Featured 1 week ago

The maximum and minimum of any

Multiply these by

The maximum is therefore

To find at which point this occurs, work backwards.

Divide both sides by

Do a backwards

Now we have our

Featured 1 week ago

Use the double angle formula for cosine to expand

Let

Featured 1 month ago

Here is the graph of the function (it's not drawn to scale):

I will explain how I graph it below:

Your given equation is

I would write the given equation in the standard form according to

You get:

This way, it is easier to define your amplitude, phase shift, vertical shift, and period multiplier.

In order to graph, we need to know the following:

**Amplitude**, which is#|A|# **Period Multiplier**, which is**B****Phase Shift**, which is**C****Vertical Shift,**which is**D**- Period, which is
#Period = (2pi)/B#

Let's get all our required values from our equation:

**Amplitude**=#|5|# =**5****Period Multiplier**= B =**1****Phase Shift**=#-pi# =#pi# units to the right**Vertical Shift**=#+1# =**1 units up****Period**=#Period = (2pi)/1# =#2pi#

It is important to note that B is not apparent in this equation because it is equal to 1. The actual equation should really be

Our period multiplier is important for determining our period, and it also affects our phase shift. However, we do not have to worry about it for this equation since it's just equal to 1.

Since we have both an amplitude and a vertical shift, we get new minimum and maximum y values for the graph:

**Maximum Y Value** = Vertical Shift + Amplitude =

**Minimum Y Value** = Vertical Shift - Amplitude =

Most importantly, this graph is a positive cosine graph because our equation is 5cos.. and not -5cos.. . A positive cosine graph always starts at the maximum value.

Also, our vertical shift is +1, so our new x-axis start has shifted up by 1. We are starting at y = 1 and not y = 0.

As with our phase shift, it is

So.. with all this information, we can now actually graph the equation of

Here are some visual cues.

My art is pretty bad, sographing the graph on a website should be much more different.

I hope this helps.

Featured yesterday

Taking

So

Featured 23 hours ago

Now we know from this handy "CAST" nmemonic that cos is negative in Q2+Q3:

We can even draw the triangles:

So in Q1:

The corresponding angle in Q2 is

The corresponding angle in Q3 is

Now because it is

After spinning round a full

The second corresponding angle in Q2 is

Featured 23 hours ago

I like getting rid of the phase shift (the

#cos(A + B) = cosAcosB - sinAsinB# .

We have:

#y = 3(cosxcos(pi) - sinxsinpi) - 3#

#y = 3(cosx(-1) - 0) - 3#

#y = -3cosx - 3#

Now you need a little bit of knowledge on the basic cosine function,

graph{y = cosx [-10, 10, -5, 5]}

Whenever there is a coefficient

In the graph of

The

Finally, the

graph{y = -3cosx - 3 [-10, 10, -5, 5]}

Hopefully this helps!

Featured 23 hours ago

Period of

#f(x) = a \sin(bx+c) + d# ,

where

Ignoring the shifts along axes i.e. assuming

#f(x) = y=1+4 \sin(2x)#

#\Rightarrow y -1 = 4 \sin(2x)#

#\Rightarrow g(x) = 4 \sin(2x)#

We know from fundamental trigonometric ideas that the period of sine function is

#\sin(x) = \sin(2n \pi +x)# ,

Let,

#\sin(u) = \sin(2n \pi + u)#

Then for

#\sin(2x) = \sin(2n \pi +2x)#

Multiplying both sides by

#4 \sin(2x) = 4\sin 2(n \pi + x)#

#\equiv g(x) = g(n\pi+x)#

Hence period is

In general the amplitude of sine function is given as;

If

graph{y=\sin(x)[-3,3,-1.5,1.5]}

Our equation is a sort of

Which becomes;

#y= a[\sin(bx)]#

And that expands the graph of ordinary sine function i.e.

graph{y=4\sin(2x)}

Therefore our amplitude is

The graph of our real function

graph{y=1+4 \sin(2x)}

But amplitude remains same.

Featured 23 hours ago

(a)

[Taking

Where

(b)

(i)

The value of

So

(ii)Here

Featured yesterday

Here is the graph of

Step 1: Sketch the graph of

Step 2: Sketch the line

Step 3: Reflect the graph of

Featured yesterday

Because

#csc(2arctan(3/4))=1/sin(2arctan(3/4))#

Using the double angle identity

#=1/(2color(blue)(sin(arctan(3/4)))color(red)(cos(arctan(3/4)))#

We can find the values of

Note that when

We then see that:

#color(blue)(sin(arctan(3/4)))=sin(theta)="opposite"/"hypotenuse"=3/5#

#color(red)(cos(arctan(3/4)))=cos(theta)="adjacent"/"hypotenuse"=4/5#

So the original expression is:

#=1/(2color(blue)((3/5))color(red)((4/5)))#

#=25/24#