Finding Relative Extrema In Exercises 53-62, find the relative extrema of the trigonometric function in the interval ( 0 ,   2 π ) . Use a graphing utility to confirm your results.

See Examples 6 and 7.

y = 2 sin x + sin 2 x

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

Given Information:

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

Formula used:

Sine derivative Rule:

ddx[sinu]=cosududx

General power derivative rule:

ddxxn=nxn−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=2sinx+sin2x

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

dydx=ddx[2sinx+sin2x]y′(x)=2ddx[sinx]+ddx[sin2x]=2cosx+cos2xddx[2x]=2cosx+cos2x[2]

Further solve.

y′(x)=2cosx+2cos2x

Apply dydx=0 in order to find critical points.

2cosx+2cos2x=02cosx+2(2cos2x−1)=04cos2x+2cosx−2=02(2cos2x+cosx−1)=0

Further solve.

2cos2x+2cosx−cosx−1=02cosx(cosx+1)−1(cosx+1)=0(2cosx−1)(cosx+1)=0

Further solve.

2cosx−1=02cosx=1cosx=12

And,

cosx+1=0cosx=−1

Since, the value of the cosθ is 12 at θ=π3 and θ=5π3, and value of cosθ is −1 at θ=π over the interval [0,2π]. So,

cosx=12cosx=cosπ3x=π3,

cosx=12cosx=cos5π3x=5π3,

And,

cosx=−1cosx=cosπx=π

Thus, critical points are x=π3, x=π and x=5π3.

Now, rewrite the first derivative of the function.

y′(x)=2cosx+2cos2x

At x=π6.

y′(π6)=2cosπ6+2cos2(π6)=2.732=+ve>0

At x=5π6.

y′(5π6)=2cos5π6+2cos2(5π6)=−0.732=−ve<0

At x=4π3.

y′(4π3)=2cos4π3+2cos2(4π3)=−2=−ve<0

At x=11π6