Featured 1 month 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 1 month 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 1 month 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 1 month ago

(a)

[Taking

Where

(b)

(i)

The value of

So

(ii)Here

Featured yesterday

Given:

Use this reference for Fundamental Frequency

Let

Using the fact that

The fundamental frequency is the greatest common divisor of the two frequencies:

Here is a graph:

graph{y = sin(4x) - cos(7x) [-10, 10, -5, 5]}

Please observe that it repeats every

Featured 1 month ago

Here is the graph of the original equation

Use the identity

Substitute the

Here is the graph of that equation:

graph{sqrt(x^2+y^2)=x^2/(x^2+y^2)-y^2/(x^2+y^2) [-10, 10, -5, 5]}

It is only half of the loops

Square both sides:

Here is the graph of that equation:

graph{x^2+y^2=((x^2-y^2)/(x^2+y^2))^2 [-10, 10, -5, 5]}

This looks like the polar equation.

Featured 4 weeks ago

We can construct this diagram from the given information.

Let length of

It is given that **1.**

Also, observe that **2.**

**We have to find #angle CAE# which will give us elevation of C from A.**

**Let #angle CAE# be #theta#.**

From

**2.**, **3.**

**4.**

**From #triangle BCD# we find:-**

**5.**

**2.**, **6.**

Now,

Substituting values of **3., 4., 5., 6.**

**Dividing the numerator and deniminator by #l_2#**

Substituting value of **1.**

**which is the angle of elevation of #C# from #A#.**

Featured yesterday

Please see the explanation.

Given:

Here is a graph of the Cartesian equation:

Expand the square:

Substitute

Substitute

There is a common factor of

There is a common factor of r in the next 2 terms:

The above is a quadratic equation in standard form where:

r in the independent variable

We use the identity #cos^2(theta) + sin^2(theta) = 1 to simplify a:

We can use the quadratic formula to write r and a function of

Substitute for every a:

Substitute for c:

Substitute for every b:

Here is a graph of the polar equation:

The graphs are identical, therefore, the conversion is complete.

Note: I am sure that the expression under the radical can be simplified but I will leave that to you.

Featured yesterday

The given function representing the height

(a) h is a cosine function of t, So it will have maximum value when

and the maximum height of the swing becomes

(b) it takes

c) The minimum height of the swing will be achieved when

Minimum height

(d). Again

Here

e) For

when

So in

f) The height of the swing at 10 s can be had by inserting

Featured yesterday

Strategy: Rewrite this equation as a quadratic equation using

Step 1. Rewrite this equation as a quadratic using

You are given

Replace

Step 2. Factor the quadratic equation.

Solving gives us

Step 3. Replace

These answer work because we were asked to find the solutions in

However, the graph of

Which shows **four** solutions in the interval

We must find the other solutions by recognizing that sine functions are periodic with respect to

So,

The other two solutions are