What is the equation of the tangent line of #f(x)=sinx-cosx/2# at #x=pi/4#?

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Answer 1 · 2017-09-12T12:46:40

The equation of tangent is #y= 3/(2sqrt2)x +1/(2sqrt2) (1- (3pi)/4)#

#f(x) = sinx - 1/2 cosx :. f(pi/4) = sin (pi/4) -1/2 cos (pi/4)# or

# f(pi/4) = 1/sqrt2 -1/2 *1/sqrt2 = 1/(2sqrt2)# . The point at which

tangent equation to be found is # (pi/4, 1/(2sqrt2))#

#f(x) = sinx - 1/2 cosx : f^'(x) = cosx + 1/2 sinx# or

# f^'(pi/4) = cos(pi/4) + 1/2 sin(pi/4) = 1/sqrt2 +1/2*1/sqrt2# or

# f^'(pi/4) = 1/sqrt2(3/2) or 3/(2sqrt2)# , So slope at the point is

# m= 3/(2sqrt2)# The equation of tangent at # (pi/4, 1/(2sqrt2))#

is #y -1/(2sqrt2) = 3/(2sqrt2) (x- pi/4) # or

#y= 3/(2sqrt2)x +1/(2sqrt2) (1- (3pi)/4)# [Ans]