Question

solve Q26 iii)......

Answer 1

Tn = sinⁿ∅ + cosⁿ∅

T1 = sin∅ + cos∅
T3 = sin³∅ + cos³∅
T5 = sin^5∅ + cos^5∅
T7 = sin^7∅ + cos^7∅

(a) LHS = ( T3 - T5)/T1

= { sin³∅+ cos³∅ - sin^5∅- cos^5∅}/( sin∅ + cos∅)

= { sin³∅( 1- sin²∅) + cos³∅(1- cos²∅)}/( sin∅ + cos∅)

=sin³∅.cos²∅ + sin²∅.cos³∅}/( sin∅+ cos∅)

= sin²∅.cos²∅( sin∅+ cos∅)/( sin∅ + cos∅)

= sin²∅.cos²∅

RHS = (sin^5∅ + cos^5∅ - sin^7∅ - cos^7∅}/( sin³∅ + cos³∅)

= { sin^5( 1-sin²∅) + cos^5( 1- cos²∅)}/( sin³∅ + cos³∅)

= sin²∅.cos²∅( sin³∅ + cos³∅)/( sin³∅+ cos³∅)

= sin²∅.cos²∅

LHS = RHS
hence proved
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(b)
T6 = sin^6∅ + cos^6∅
T4 = sin⁴∅ + cos⁴∅

LHS = 2T6 -3T4 +1

= 2( sin^6∅ + cos^6∅) -3( sin⁴∅+ cos⁴∅) +1

= 2( sin²∅ + cos²∅)³ -3sin²∅.cos²∅( sin²∅ + cos²∅) -3{( sin²∅+ cos²∅)² -2sin²∅.cos²∅} + 1

= 2 - 6sin²∅.cos²∅ -3 +6 sin²∅.cos²∅ +1

=3 -3 = RHS

iii)6T10 -15T8 +10T6 -1

=6( sin^10∅ + cos^10∅) -15( sin^8∅+ cos^8∅) +10(sin^6∅+ cos^6∅) -1


sin^10∅ + cos^10∅ =
{(sin²∅)^5 +(cos²∅)^5 }

=(sin⁴∅+ cos⁴∅)(sin^6∅+ cos^6∅) -sin⁴∅.cos^6∅-cos^4∅.sin^6∅

=(1-2sin²∅.cos²∅)(1--3sin²∅.cos²∅) -sin⁴∅.cos⁴∅(sin²∅ + cos²∅)

= 1-5sin²∅.cos²∅+6sin⁴∅.cos⁴∅-sin⁴∅.cos⁴∅

= 1- 5sin²∅.cos²∅+5sin⁴∅.cos⁴∅


sin^8∅+ cos^8∅ = ( sin²∅)⁴ +( cos²∅)⁴

=(sin⁴∅+cos⁴∅)² -2sin⁴∅.cos⁴∅

=1+2sin⁴∅.cos⁴∅ -4sin²∅.cos²∅

sin^6∅+cos^6∅ =
(1 -3sin²∅.cos²∅)

now put this in above

=6( 1-5sin²∅.cos²∅+5sin⁴∅.cos⁴∅)-15(1+2sin⁴∅.cos⁴∅-4sin²∅.cos²∅)
+10(1-3sin²∅.cos²∅)-1

= 0 = RHS