Class 11 Maths Ncert Solutions Chapter 3 Ex 3.5 Trigonometric Functions PDF

Class 11 Maths Ncert Solutions Ex 3.5

Class 11 Maths Ncert Solutions Chapter 3 Ex 3.5

 Q.1: Prove that:

2cosπ13cos9π13+cos3π13+cos5π13=0

 

Sol:

Taking L.H.S.

2cosπ13cos9π13+cos3π13+cos5π13 :

= 2cosπ13cos9π13+2cos(3π13+5π132)cos(3π135π132)

Since, [cosx+cosy=2cos(x+y2)cos(xy2)]

= 2cosπ13cos9π13+2cos4π13cos(π13)

= 2cosπ13cos9π13+2cos4π13cosπ13

= 2cosπ13[cos9π13+cos4π13]

= 2cosπ13[2cos(9π13+4π132)cos(9π134π132)]

= 2cosπ13[2cosπ2cos5π26]

= 2cosπ13×2×0×cos5π26

= 0

= R.H.S.

Therefore, 2cosπ13cos9π13+cos3π13+cos5π13=0

 

 

Q.2: Prove that:

(sin3x+sinx)sinx+(cos3xcosx)cosx=0

 

Sol:

Taking L.H.S.

(sin3x+sinx)sinx+(cos3xcosx)cosx :

= sin3xsinx+sin2x+cos3xcosxcos2x

= cos3xcosx+sin3xsinx(cos2xsin2x)

= cos(3xx)cos2x

Since, [cos(AB)=cosAcosB+sinAsinB]

= cos2xcos2x

= 0

= R.H.S.

Therefore, (sin3x+sinx)sinx+(cos3xcosx)cosx=0

 

 

Q-3: Prove that:

(cosx+cosy)2+(sinxsiny)2=4cos2x+y2

 

Sol:

Taking L.H.S.

(cosx+cosy)2+(sinxsiny)2 :

= cos2x+cos2y+2cosxcosy+sin2x+sin2y2sinxsiny

= (cos2x+sin2x)+(cos2y+sin2y)+2(cosxcosysinxsiny)

= 1+1+2cos(x+y)

Since, [cos(A+B)=cosAcosBsinAsinB]

= 2+2cos(x+y)

= 2[1+cos(x+y)]

= 2[1+2cos2(x+y2)1]

Since, [cos2A=2cos2A1]

= 4cos2(x+y2)

= R.H.S.

Therefore, (cosx+cosy)2+(sinxsiny)2=4cos2x+y2

 

 

Q-4: Prove that:

(cosxcosy)2+(sinxsiny)2=4sin2xy2

 

Sol:

Taking L.H.S.

(cosxcosy)2+(sinxsiny)2 :

= cos2x+cos2y2cosxcosy+sin2x+sin2y2sinxsiny

= (cos2x+sin2x)+(cos2y+sin2y)2(cosxcosysinxsiny)

= 1+12[cos(xy)]

Since, [cos(AB)=cosAcosB+sinAsinB]

= 2[1cos(xy)]

= 2[1{12sin2(xy2)}]

Since, [cos2A=12sin2A]

= 4sin2xy2

= R.H.S.

Therefore, (cosxcosy)2+(sinxsiny)2=4sin2xy2

 

 

Q-5: Prove that:

sinx+sin3x+sin5x+sin7x=4cosxcos2xcos4x

 

Sol:

Taking L.H.S.

sinx+sin3x+sin5x+sin7x :

= (sinx+sin5x)+(sin3x+sin7x)

= 2sin(x+5x2)cos(x5x2)+2sin(3x+7x2)cos(3x+7x2)

Since, [sinA+sinB=2sin(A+B2).cos(AB2)]

= 2sin3xcos(2x)+2sin5xcos(2x)

= 2sin3xcos2x+2sin5xcos2x

= 2cos2x[sin3x+sin5x]

= 2cos2x[2sin(3x+5x2).cos(3x5x2)]

= 2cos2x[2sin4x.cos(x)]

= 4cos2xsin4xcosx

= R.H.S.

Therefore, sinx+sin3x+sin5x+sin7x=4cosxcos2xcos4x

 

 

Q-6: Prove that:

(sin7x+sin5x)+(sin9x+sin3x)(cos7x+cos5x)+(cos9x+cos3x)=tan6x

 

Sol:

Since, sinA+sinB=2sin(A+B2).cos(AB2)

And, cosA+cosB=2cos(A+B2).cos(AB2)

Taking L.H.S.

(sin7x+sin5x)+(sin9x+sin3x)(cos7x+cos5x)+(cos9x+cos3x) :

= [2sin(7x+5x2).cos(7x5x2)]+[2sin(9x+3x2).cos(9x3x2)][2cos(7x+5x2).cos(7x5x2)]+[2cos(9x+3x2).cos(9x3x2)]

= [2sin6x.cosx]+[2sin6x.cos3x][2cos6x.cosx]+[2cos6x.cos3x]

= 2sin6x[cosx+cos3x]2cos6x[cosx+cos3x]

= tan6x

= R.H.S.

Therefore, (sin7x+sin5x)+(sin9x+sin3x)(cos7x+cos5x)+(cos9x+cos3x)=tan6x

 

 

Q-7: Show that: sin3y+sin2ysiny=4sinycosy2cos3y2

 

Sol:

Here, L.H.S = sin3y+sin2ysiny

= sin3y+(sin2ysiny)=sin3y+[2cos(2y+y2)sin(2yy2)]

Since,  (sinxsiny)=[2cos(x+y2)sin(xy2)]

= sin3y+[2cos(3y2)sin(y2)]

=[2sin(3y2)cos(3y2)]+[2cos(3y2)sin(y2)]

Since, sin 2x = 2 sin x cos x

=2cos3y2[sin 3y2+siny2]

=2cos3y2[sin (3y2+y2)2][cos (3y2y2)2]

Since, sin x + sin y = 2sin(x+y2)cos(xy2)

=2cos(3y2)2sinycos(y2) =4sinycos(y2)cos(3y2)

= R.H.S

 

 

Q-8: The value of tany=43 where y in in 2nd quadrant then find out the values of siny2,cosy2andtany2.

 

Sol:

Here, y is in 2nd quadrant.

So, π2<y<ππ4<y2<π2

Thus, siny2,cosy2andtany2 lies in 1st quadrant.

Now,

tany=43 sec2y=1+tan2y=1+(43)2=1+169=259

So, cos2y=925

cosy=±35

As y is in 2nd quadrant, cos y is negative.

cosy=35

So, cos y = 2cos2y21

35=2cos2y212cos2y2=1352cos2y2=252cos2y2=152cos2y2=35  [Since, cosy2 is positive]

cosy2=55sin2y2+cos2y2=1sin2y2+cos215=1sin2y2=115=45sin2y2=25  [Since,  siny2 is positive]

sin2y2=255tany2=siny2cosy2=2515=2

 

 

Q-9: The value of cosy=13 where y in in 3rd quadrant then find out the values of siny2,cosy2andtany2.

 

Sol:

Here, y is in 3rd  quadrant.

So, π<y<3π2π2<y2<3π4

Thus, cosy2andtany2 are negative and siny2 is positive.

Now,  cosy=13  [Given]

cosy=12sin2y2sin2y2=1cosy2sin2y2=1132=1+132=23

siny2=23×33=63   [Since, siny2 is positive]

Now, cosy=2cos2y21

2cos2y2=1+cosy2=1132=13

   cosy2=13=13×33=13

Therefore, tany2=siny2cosy2=2313=2

 

 

Q-10: The value of siny=14 where y in in 2nd quadrant then find out the values of siny2,cosy2andtany2.

 

Sol:

Here, y is in 2nd quadrant.

So, π2<y<ππ4<y2<π2

Thus, siny2,cosy2andtany2 lies in 1st quadrant.

Now, siny=14

cos2y=12sin2y=1(14)2=1116=1516cosy=154  [Since, cosy is negative]

sin2y2=1(154)2=4+158

   siny2=4+158  [ Since, siny2 is positive ]

=4+158×22=8+21516=8+2154

cos2y=1+cosy2=1+(154)2=4158

Therefore, cosy2=4158 = 4158×22

= 821516=82154

Now, tany2=siny2cosy2=(8+2154)(82154)

= 8+2158215=8+2158215×8+2158+215

= (8+215)26460=8+2152=4+15

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