Question

Sa se calculeze cos(a+b), cos(a-b), daca b) tga =3/4 si cos b=5/13

Answer 1

tg a=3/4
sin a/cos a=3/4
Ridicand la patrat obtinem:
sin²a/cos²a=9/16
Pe sin²a il putem scrie ca: 1-cos²a.
1-cos²a/cos²a=9/16
1/cos²a-cos²a/cos²a=9/16
1/cos²a=9/16+1
1/cos²a=9/16+16/16
1/cos²a=25/16
25cos²a=16
cos²a=16/25
cos a=4/5
sin²a=1-cos²a
sin²a=1-16/25
sin²a=9/25
sin a=3/5

cos b=5/13
sin²b=1-cos²b
sin²b=1-25/169
sin²b=169-25/169
sin²b=144/169
sin b=12/13

cos (a+b)=cos a*cos b-sina*sinb=4/5*5/13 - 3/5*12/13=4*5/5*13 - 3*12/5*13=20-36/65=-16/65

cos (a-b)=cosa*cosb+sina*sinb=4/5*5/13 + 3/5*12/13=4*5/5*13 + 3*12/5*13=20+36/65=56/25