Jika diketahui cos α=54​ dan cos β=1312​, hitunglah bentuk berikut : a. cos (2π​−α), untuk 0<α< 2π​  b. cos (2π​+α), untuk 0<α< 2π​

Pertanyaan

Jika diketahui cos space alpha equals 4 over 5 dan cos space beta equals 12 over 13, hitunglah bentuk berikut :

acos space open parentheses pi over 2 minus alpha close parentheses comma space text untuk end text space 0 less than alpha less than space pi over 2 space

b. cos space open parentheses pi over 2 plus alpha close parentheses comma space text untuk end text space 0 less than alpha less than space pi over 2 space

A. Salim

Master Teacher

Mahasiswa/Alumni Universitas Pelita Harapan

Jawaban terverifikasi

Jawaban

 table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses straight pi over 2 plus alpha close parentheses end cell equals cell negative 3 over 5 end cell end table.

Pembahasan

Ingat :

1. Rumus pythagoras pada segitiga siku-siku

z squared equals x squared plus y squared

2. Perbandingan trigonometri pada segitiga siku-siku

sin space theta equals y over z cos space theta equals x over z

3. Rumus kosinus selisih dua sudut

cos space open parentheses alpha minus beta close parentheses equals cos space alpha space cos space beta plus sin space alpha space sin space beta

4. Rumus kosinus jumlah dua sudut

cos space open parentheses alpha plus beta close parentheses equals cos space alpha space cos space beta minus sin space alpha space sin space beta

5. cos space straight pi over 2 equals 0 

6. sin space straight pi over 2 equals 1

Dari soal diketahui cos space alpha equals 4 over 5 dan cos space beta equals 12 over 13.

Berdasarkan rumus pythagoras maka :

table attributes columnalign right center left columnspacing 0px end attributes row cell z squared end cell equals cell x squared plus y squared end cell row cell y squared end cell equals cell z squared minus x squared end cell row cell y squared end cell equals cell 5 squared minus 4 squared end cell row cell y squared end cell equals cell 25 minus 16 equals 9 end cell row y equals cell square root of 9 equals 3 end cell end table

Berdasarkan konsep di atas yaitu perbandingan trigonometri pada segitiga siku-siku maka :

sin space alpha equals y over z sin space alpha equals 3 over 5

Berdasarkan rumus pythagoras maka :

table attributes columnalign right center left columnspacing 0px end attributes row cell z squared end cell equals cell x squared plus y squared end cell row cell y squared end cell equals cell z squared minus x squared end cell row cell y squared end cell equals cell 13 squared minus 12 squared end cell row cell y squared end cell equals cell 169 minus 144 equals 25 end cell row y equals cell square root of 25 equals 5 end cell end table

Berdasarkan konsep di atas yaitu perbandingan trigonometri pada segitiga siku-siku maka :

sin space beta equals y over z sin space beta equals 5 over 13

acos space open parentheses pi over 2 minus alpha close parentheses comma space text untuk end text space 0 less than alpha less than space pi over 2 space

Berdasarkan konsep kosinus selisih dua sudut :

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses alpha minus beta close parentheses end cell equals cell cos space alpha space cos space beta plus sin space alpha space sin space beta end cell row cell cos space open parentheses straight pi over 2 minus alpha close parentheses end cell equals cell cos space straight pi over 2 space cos space alpha plus sin space straight pi over 2 space sin space alpha end cell row blank equals cell open parentheses 0 cross times 4 over 5 close parentheses plus open parentheses 1 cross times 3 over 5 close parentheses equals 3 over 5 end cell end table

Dengan demikian, table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses straight pi over 2 minus alpha close parentheses end cell equals cell 3 over 5 end cell end table.

b. cos space open parentheses pi over 2 plus alpha close parentheses comma space text untuk end text space 0 less than alpha less than space pi over 2 space

Berdasarkan konsep kosinus selisih dua sudut :

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses alpha plus beta close parentheses end cell equals cell cos space alpha space cos space beta minus sin space alpha space sin space beta end cell row cell cos space open parentheses straight pi over 2 plus alpha close parentheses end cell equals cell cos space straight pi over 2 space cos space alpha minus sin space straight pi over 2 space sin space alpha end cell row blank equals cell open parentheses 0 cross times 4 over 5 close parentheses minus open parentheses 1 cross times 3 over 5 close parentheses equals negative 3 over 5 end cell end table

Dengan demikian, table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses straight pi over 2 plus alpha close parentheses end cell equals cell negative 3 over 5 end cell end table.

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