PROVING TRIGONOMETRIC IDENTITIES SAMPLE PROBLEMS

Problem 1 :

If (cos4α/cos2β) + (sin4α/sin2β)  =  1, prove that

(i) sin4α + sin4β  =  2sin2αsin2β

(ii) (cos4β/cos2α) + (sin4β/sin2α)  =  1

Solution :

Given that :

(cos4α/cos2β) + (sin4α/sin2β)  =  1

((cos4α sin2β + sin4α cos2β) / cos2β sin2β)  =  1

((cos4α sin2β + sin4α cos2β) / cos2β sin2β)  =  1

cos4α sin2β + sin4α cos2β  =  cos2β sin2β

cos4α (1- cos2β) + cos2β (1 - cos2α)2  =  cos2β sin2β

cos4α (1- cos2β) + cos2β(1 + cos4α - 2cos2α) = cos2β sin2β

cos4α - cos4α cos2β + cos2β + cos2β cos4α - 2cos2β cos2α 

=  cos2β(1 - cos2β)

cos4α - cos4α cos2β + cos2β + cos2β cos4α - 2cos2β cos2α  - cos2β + cos4β  =  0

cos4α - 2cos2β cos2α + cos4β  =  0

(cos2α - cos2β)2  =  0

cos2α  =  cos2β

1 - sin2α  =  1 - sin2β

sin2α  =  sin2β

(i)  sin4α + sin4β  =  2sin2α sin2β

L.H.S :

  =  sin4α + sin4β

  =  (sin2α - sin2β)2 + 2sin2α sin2β

  =  (sin2 α - sin2 α)2 + 2sin2 α sin2 β

  =  2sin2α sin2β  ----> R.H.S

(ii) (cos4β/cos2α) + (sin4β/sin2α)  =  1.

  =  (cosβ cosβ/cos2α) + (sin2β sin2β/sin2α)

  =  (cos2β cos2α/cos2α) + (sin2β sin2α/sin2α)

  =  cos2β + sin2β

  =  1

= R.H.S

Problem 2 :

If y = 2 sinα/(1 + cosα + sinα), then prove that

(1 − cosα + sinα)/(1 + sinα)  =  y

Solution :

(1 − cos α + sin α)/(1 + sin α)

Multiply both numerator and denominator by (1 + cosα + sinα).

=   2sinα/(1 + cosα + sinα)

Hence proved.

Problem 3 :

Solution :

x  = 1 + cos2θ + cos4θ + cos6θ + .............

x  =  1 / (1 - cos2θ)  =  1/sin2θ 

y  = 1 + sin2θ + sin4θ + sin6θ + .............

y  =  1 / (1 - sin2θ)  =  1/cos2θ

y  = 1 + cos2θ sin2θ + cos4θ sin4θ +  .............

z  =  1 / (1 - sin2θ cos2θ)  =  1/(1 - sin2θcos2θ)

L.H.S 

xyz  =  (1/sin2θ) (1/cos2θ) (1/(1 - sin2θ cos2θ))

=  1/[sin2θ cos2θ(1 - sin2θ cos2θ)]

R.H.S

x + y + z  =  (1/sin2θ) + (1/cos2θ) + (1/(1 - sin2θ cos2θ)) 

Hence proved.

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