from the given function #f(x)=sin^2 x - sin 2x# and the given value of abscissa #x=pi/4#, find the value of y.
look for the value of #f(pi/4)#
#f(pi/4)=(sin (pi/4))^2 - sin 2(pi/4)=(sin (pi/4))^2 - sin (pi/2)#
it follows;
#f(pi/4)=(1/sqrt(2))^2-1#
and #f(pi/4)=1/2-1#
#f(pi/4)=-1/2#
the required point is at #(pi/4, -1/2)#
The slope m is now required to solve for the tangent line . Find the first derivative of the given function #f(x)# and then evaluate it at #x=pi/4# to obtain the value of m.
The first derivative of #f(x)#:
#f(x)=sin^2 x-sin 2x#
#f '(x)=2*sin x*cos x*d/dx(x) - cos 2x*d/dx(2x)#
#f '(x)=2*sin x*cos x - cos 2x*2#
#f' (x)=sin 2x-2*cos 2x#
compute slope #m# at #x=pi/4#
#m=f' (pi/4)=sin 2(pi/4)-2*cos 2(pi/4)#
#m=f' (pi/4)=sin (pi/2) - 2* cos (pi/2)#
#m=f' (pi/4) = 1-2(0)#
#m= 1#
Solve for the tangent liine using the POINT-SLOPE form:
#y-y_1=m*(x-x_1)#
#y-(-1/2)=1*(x-pi/4)#
#y+1/2=x-pi/4#
#y=x-pi/4-1/2#
#y=x-pi/4-2/4#
#y=x-(pi+2)/4# The required Tangent Line.
