# y= sin^2  x – 2cos^2  x  Analyze the function using graph sketching techniques(listed below) and then sketch the graph  Hint: Maybe show first that the function can be converted to a linear...

y= sin^2  x – 2cos^2  x

Analyze the function using graph sketching techniques(listed below) and then sketch the graph

Hint: Maybe show first that the function can be converted to a linear expression with cos(2x). You may use these two formulas: sin^2  x= (1- cos 2x)/2 and  cos^2  x= (1+ cos 2x)/2.

The graph sketching techniques that need to be used to analyze the function & sketch a graph.

1. Domain. restricting your attention only to values of x for which f(x) is defined.
2. Intercepts. Find intersections of the graph with the axes. The Y-intercept is (0, f(0)). To obtain the X-intercept (c, 0) is harder and sometimes impossible. You need to find c such that f(c)= 0
3. Symmetry. If f(x)=f(-x) , for example cos x = cos (-x), then the graph is symmetric with respect to the Y-axis. If f(x)= -f(-x), for example sin x= - sin (-x), then the graph is symmetric with respect to the origin.
4. Asymptotes. Look for infinite limits and limits at infinity.
5. Intervals of increase or decrease and local maxima and minima Reminder: Solve f’(x)> 0 and f’(x)<0 for monotonicity.
6. Convavity and points of inflection Reminder: If not too hard, solve f’’(x)> 0 and f’’ (x)< 0.

Sketch the curve Use all information gathered above to mark first points where the graph changes its behavior (intercepts, a minimum, a maximum, an inflection point) and then connect the points.

embizze | High School Teacher | (Level 1) Educator Emeritus

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Graph `y=sin^2x-2cos^2x` :

First we can rewrite in terms of `cos2x` : Use `cos2x=2cos^2x-1` ; also we can solve for `cos^2x` so that `cos^2x=(cos2x+1)/2`

`sin^2x-2cos^2x=-(2cos^2x-sin^2x)`

`=-(2cos^2x-(1-cos^2x))` using the Pythagorean identity

`=-(cos2x+cos^2x)`

`=-(3/2cos2x+1/2)`

`=-3/2cos2x-1/2`

So we are looking at `y=-3/2cos2x-1/2`

(1) The domain is all real numbers. The range is `-2<= y <= 1` . (This is a sinusoid of the form `y=AcosB(x-h)+k` : `A=-3/2` indicates that the cos function is reflected over the x-axis with an amplitude of `-3/2` , thus `("max"-"min")/2="Amplitude"` . The midline is determined by `k=-1/2` , so the maximum is `-1/2+3/2=1` and the minimum is `-1/2-3/2=-2` . Since `h=0` the function is not translated horizontally. And since `B=2` we have the period is `(2pi)/2=pi` )

(2) The y-intercept is at `-3/2cos(0)-1/2=-3/2-1/2=-2` .

To find the x-intercepts:

`-3/2cos2x-1/2=0`

`cos2x=-1/3`

`2cos^2x-1=-1/3`

`cos^2x=1/3`

`cosx=+-sqrt(3)/3`

`x=cos^(-1)sqrt(3)/3+npi~~.955+npi` or `x=cos^(-1)(-sqrt(3)/3)+npi~~2.186+npi` (Where `n in ZZ` )

** Notice we use the period of `pi` to find all intercepts **

(3) Since `cos(a)=cos(-a)` the function is symmetric about the y-axis. The function is not symmetric about the origin

(4) There are no asymptotes. There is no limit as the function is periodic.

(5) `y=-3/2cos2x-1/2`

`y'=-3/2(-sin2x)(2)=3sin(2x)`

Setting the first derivative equal to zero we find:

`3sin2x=0==>sin2x=0==>x=0+npi,pi/2+npi` .

`3sin2x>0` on `(0,pi/2)` so the function is increasing on `(0+npi,pi/2+npi)`

`3sin2x<0` on `(pi/2,pi)` so the function is decreasing on `(pi/2+npi,pi+npi)`

By the first derivative test, the function has maxima at `pi/2+npi` and minima at `npi`

(6) `y'=3sin(2x)` so `y''=3(cos2x)(2)=6cos2x`

`6cos2x=0==>x=pi/4+npi,(3pi)/4+npi` Points of inflection

`6cos2x>0` on `(0+npi,pi/4+npi),((3pi)/4+npi,pi+npi)` Concave up

`6cos2x<0` on `(pi/4+npi,(3pi)/4+npi),((5pi)/4+npi,(7pi)/4+npi)` Concave down

The graph:

`sin^2x-2cos^2x:`

`y=-3/2cos2x-1/2:`

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