# 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.

**Domain.**restricting your attention only to values of x for which f(x) is defined.**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**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.**Asymptotes.**Look for infinite limits and limits at infinity.**Intervals of increase or decrease and local maxima and minima**Reminder: Solve f’(x)> 0 and f’(x)<0 for monotonicity.**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.

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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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