.S

Question-89

If α and β be two distinct real numbers satisfying the equation acosx + bsinx = c, prove that:
          (i)sin(α + β ) =
         (ii)tan(α + β )=

Solution:
acosα + bsinα = c ………………….(1)
acosβ + bsinβ = c ………………….(2)

[Using the formula sin (θ + φ ) =sinθ cosφ + cosθ sinφ ]
a(cosα - cosβ ) + b(sinα - sinβ ) = 0
or –2asinsin+ 2bcossin= 0
or –2sin= 0
Since α - β 2nπ , therefore
tan=b/a

(i)R.H.S =
          =
         = sin(α + β )
         = L.H.S

(ii) R.H.S =
             =
           = cos(α + β )
           = L.H.S

Question-90

If A + B +C = π , prove that following:
          ++= 2

Solution:
= 2
Multiplying both sides by 2 we get
sin2A + sin2B + sin2B = 4sinAsinBsinC

Hence it is enough if we prove the above result. The proof of the above result is same as in Q.62.

Question-91

If A + B +C = π/2 , prove that following:
  cos4A + cos4B + cos4C = -1 - 4cos2Acos2Bcos2C

Solution:
L.H.S = cos4A + cos4B + cos4C
[Using the formula cosx + cosy = 2 cos
cos]
= 2 cos(2A + 2B) cos (2A – 2B) + cos4C
= 2 cos(π - 2C) cos 2(A – B) + cos4C
= -2 cos 2C cos 2(A – B) + 2cos22C - 1
= -2 cos 2C [cos 2(A – B) - cos2C] - 1
= -2 cos 2C [cos 2(A – B) + cos2(A + B)] – 1
= -2 cos 2C [2cos2Acos(-2B)] – 1
= - 4cos2Acos2Bcos2C – 1

Question-92

Using the values of trigonometric functions for 18o and 36o, prove each of the following:
          sin2 72o – sin260o =

Solution:

sin72o = sin(90 - 18o) = cos18o
cos218o = 1 – sin218o=
L.H.S = sin2 72o – sin260o
       
= cos2 18o – sin260o
       
=
       =
      =
      =
    
= R.H.S

Question-93

Using the values of trigonometric functions for 18o and 36o, prove each of the following:
          sinsinsinsin=

Solution:
L.H.S = sinsinsinsin
        = sinsinsinsin
        = sinsinsinsin
        =
        = (sin36osin72o)2
        = (sin36ocos18o)2
        =
        =
        = 80/256
        = 5/16
        = R.H.S

Question-94

Prove that (cosα+cosβ)2  + (sinαβ)2  =  4cos2.

Solution:
L.H.S   = cos2α + cos2β + 2cosαcosβ + sin2α + sin2β + 2sinαsinβ 

          = 1+1+2(cosαcosβ + sinαsinβ)

          = 2+2 cos (αβ)

          = 2(1 + cos (αβ))

          =
          =

Question-95

Prove that   cos2A +  cos2(A+120o)  +  cos2(A-120o) =3/2.

Solution:
L.H.S   =  cos2A +  cos2(A+120o)  +  cos2(A-120o

          =   [2cos2A + 2 cos2(A+120o)  + 2 cos2(A-120o)]
          =   [1 + cos2A + 1 + cos(2A+240o)  + 1 + cos(2A-240o)]
          =   [3 + cos2A + cos(2A+240o)  + cos(2A-240o)]

[... cos(A+B) + cos(A-B)]

          =   (3 + cos2A + 2 cos2Acos240o)

          =   [3 + cos2A+2cos2A (-)]
 
          =   (3 + cos2A - cos2A)

          =   = R.H.S

Hence proved.

Question-96

Prove that sin12A = sin6A[64cos6A - 96cos4A + 36cos2A-2]

Solution:
sin12A = 2sin6Acos6A
cos6A = cos
3×2A
cos3×2A       = 4 cos32A - 3 cos2A

                    = 4(2cos2A-1)3 - 3(2cos2A-1)

                    = 4[(2cos2A)3 - 3×(2cos2A)2×1 + 3 × 2 cos2A×1 - 13]-6cos2A + 3

                    = 32cos6A - 48cos4A + 18cos2A -1

Hence sin12A =2sin6A[32cos6A - 48cos4A + 18cos2A -1]
                    =
sin6A[64cos6A - 96cos4A + 36cos2A-2] 
                     
Hence proved

Question-97

Prove that (sin3A + sin A) sinA + (cos 3A - cosA) cosA = 0.

Solution:
L.H.S. = sin3AsinA + sin2A + cos3AcosA - cos2A

         = sin3AsinA + cos3AcosA + sin2A - cos2A

[cosA cosB + sinA sinB = cos(A - B)]

         = cos(3A - A) - cos2A

[cos2A - sin2A = cos 2A]

         = cos2A - cos2A

        = 0

       = R.H.S

Hence proved. 

Question-98

Prove that sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A = cos A

Solution:
cosA cosB + sinA sinB = cos(A-B)

... L.H.S = cos(n+2)A cos (n+1)A + sin(n+1)A sin(n+2)A

            = cos [(n+2)A - (n+1)A]

            = cos A

            = R.H.S

Hence proved

Question-99

If A+B+C = π show that cosA + cosB + cosC = .

Solution:
L.H.S   =   cosA + cosB + cosC

         
          =
         




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