if cos 45°=1/root2, find the exact value of a.cos 315° b.sin 270° c. sin 210° d. tan 210°?

Emmyrosh

4 years ago

cos45=1/√2; sin45=√(1-cos^2(45))=√(1-1/2)=√(1/2)=1/√2=cos45;

sin90=sin(2*45)=2sin45cos45=2*1/√2*1/√2=1; cos90=cos(2*45)=cos45cos45-sin45sin45=1/2-1/2=0;

sin180=sin(2*90)=2sin90cos90=0; cos180=cos(2*90)=cos90cos90-sin90sin90=0-1=-1;

cos(X)=cos(-X), so cos45=cos(-45)=cos(360-45)=cos(315)=1√2.

sin270=sin(180+90)=sin180cos90+cos180sin90=0-1=-1.

sin3X=sin(2X+X)=sin2XcosX+cos2XsinX=2sinXcos^2(X)+cos^2(X)sin(X)-sin^3(X), so if 3X=90:

1=2sin30(1-sin^2(30))+(1-sin^2(30))sin(30)-sin^3(30)=2sin30-2sin^3(30)+sin(30)-sin^3(30)-sin^3(30)=

3sin30-4sin^(30); so 4sin^3(30)-3sin30+1=0=(2sin30-1)^2(sin30+1), so sin30=1/2 or -1; but sin270=-1 so sin30=1/2.

cos30=√(1-sin^2(30))=√(3/4)=√3/2. (3*270=810=2*360+90; sin810=sin90=1; hence the -1 solution of the cubic.)

sin210=sin(180+30)=sin180cos30+cos180sin30=0-1/2=-1/2.

cos210=cos(180+30)=cos180cos30-sin180sin30=-√3/2; tan210=sin210/cos210=(-1/2)/-√3/2)=1/√3.