Remember .
I use the quotient rule to differentiate .
(1)
If then from equation
(2)
(3)
Remember the Pythagorean identity
(4)
Hence
(5)
Remember .
I use the quotient rule to differentiate .
(1)
If then from equation
(2)
(3)
Remember the Pythagorean identity
(4)
Hence
(5)
In the last post we looked at two trig limits:
(1)
(2)
We are going to use these two limits to differentiate sine and cosine functions from first principals.
Use the trig identity
Evaluate the limits
Hence, .
Now we are going to do the same for .
Use the trigonometric identity
Evaluate the limits
Hence
Filed under Calculus, Differentiation, Identities, Trigonometry, Year 12 Mathematical Methods
Remember , hence
and the co-ordinate of
is
.
, hence
and the co-ordinate of
is
And from the definition of we know
is the point
Consider the areas of triangle , sector
, and triangle
.
We know from inspection of the above diagram that
Area Area
Area
Which means,
We can ignore all of the halves.
Remember
Divide everything by (as we are in the first quadrant we know
, so we don’t need to worry about the inequality)
Invert everything and change the direction of the inequalities)
I am going to rewrite it as follows
because I like to use less thans rather than greater thans.
Now what happens as tends to
?
Hence by the squeeze theorem
Now we know this limit, we are going to use it to find
Multiply by
If we evaluate the limits,
Hence,
In the next post we are going to use these limits to differentiate sine and cosine functions.
, find
.
I came across this sum in An Imaginary Tale by Nahin and I was fascinated.
Let and
.
Remember![]() Hence, ![]() ![]() ![]() Therefore, ![]() and ![]() ![]() ![]() |
Which means,
Let’s try a few partial sums
Hence,
What happens as ?
Because we know is undefined.
Filed under Identities, Interesting Mathematics, Puzzles, Sequences, Trigonometry
Remember
(1)
We know that for odd integer multiples of
, i.e.
, which is
for
Hence,
for
We can factorise our expansion
We know
We are going to use De Moivre’s theorem to prove trigonometric identities.
Remember, De Moivre’s Theorem
If , then
Or a shorter version , then
Now, let , find
Remember and
It is the same for
Prove ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
We can do something similar with sine.
Hence
Prove ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
Let’s find an identity for
And ?
Filed under Complex Numbers, Identities, Trig Identities, Trigonometry
I worked on this question with one of my students (I don’t know where it is from).
Mike leaves the rose bush he was examining and walks 35m in the direction S ![]() From there he walks 70m towards a rotunda. Mike is now 100m from the rose bush. Find the bearing of the rotunda from the pond. |
Let’s try to draw a diagram
Because we don’t the direction Mike walked from the pond, I have drawn a circle with radius 70m centred at the pond.
We know Mike is now 100m from the rose bush. As we don’t know the direction, I have drawn another circle with radius 100m centred at the rose bush. Where the two circles intersect are the possible locations of the rotunda.
Use the cosine rule to find the angle
Using the fact that alternate angles in parallel lines are congruent, we can see
that the bearing from the pond to the rotunda is
It is the same triangle, so
This time the bearing is
Hence, the two possible bearings of the rotunda from the pond are or
.
Filed under Bearings, Non-Right Trigonometry, Trigonometry
As I have done a cosine rule proof, I thought I should also do a sine rule proof.
From the above diagram, we can find in two ways.
Set equation 1 equal to equation 2
or
We could have put the altitude of the triangle from vertex A
Following the same process as above
Set equation 3 equal to equation 4.
Now therefore
or
Filed under Non-Right Trigonometry, Right Trigonometry, Trigonometry
From the diagram above
(1)
From the diagram above, we can see
(2)
Substitute equation 2 into equation 1
It can also be handy to have the angle version
Filed under Non-Right Trigonometry, Right Trigonometry, Trigonometry, Uncategorized