Page 52 - E-Modul Strukbar Berbasis Case Method

Penyelesaian:

Ambil sebarang unsur = + dan = + sehingga:
1
2
2 ,
1
1
2
+ −
| 1 | = | 1 1 ⋅ 2 2 |
2 + 2 − 2
2
2
− + −
= | 1 2 1 2 2 1 1 2 |
2
+ 2 2
2
+ −
= | 1 2 1 2 + 2 1 1 2 |
2
2
+ 2 2 + 2 2
2
2
2 2
+ −
= √( 1 2 1 2 ) + ( 2 1 1 2 )
2
2
+ 2 2 + 2 2
2
2
2 2
2 2
2 2
2 2
+ 2 + + − 2 +
1 2 1 2
1 2 1 2
= √ 1 2 2 1 2 1 1 2
2
2 2
( + )
2
2
2 2
2 2
2 2
2 2
( + )+( + )
1 2
2 1
1 2
1 2
= √ 2
2
2
( + )
2
2
2
2
2
2
2
2
( + ) + ( + )
2
2
1
= √ 2 1 2 2
2
2
( + )
2
2
2
2
2
2
( + ) ( + )
= √ 1 1 ⋅ 2 2
2
2
2 2
2 2
( + ) ( + )
2
2
2
2
2
√ + 2
1
1
=
2
√ + 2
2
2
| |
1
=
| |
2
Latihan nomor 1,3,4,5.
2) Akar pangkat dari bilangan satuan

Jika = 1 maka:
cos 2 sin 2
= + dimana = {1,2,3, … , }

= {1,2,3, … , }
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