Trigonometry Functions – Class XI – Exercise 3.3

  1. sin2 π/6 + cos2 π/3 – tan2 π/4 = –1/2

Solution:

L.H.S = sin2 π/6 + cos2 π/3 – tan2 π/4

= (1/2)2 + (1/2)2 – (1)2

= 1/4 + 1/4 – 1 = –1/2

= R.H.S


  1. Prove that 2sin2 π/6 + coses2 7π/6 cos2 π/3 = 3/2

Solution:

L.H.S = 2sin2 π/6 + coses2 7π/6 cos2 π/3

= 2(1/2)2 + cosec2 (π + π/6)(1/2)2

= 2(1/4) + (-cosec π/6)2(1/4)

= 1/2 + (-2)2(1/4)

= 1/2 + (4/4) = = 1/2 + 1 = 3/2

= R.H.S


  1. Prove that cot2 π/6 + cosec /6 + 3tan2 π/6

Solution:

L.H.S = cot2 π/6 + cosec /6 + 3tan2 π/6

= (√3)2 + cosec(π- π/6) +3(1/√3)2

= 3 + cosec π/6 + 3 x 1/3

= 3 + 2 + 1

= 6

= R.H.S


  1. Prove that 2 sin2 (3π/4) + 2cos2( π/4) +2sec2(π/3) = 10

Solution:

L.H.S = 2 sin2 (3π/4) + 2cos2( π/4) +2sec2(π/3)

= 2{sin(π – π/4)}2 + 2(1/√2)2 + 2(2)2

=2{sinπ/4}2 + 2 x 1/2 + 8

= 2(1/√2)2 + 1+ 8

= 1 + 1 + 8

= 10

= R.H.S


  1. Find the value of

(i) sin 75˚

(ii)tan15˚

Solution:

(i) sin 75˚

= sin 75˚ cos 30˚ + cos 45˚ sin 30˚

[sin(x + y) = sin x cos y + cos x sin y]

= (1/√2) x(√3/2) + (1/√2) x(1/2)

= (√3/2√2)+(1/2√2)

=√3+1/2√2

 

(ii) tan15˚

= tan(45˚ – 30˚)

= tan45˚-tan30˚/1+tan45˚-tan30˚

= [1-(1/√3)]/[1+1(1/ √3)]

= [(√3-1)/√3]/[(√3+1)/√3]

= (√3-1)/(√3+1)

= (√3-1)2/[(√3+1)(√3-1)]

= 3+1-2√3/(√3)2-(1)2

= (4 – 2√3)/3-1 = 2 – √3


  1. Prove that cos (π/4 – x) cos(π/4 – y) – sin(π/4 – x)sin(π/4 – y) = sin(x+y)

Solution:

L.H.S = cos (π/4 – x) cos(π/4 – y) – sin(π/4 – x)sin(π/4 – y)

 

= 1/2 [2cos (π/4 – x) cos(π/4 – y)] +1/2[ –2 sin(π/4 – x)sin(π/4 – y)]

=1/2[cos{( π/4 – x)+( π/4 – y)}+cos{( π/4 – x)-( π/4 – y)}] + 1/2[cos{( π/4 – x)+( π/4 – y)} – cos{( π/4 – x)-( π/4 – y)}]

[2cosAcosB = cos(A+B)+cos(A-B)

-2sinAsinB = cos(A+B)-cos(A-B)]

= 2 x 1/2[cos{( π/4 – x)+( π/4 – y)}]

=cos [π/2 –(x + y)]

= sin(x+y)

= R.H.S


  1. Prove that tan[(π/4) +x]/ tan[(π/4) -x] = [1+tanx/1-tanx]2

Solution:

tan(A+B) = tanA+tanB/1-tanAtanB and tan(A-B) = tanA-tanB/1+tanAtanB

L.H.S = tan[(π/4) +x]/ tan[(π/4) -x] = [tan(π/4) +tanx]/[1-tan (π/4)tanx]/ [tan(π/4) -tanx] [1+tan (π/4)tanx]

= [(1+tanx)/(1-tanx)]/ [(1-tanx)/(1+tanx)]

= [(1+tanx)/(1-tanx)]2

= R.H.S


  1. Prove that cos(π+x)cos(-x)/sin(π-x)cos((π/2)+x) = cot2x

Solution:

L.H.S = cos(π+x)xos(-x)/sin(π-x)cos((π/2)+x)

= [-cosx][cosx]/[sinx][-sinx]

= [-cos2x]/[-sin2x]

=cot2x

= R.H.S


  1. cos(/2 + x)cos(2π+x)[cot (/2 – x)cot(2π+x)] = 1

Solution:

L.H.S = cos(/2 + x)cos(2π+x)[cot (/2 – x)cot(2π+x)]

= sinx cosx[tanx + cotx]

= sinx cosx (sinx/cosx + cosx/sinx)

=(sinx cosx)[(sin2x + cos2x)/(sinx cosx)]

=1

=R.H.S


  1. Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

Solution:

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

= 1/2(2sin(n+1)x sin(n+2)x + 2cos(n+1)x cos(n+2)x]

=1/2[cos{(n+1)x-(n+2)x}-cos{(n+1)x+(n+2)x}+cos{(n+1)x+(n+2)x}+cos{(n+1)x-(n-2)x}]

[-2sinAsinB = cos(A+B)-cos(A-B)

2cosAcosB = cos(A+B)+cos(A-B)]

=1/2 x 2cos{(n+1)x-(n+2)x}

=cos(-x)

=cos x

=R.H.S


  1. Prove that cos(3π/4 + x)-cos(3π/4 + x) = -√2.sinx

Solution:

We know that cosA – cosB = -2sin(A+B/2).sin(A-B/2)

L.H.S = cos(3π/4 + x)-cos(3π/4 + x)

= -2sin[(3π/4 + x)+(3π/4 – x)/2].sin[(3π/4 + x)-(3π/4 – x)/2]

= -2 sin(3π/4) sinx

= -2 sin(π – π/4) sinx

= -2 sin π/4.sinx

= -√2.sinx


  1. Prove that sin26x – sin24x = sin 2x sin 10x

Solution:

We know that,

sinA+sinB = 2sin[A+B/2]cos[A-B/2] ;

sinA-sinB = 2cos[A+B/2]sin[A-B/2]

L.H.S. = sin26x – sin24x

= (sin6x+sin4x)(sin6x-sin4x)

=[  2sin[6x+4x/2]cos[6x-4x/2] [2cos[6x+4x/2]sin[6x-4x/2]

= (2sin5xcosx)(2cos5xsinx)

= (2sin5xcos5x)(2sinxcosx)

= sin10x sin2x

= R.H.S


  1. Prove that cos22x – cos26x = sin 4x sin 8x

Solution:

We know that

cosA+cosB = 2cos[A+B/2]cos[A-B/2];

cosA – cosB = -2sin[A+B/2]sin[A-B/2];

L.H.S = cos22x – cos26x

= (cos 2x + cos 6x)(cos 2x – cos 6x)

= {2cos[2x+6x/2]cos[2x-6x/2]}{ -2sin[2x+6x/2]sin[2x-6x/2]

= [2cos4xcos(-2x)][-2sin4xsin(-2x)]

=[2cos4x cos2x][-2 sin4x (-sin2x)]

=[2 cos 4x cos2x][-2 sin4x(-sin2x)]

= (2 sin 4x cos 4x) (2 sin 2x cos 2x)

= sin 8x sin 4x

= R.H.S.


14: Prove that sin 2x + 2sin 4x + sin 6x = 4cos2x sin 4x

Solution:

L.H.S. = sin 2x + 2 sin 4x + sin 6x

= [sin 2x + sin 6x] + 2 sin 4x

=[2sin(2x+6x/2)(2x-6x/2)]+2sin4x

[sinA+sinB = 2sin(A+B/2)cos(A-B/2)]

= 2 sin4x cos(– 2x) + 2 sin4x

= 2 sin4x cos2x + 2 sin4x

= 2 sin4x (cos2x + 1)

= 2 sin4x (2 cos2x – 1 + 1)

= 2 sin4x (2 cos2x)

= 4cos2x sin 4x

= R.H.S.


15: Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Solution:

L.H.S = cot 4x (sin 5x + sin 3x)

=cot4x/sin4x [2sin(5x+3x/2)cos(5x-3x/2)]

[sinA + sinB = 2sin(A+B/2)cos(A-B/2)]

=[cos4x/sin4x ][2sin4xcosx]

= 2 cos 4x cos x

R.H.S. = cot x (sin 5x – sin 3x)

=cosx/sinx [2cos(5x+3x/2)sin(5x-3x/2)

[sinA – sinB = 2cos(A+B/2)sin(A-B/2)]

= cosx/sinx [2cos4x sinx]

= 2 cos 4x. cos x

L.H.S. = R.H.S.


  1. Prove that cos9x – cos5x/sin17x – sin3x = –sin2x/cos10x

Solution:

We know that,

cosA – cosB = -2sin[A+B/2]sin[A-B/2];

sinA – sinB = 2cos[A+B/2]sin[A-B/2];

L.H.S = cos9x – cos5x/sin17x – sin3x

={-2sin[9x+5x/2]sin[9x-5x/2]} / {2cos[17x+3x/2]sin[17x-3x/2]}

= {-2sin7x.sin2x}/{2cos10x.sin7x}

= – sin2x/cos10x

= R.H.S.


  1. Prove that sin5x+sin3x/cos5x+cos3x = tan4x

Solution:

We know that,

sinA+sinB = 2sin(A+B/2)cos(A-B/2);

cosA+cosB = 2cos(A+B/2)cos(A-B/2);

L.H.S. = sin5x+sin3x/cos5x+cos3x

= {2sin(5x+3x/2)cos(5x-3x/2)} / {2cos(5x+3x/2)cos(5x-3x/2)

= 2sin4x.cosx/2cos4x.cosx

= sin4x/cos4x

= tanx

= R.H.S.


  1. Prove that sinx-siny/cosx+cosy = tanx-y/2

Solution:

We know that

sinA – sinB = 2cos(A+B/2)sin(A-B/2);

cosA+cosB = 2cos(A+B/2)cos(A-B/2);

L.H.S. = sinx-siny/cosx+cosy

= {2cos(x-y/2)sin(x-y/2)} / {2cos(x+y/2)cos(x-y/2)}

= {sin(x-y/2)} / {cos(x-y/2)}

= tan (x-y/2)

= R.H.S.


  1. Prove that sinx+sin3x/cosx+cos3x = tan 2x

Solution:

We know that

sinA+sinB = 2sin(A+B/2)cos(A-B/2);

cosA+cosB = 2cos(A+B/2)cos(A-B/2);

L.H.S. = sinx+sin3x/cosx+cos3x

= {2sin(x+3x/2)cos(x-3x/2)} / {2cos(x+3x/2)cos(x-3x/2)}

= sin2x/cos2x

= tan2x

=R.H.S


  1. Prove that {sinx-sin3x}/{sin2x-cos2x} = 2sinx

Solution:

We know that,

sinA – sinB = 2cos(A+B/2)sin(A-B/2);

cos2A – sin2A = cos 2A

L.H.S = {sinx-sin3x}/{sin2x-cos2x}

= {2cos(x+3x/2)sin(x-3x/2)} / {-cos2x}

=2cos2xsin(-x)/-cos2x

=-2(-sinx)

=2sinx

=R.H.S.


  1. Prove that cos4x+cos3x+cos2x/sin4x+sin3x+sin2x = cot3x

Solution:

L.H.S. = cos4x+cos3x+cos2x/sin4x+sin3x+sin2x

= (cos4x+cos2x)+cos3x/(sin4x+sin2x)+sin3x

= {2cos(4x+2x/2)cos(4x-2x/2)+cos3x} / {2sin(4x+2x/2)cos(4x-2x/2)+sin3x}

[cosA+cosB = 2cos(A+B/2)cos(A-B/2);

sinA+sinB = 2sin(A+B/2)cos(A-B/2)]

={2cos3x cosx + cos3x}/{2sin3x cosx + sin3x}

=cos3x(2cosx+1)/sin3x(2cosx+1)

=cot3x

=R.H.S.


  1. Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Solution:

L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – cot (2x + x) (cot 2x + cot x)

= cotx cot2x – [cot2xcotx-1/cotx+cot2x](cot2x+cotx)

[cot(A+B) = cotAcotB-1/cotA+cotB]

= cot x cot 2x – (cot2x cotx – 1)

=1

=R.H.S.


  1. Prove that tan4x = {4tanx(1-tan2x)} / {1-6tan2x +tan4x}

Solution:

We know that,

tan2A = 2tanA/1-tan2A

L.H.S. =tan4x = tan 2(2x)

 

 Trigonometry Functions - Class XI - Exercise 3.3


  1. Prove that: cos 4x = 1 – 8sin2x cos2x

Solution:

L.H.S. = cos 4x

= cos 2(2x)

= 1 – 2 sin2 2x

[cos 2A = 1 – 2 sin2 A]

= 1 – 2(2 sin x cos x)2

[sin2A = 2sin A cosA]

= 1 – 8 sin2x cos2x

= R.H.S.


25: Prove that: cos 6x = 32 cos6x – 48 cos4x + 18 cos2x – 1

Solution:

L.H.S. = cos 6x

= cos 3(2x)

= 4 cos3 2x – 3 cos 2x

[cos 3A = 4 cos3A – 3 cos A]

= 4 [(2 cos2x – 1)3 – 3 (2 cos2x – 1)

[cos 2x = 2 cos2x – 1]

= 4 [(2 cos2x)3 – (1)3 – 3 (2 cos2 x) 2 + 3 (2 cos2 x)] – 6cos2 x + 3

= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3

= 32 cos6x – 4 – 48 cos4x + 24 cos2 x – 6 cos2x + 3

= 32 cos6x – 48 cos4x + 18 cos2x – 1

= R.H.S.


 

 

 

 

 

 

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