1.已知cos(75度+α)等于三分之一,α是第三象限角,求sin(105度-α)+cos(α-105度)2.在三角形ABC中,sinA等于五分之三,cosB等于十三分之五,求cosC的值

Question

1.已知cos(75度+α)等于三分之一,α是第三象限角,求sin(105度-α)+cos(α-105度)
2.在三角形ABC中,sinA等于五分之三,cosB等于十三分之五,求cosC的值

Answer 1

1.
sin(105-a)+cos(a-105)
=sin(180-(75+a))+cos((a+75)-180)
=sin180 cos(75+a)-cos180 sin(75+a)+cos(a+75)cos180+sin(a+75)sin180
=(0)cos(75+a)-(-1)sin(75+a)+(-1)cos(a+75)+(0)sin(a+75)
=sin(75+a)-cos(a+75)
=√(1-cos^2(75+a))-cos(a+75)
=√(1-1/9)-1/3
=(√(8)-3)/9
2.
cosC
=cos(180-(A+B) )
=cos180 cos(A+B)+sin180 sin(A+B)
=-cos(A+B)
=-cosA cosB+sinA sinB
=-√(1-sin^2 A) cosB+sinA √(1-cos^2 B)
=-√(1-(3/5)^2) (5/13)+(3/5)√(1-(5/13)^2)
=-(4/5)(5/13)+(3/5)(12/13)
=-16/65