Evaluate a using the halfangle formula and b using the power

Evaluate (a) using the half-angle formula, and (b) using the power-reducing formula. Simplify your answers. tan (-17 p i/12) tan^2 9 pi/8

Solution

a) tan(-17pi/12) = -tan(17pi/12) = -tan( pi + 5pi/12)

= -tan(5pi/12)

= -tan(pi -pi/12) = tan(pi/12)

Now sin(pi12) = sqrt{( 1-cospi/6)/2} = sqrt{ 1- sqrt3/2)/2}

= sqrt(2-sqrt3)/2

We know :cosx = sqrt( cos2x +1)/2

cos(pi/12) = sqrt{ (1+cospi/6)/2 } =  sqrt{ 1+sqrt3/2)/2}

tan(pi/12) = sin(pi/12)/cos(pi/12) = sqrt( 2-sqrt3)/(sqrt(2+sqrt3)

b) tan^2(9pi/8)

= sec^2(9pi/8) -1

sec(9pi/8) = 1/cos(9pi/8)

cos9pi/8 = cos(pi +pi/8) = -cospi/8

use the formula : cosx = sqrt( cos2x +1)/2

cospi/8 = sqrt{( cospi/4 +1)/2}

=sqrt{(1/sqrt2 +1)/2 }

sqrt{(1+sqrt2)/2sqrt2}

sec pi/8 = sqrt{2sqrt2/(1+sqrt2)}

tan^2(9pi/8)

= sec^2(9pi/8) -1

= 2sqrt2/(1+sqrt2) -1

= (2sqrt2 -1 -sqrt2)/(1+sqrt2)

= (sqrt2-1)/(1+sqrt2)