Difference between revisions of "Proofs of trig identities"
Line 288: | Line 288: | ||
Then, <math>u=\frac{-4y\pm\sqrt{16y^2-1}}{8}</math> | Then, <math>u=\frac{-4y\pm\sqrt{16y^2-1}}{8}</math> | ||
− | The solutions are <math>\sqrt[3]{4y+\sqrt{16y^2-1}}</math>, <math>\frac{\sqrt[3]{4y+\sqrt{16y^2-1}}+\sqrt[3]{4y-\sqrt{16y^2-1}}}{2}</math>, and <math>\sqrt[3]{4y-\sqrt{16y^2-1}}</math>. | + | The solutions are <math>\sqrt[3]{-4y+\sqrt{16y^2-1}}</math>, <math>\frac{\sqrt[3]{-4y+\sqrt{16y^2-1}}+\sqrt[3]{4y-\sqrt{16y^2-1}}}{2}</math>, and <math>\sqrt[3]{-4y-\sqrt{16y^2-1}}</math>. |
A tiny adjustment gives us the cosine third-angle formulas: | A tiny adjustment gives us the cosine third-angle formulas: | ||
Line 374: | Line 374: | ||
<math>\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}</math> | <math>\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}</math> | ||
+ | ==Thirds== | ||
+ | <math>\sin\theta=\sqrt[3]{-4\sin 3\theta+\sqrt{16\sin^2(3\theta)-1}},\frac{\sqrt[3]{-4\sin 3\theta+\sqrt{16\sin^2(3\theta)-1}}+\sqrt[3]{-4\sin 3\theta-\sqrt{16\sin^2(3\theta)-1}}}{2}, \\ \text{or }\sqrt[3]{4\sin 3\theta-\sqrt{16\sin^2 3\theta-1}}</math> | ||
+ | |||
+ | <math>\cos\theta=\sqrt[3]{4\cos 3\theta+\sqrt{16\cos^2(3\theta)-1}},\frac{\sqrt[3]{4\cos 3\theta+\sqrt{16\cos^2(3\theta)-1}}+\sqrt[3]{4\cos 3\theta-\sqrt{16\cos^2(3\theta)-1}}}{2}, \\ \text{or }\sqrt[3]{4\cos 3\theta-\sqrt{16\cos^2 3\theta-1}}</math> | ||
+ | |||
+ | <math>\tan\theta=\frac{\sqrt[3]{-4\sin 3\theta+\sqrt{16\sin^2(3\theta)-1}}}{\sqrt[3]{4\cos 3\theta+\sqrt{16\cos^2(3\theta)-1}}},\frac{\sqrt[3]{-4\sin 3\theta+\sqrt{16\sin^2(3\theta)-1}}+\sqrt[3]{-4\sin 3\theta-\sqrt{16\sin^2(3\theta)-1}}}{\sqrt[3]{4\cos 3\theta+\sqrt{16\cos^2(3\theta)-1}}+\sqrt[3]{4\cos 3\theta-\sqrt{16\cos^2(3\theta)-1}}}, \\ \text{or }\frac{\sqrt[3]{-4\sin 3\theta-\sqrt{16\sin(3\theta)^2-1}}}{\sqrt[3]{4\cos 3\theta-\sqrt{16\cos^2(3\theta)-1}}}</math> |
Revision as of 21:53, 20 January 2024
Contents
Introduction
and
are easy to define. I prefer the unit circle definition as it makes these proofs easier to understand.
Next, we define some other functions:
Note: I've omitted because it's unnecessary and might clog things up a little.
With a bit of ingenuity, we can create the following diagram:
We can note that the functions are correct by similar triangles.
Symmetric identities
If we draw a few copies of the triangle, we get:
The other three can be derived by taking the reciprocals of these three.
Pythagorean identities
Pythagorean identities are easy and there's no algebra involved. In fact, the name Pythagorean is a giveaway of what we should do!
The proof here is very straightforward. We use the pythagorean theorem on giving us
or
.
Same story here. Applying pythagorean to gives us
or
.
Same. Pythagorean on gives
or
.
Conclusion
Even though with the first one and the definitions, we can make the rest from algebra, having a geometric meaning is nice when we want to know what it actually means.
Angle addition and subtraction
where
The diagram illustrates the identities nicely.
The diagram shows the height of point is
.
However, the length of
is
. To compensate, we must divide by
to make it the sine. After some *easy* algebra, we arrive at
.
The diagram says that it is , but we need to divide by
again. We arrive at
.
This time we can't get it from our diagram. We need to go back to the original definition of tangent. This is the summary of my algebra:
Double angle formulas
This is a breeze. Just sub in for sum:
Variations
Since , we can edit the double angle cosine formula a bit. Here are the three most helpful variants:
We can also solve for other expressions:
Sum to Product to Sum
Our angle addition formulas look nasty. Let's try to cancel something.
Chapter 1
Let's start with the formula .
Both terms are symmetrical, let's try to cancel out the first one.
and
might give us an idea.
Therefore, .
To put this in a nicer form, where:
Chapter 2
We did all we could. Now let's try doing something to the other formula: . Let's try to cancel the first one first.
is a lot easier to cancel than
so we have to subtract.
So
Or, (same conversions of
)
Chapter 3
Next: . Now let's cancel the second one.
So
Or, (same conversions of
)
Bonus: Product identity
This is a special identity. I hope this helps you.
and
There's something we can cancel.
If , then it simplifies to
Notice . If we let
:
Halved angles
Starting with the identities from the double section:
We take the square root to obtain:
For tangent:
There are two nice variations to know.
Triple angles and more
Triple sums
Triple angles
Third angles
Let and
. We get this depressed cubic:
First, divide both sides by -4 and rearrange: . The discriminant
Then,
The solutions are ,
, and
.
A tiny adjustment gives us the cosine third-angle formulas:
,
, and
.
For tangent:
,
, and
All identities
Definition
Symmetric
Pythagorean
Sum
Double
Sum
Product
Product
Halves
3 Sums
Triple
Thirds