Question

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Answer 1

1. 2(4cos^3 pi/9 - 3 cos pi/9)

=2(cos(3*pi/9))

=2*cos(pi/3)

=2*1/2

=1


2.

To prove it we need a formula for

sin(3x) = sin2x cos x - sin x cos 2x
= 2 sin x cos^2 x - sin x ( 1 - 2 sin ^2 x) 
2 sin x (1-sin^2 x) - sin x ( 1 - 2 sin ^2 x) 

Sin3x= 3sin x - 4 sin ^3 x 

and

cos 3x = cos 2x cos x - sin 2x sin x = 
(2 cos^2 x - 1) cos x - 2 sin ^2 x cos x 
(2 cos^2 x - 1) cos x - 2 (1 - cos^2 x) cos x= 

Cos3x = 4 cos ^3 x - 3 cos x 


(3sin x - 4 sin ^3 x)/sin x - (4 cos ^3 x - 3 cos x)/cos x

= 3 - 4 sin ^2 x + 4 cos ^2 x +3

= 6 -4(sin^2 x + cos^2 x) 
= 6 - 4*1
= 2


3. sin(a)sin(60 - a)sin(60 + a)

=> sin(a) * (sin(60)cos(a) - sin(a)cos(60)) * (sin(60)cos(a) + sin(a)cos(60))

=> sin(a) * (sin(60)^2 * cos(a)^2 - sin(a)^2 * cos(60)^2)

=> sin(a) * ((3/4) * cos(a)^2 - (1/4) * sin(a)^2)

=> (1/4) * sin(a) * (3cos(a)^2 - sin(a)^2)

=> (1/4) * sin(a) * (3 - 3sin(a)^2 - sin(a)^2)

=> (1/4) * (3sin(a) - 4sin(a)^3)

=> (1/4) * sin(3a) 


5.
 
It is very simple ..

Just expand the function ..

 

Tan A +  (tan 60 + tan A )/(1-tan60.tanA) +  (tan (120) + tan A )/(1-tan120.tanA)

 

tan A +  [root(3) + tanA]/[1-root(3).tanA] + [ tan A - root(3) ] /[1 + root(3).tanA]

 

tanA +  [ root(3) + 3.tanA + tanA + root(3).tan^2A  + tanA - root(3) -root(3).tan^2A + 3.tanA ] / [ 1 - 3tan^2A ]

 

tanA + [ 8tanA] / [ 1-3tan^2A ]

 

[9tanA - 3.tan^3A ]/[1-3tan^2A]

 

= 3.tan3A   ( eqaul to RHS )

 

Hence proved




6.


LHS = cos3A cos3A + sin3A sin3A 
       = cos​3A (4cos3A - 3cosA) + sin3A (3sinA - 4sin3A)      (using identities of cos3A and sin3A)
      = 4cos6A - 3cos4A + 3sin4A - 4sin6A 
     = 4 (​cos6A - ​sin6A) + 3 (​sin4A - ​cos4A)     
     = 4 ( (cos2A)3 - (sin2A)3) + 3 ( (sin2A)2 - (cos2A)2 )
    = 4 [ (​cos2A - ​sin2A) (​cos4A + ​cos2A​sin2A + ​sin4A)] + 3 [(sin2A+cos2A)(sin2A - cos2A)]  

                                                                                                        using a3 - b3 = (a-b)(a2+ab+b2) and a2-b2=(a-b)(a+b)

  = 4 [ cos2A ((sin2A+cos2A)2 - cos2A​sin2A) ] + 3 [ 1 (-cos2A)]
 = 4 [cos2A (1 - ​​cos2A​sin2A )] - 3 cos2A
= 4cos2A - 4cos2A ​cos2A​sin2A - 3cos2A                                       
= cos2A - 4cos2A  ​cos2A​sin2A                                                         
= cos2A (1 - 4​cos2A​sin2A )
= cos2A (1 - sin22A)                                         ​  (Since 2sinA cosA = sin2A )
= cos2A cos22A
= cos32A
= RHS

Hence proved




7.


sin^3 x + sin^3 (x + 2π/3) + sin^3 (x + 4π/3) 
= sin^3 x + sin^3 (π + x - π/3) + sin^3 (π + x + π/3) 
= sin^3 x + sin^3 (x - π/3) + sin^3 (x + π/3) 
= (1/4) (3sinx - sin3x) + (1/4) [3sin(x - π/3) - sin(3x - π)] + (1/4) [3sin(x + π/3) - sin(3x + π)] 
= (1/4) [3sinx - sin3x + 3sin(x - π/3) + sin3x + 3sin(x + π/3) - sin3x] 
= (3/4) [sinx - sin3x + sin(x - π/3) + sin(x + π/3)]
= (3/4) [sinx - sin3x - 2 sinx sin(π/6)] 
= (3/4) (sinx - sin3x - sinx) 
= - (3/4) sin3x. 

Note:- The following formulae used in proving the above identity: 
1) sin^3 x = (1/4) (3sinx - sin3x) 
2) sinA + sinB = 2sin (A+B)/2 sin (A-B)/2



8.

sin 5A

=sin(2A +3A)

= sin2A cos3A +cos2A sin3A

= 2sinAcosA (4cos cub.A - 3cosA) + (1-2sin sq.A)(3sin A-4sin cub.A)

=2sinA cos sq.A(4cos sq.A-3) + sinA (1-2sin sq.A)(3-4sin sq.A)

=2 sinA (1- sin sq. A)[4-(1-sin sq.A)-3] + 3sin A -10sin cub.A + 8sin (to the power 5)A

=2sinA-2sin cub.A)(4-4sin sq.A-3) +

3sin A -10sin cub.A + 8sin (to the power 5)A

=2sinA - 2sin cub.A -8sin cub.A +8sinA+3sin A -10sin cub.A + 8sin (to the power 5)A

= 5sinA- 20sin cub.A +16 sin (to the power 5)A

=RHS



9.


Find the value of, tan^6 20° - 33 tan^4 20° + 27 tan^2 20° =?

Answer

13

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2 ANSWERS



Yuvraj Pokharna, Mathematics Faculty at IITeasy (2014-present)

Answered Jul 13, 2017

Use triple angle formula:

tan(3x) = (3 tan(x) − tan³(x)) / (1 − 3 tan²(x))

(3 tan(π/9) − tan³(π/9)) / (1 − 3 tan²(π/9)) = tan(π/3)

(3 tan(π/9) − tan³(π/9)) / (1 − 3 tan²(π/9)) = √3

(3 tan(π/9) − tan³(π/9)) = √3 (1 − 3 tan²(π/9))

Square both sides:

(3 tan(π/9) − tan³(π/9))² = 3 (1 − 3 tan²(π/9))²

tan⁶(π/9) − 6 tan⁴(π/9) + 9 tan²(π/9) = 3 (9 tan⁴(π/9) − 6 tan²(π/9) + 1)

tan⁶(π/9) − 6 tan⁴(π/9) + 9 tan²(π/9) = 27 tan⁴(π/9) − 18 tan²(π/9) + 3

tan⁶(π/9) − 6 tan⁴(π/9) − 27 tan⁴(π/9) + 9 tan²(π/9) + 18 tan²(π/9) = 3

tan⁶(π/9) − 33 tan⁴(π/9) + 27 tan²(π/9) = 3



10.


B.






We use formulas (A) - cosA=sin(90∘−A),

(B) - cos2A−sin2A=cos2A

(C) - 2sinAcosA=sin2A,

(D) - sinA+sinB=2sin(A+B/2)cos(A−B/2)and

(E) - sinA−sinB=2cos(A+B/2)sin(A−B/2)

(cos^2 33∘−cos^2 57∘) / (sin210.5∘−sin234.5∘)

= cos2^ 33∘−sin2(90∘−57∘) / (sin10.5∘+sin34.5∘)(sin10.5∘−sin34.5∘)- used A

= cos2^ 33∘−sin2^ 33∘ / −(2sin22.5∘cos12∘)(2cos22.5∘sin12∘) - used D & E

= cos66∘ / −(2sin22.5∘cos22.5∘×2sin12∘cos12∘) - used B

= −sin(90∘−66∘) / sin45∘sin24∘ - used A & C

= (−sin24∘)/ (1 /√2sin24∘)

= −√2

Answer 2

I Hope This Solution Help You Sweetheart !