Find the relative extrema of y =   1 2 s i n   2 x +   c o s x on the interval ( 0 ,   2 π ) .

To calculate: The relative extrema of the trigonometric function y=12sin2x+cosx over the interval [0,2π].

Given Information:

The provided trigonometric function is y=12sin2x+cosx over the interval [0,2π].

Formula used:

Sine derivative Rule:

ddx[sinu]=cosududx

Cosine derivative Rule:

ddx[cosu]=−sinududx

General power derivative rule:

ddxxn=nxn−1

Double angle formula for trigonometric identity.

cos2θ=1−2sin2θ=2cos2θ−1

Condition for critical points:

dydx=f′(x)=0

If function f(x) is increasing, then f′(x)>0 and if function is decreasing, then f′(x)<0.

From first derivative Test, if f′(x) changes positive to negative, then function f(x) has relative maxima and, if f′(x) changes negative to positive then function f(x) has relative minima. In order words, if function f(x) changes from increasing function to decreasing at any point critical, then function f(x) will have relative maxima at that point, and if function f(x) changes from decreasing function to increasing at any critical point, then function f(x) will have relative minima at that point.

Calculation:

Consider the provided trigonometric function:

y=12sin2x+cosx

Differentiate the above function with respect to x using Sine, Cosine and Power derivative rules.

dydx=ddx[12sin2x+cosx]y′(x)=12ddx[sin2x]+ddx[cosx]=12(cos2x)ddx[2x]+[−sinx]=12(cos2x)[2]−sinx

Further solve.

y′(x)=cos2x−sinx

Apply dydx=0 in order to find critical points.

cos2x−sinx=01−2sin2x−sinx=0−(−1+2sin2x+sinx)=02sin2x+sinx−1=0

Further solve.

2sin2x+2sinx−sinx−1=02sinx(sinx+1)−1(sinx+1)=0(2sinx−1)(sinx+1)=0

Further solve.

2sinx−1=02sinx=1sinx=12

And,

sinx+1=0sinx=−1

Since, Sine is positive in first and second quadrants, while, sine is negative in third and fourth quadrants. Thus, three are three solutions for y=12sin2x+cosx over the interval (0,2π), which can be mathematically calculated as:

For sinx=−1,

sinx=−1sinx=sin3π2x=3π2,

For sinx=12,

sinx=12sinx=sinπ6x=π6

For sinx=12

sinx=12sinx=sin5π6x=5π6

Thus, critical points are x=π6, x=5π6 and x=3π2