Question

How do you use half angle identities and the fact that sin(pi/6)=1/2 to find cos(-pi/12) and sin (-pi/12)?

Answer 1

Please see below.

Explanation:

We have , `sin(pi/6)=sin30^circ=1/2`

Now , `theta=pi/12=15^circto1^(st)Quadrant`

`"Using"color(blue)" half angle formula :"`

`color(blue)(sin^2 (theta/2)=(1-costheta)/2=(1-sqrt(1-sin^2theta))/2`

Let , `theta=30^circ=>theta/2=15^circto1^(st)Quadrant`

`:.sin^2 15^circ=(1-sqrt(1-sin^2 30^circ))/2`

`:.sin^2 15^circ=(1-sqrt(1-1/4))/2`

`:.sin^2 15^circ=(1-sqrt3/2)/2=(2-sqrt3)/4=(4-2sqrt3)/8`

`:.sin^2 15^circ=(3+1-2sqrt3)/8`=`((sqrt3)^2-2(sqrt3)(sqrt1)+(sqrt1)^2)/8`

`:.sin^2 15^circ=(sqrt3-1)^2/(4xx2)`

`:.sin 15^circ=(sqrt3-1)/(2sqrt2)......to[because1^(st)Quadrant]`

`:.sin 15^circ=(sqrt3-1)/(2sqrt2) xxsqrt2/sqrt2=(sqrt6-sqrt2)/4`

`color(green)(-pi/12=-15^circtoIV^(th)Quadrant =>sin(-pi/12) < 0`

`:.color(red)(sin(-pi/12)=-sin(pi/12)=-(sqrt6-sqrt2)/4`

For `cos(-pi/12)` ,please see next answer.

Answer 2

Please see below.

Explanation:

We have , `sin(pi/6)=sin30^circ=1/2`

Now , `theta=pi/12=15^circto1^(st)Quadrant`

`"Using"color(blue)" half angle formula :"`

`color(blue)(cos^2 (theta/2)=(1+costheta)/2=(1+sqrt(1-sin^2theta))/2`

Let , `theta=30^circ=>theta/2=15^circto1^(st)Quadrant`

`:.cos^2 15^circ=(1+sqrt(1-sin^2 30^circ))/2`

`:.cos^2 15^circ=(1+sqrt(1-1/4))/2`

`:.cos^2 15^circ=(1+sqrt3/2)/2=(2+sqrt3)/4=(4+2sqrt3)/8`

`:.cos^2 15^circ`=`(3+1+2sqrt3)/8`=`((sqrt3)^2+2(sqrt3)(sqrt1)+(sqrt1)^2)/8`

`:.cos^2 15^circ=(sqrt3+1)^2/(4xx2)`

`:.cos15^circ=(sqrt3+1)/(2sqrt2)......to[because1^(st)Quadrant]`

`:.cos15^circ=(sqrt3+1)/(2sqrt2) xxsqrt2/sqrt2=(sqrt6+sqrt2)/4`

`color(green)(-pi/12=-15^circtoIV^(th)Quadrant =>cos(-pi/12) > 0`

`:.color(red)(cos(-pi/12)=+cos(pi/12)=(sqrt6+sqrt2)/4`

For `sin(-pi/12)` , please see next answer.