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Math is amazing.
In this thread, I'll guide you through the double angle identities involving sine and cosine.
I will explain these identities using a "proof without words," similar to this figure.
See below for figure.
Math is amazing.
In this thread, I'll guide you through the double angle identities involving sine and cosine.
I will explain these identities using a "proof without words," similar to this figure.
See below for figure.
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Before we dive into this complicated figure, let's review what sin, cos, and tan even mean!
There are many ways to interpret these expressions. The simplest, most useful idea is that when we scale a triangle without changing its angles, ratios of the sides don't change!
Before we dive into this complicated figure, let's review what sin, cos, and tan even mean!
There are many ways to interpret these expressions. The simplest, most useful idea is that when we scale a triangle without changing its angles, ratios of the sides don't change!
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The ratio of the sides depends solely on the size of one of the angles, namely BAC.
Different values of this angle result in different ratios of the triangle's sides.
So the trig functions are defined as the ratio of the sides for a particular angle:
The ratio of the sides depends solely on the size of one of the angles, namely BAC.
Different values of this angle result in different ratios of the triangle's sides.
So the trig functions are defined as the ratio of the sides for a particular angle:
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To recapitulate, the functions sine, cosine, and tangent represent ratios of sides in a right triangle.
The input to these functions is an ANGLE of the triangle and the output is a RATIO of two sides of the triangle.
The easy way to remember these ratios is: SOHCAHTOA.
To recapitulate, the functions sine, cosine, and tangent represent ratios of sides in a right triangle.
The input to these functions is an ANGLE of the triangle and the output is a RATIO of two sides of the triangle.
The easy way to remember these ratios is: SOHCAHTOA.
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Now that we understand the basic trig functions, we can ask a fundamentally important question:
If we know the value of these functions at a particular angle, can we calculate the value of the function at double that angle?
These are called the double angle formulas.
Now that we understand the basic trig functions, we can ask a fundamentally important question:
If we know the value of these functions at a particular angle, can we calculate the value of the function at double that angle?
These are called the double angle formulas.
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To solve this problem, we will need a very useful fact known as the "Pythagorean identity." It relates the values of the sine and cosine functions using the Pythagorean theorem.
The figure below shows a proof of the Pythagorean identity that hopefully is intuitive.
To solve this problem, we will need a very useful fact known as the "Pythagorean identity." It relates the values of the sine and cosine functions using the Pythagorean theorem.
The figure below shows a proof of the Pythagorean identity that hopefully is intuitive.
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Now we can finally begin creating a diagram that will help to prove the double angle formulas.
Start by drawing a circle with radius 1.
A is the center, and C/B are points on the circle such that CB is a diameter of the circle.
Now we can finally begin creating a diagram that will help to prove the double angle formulas.
Start by drawing a circle with radius 1.
A is the center, and C/B are points on the circle such that CB is a diameter of the circle.
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Let's create a point D such that the angle DBA is theta.
By the well-known inscribed angle theorem, the angle DAC is 2*theta, as you can see in the figure below.
Let's create a point D such that the angle DBA is theta.
By the well-known inscribed angle theorem, the angle DAC is 2*theta, as you can see in the figure below.
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Now, the critical part.
Draw segment DC. BDC is an angle inscribed in a semicircle. Using the inscribed angle theorem, we see that angle BDC is actually 90˚! So BDC is a right triangle, our favorite kind of triangle for trigonometry.
Now, the critical part.
Draw segment DC. BDC is an angle inscribed in a semicircle. Using the inscribed angle theorem, we see that angle BDC is actually 90˚! So BDC is a right triangle, our favorite kind of triangle for trigonometry.
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Next we will do the most useful step in any geometry problem: drop an altitude (F is the foot of this altitude).
We find, using our definitions of sin and cos, that the lengths of DF and AF can be written in terms of sin, cos, and 2 theta.
Next we will do the most useful step in any geometry problem: drop an altitude (F is the foot of this altitude).
We find, using our definitions of sin and cos, that the lengths of DF and AF can be written in terms of sin, cos, and 2 theta.
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We can use this altitude to find CF in terms of cos(2theta):
We found that AF = cos(2theta) in the last step, and since segments AF and CF add to 1, CF is 1 - cos(2theta).
See below for a more rigorous explanation.
We can use this altitude to find CF in terms of cos(2theta):
We found that AF = cos(2theta) in the last step, and since segments AF and CF add to 1, CF is 1 - cos(2theta).
See below for a more rigorous explanation.
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Now we will drop one last altitude from A to BD. This will create two new congruent triangles (see proof below).
We can find the side lengths of these triangles (AE, BE, DE) in terms of cos(theta) and sin(theta).
Now we will drop one last altitude from A to BD. This will create two new congruent triangles (see proof below).
We can find the side lengths of these triangles (AE, BE, DE) in terms of cos(theta) and sin(theta).
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At this point, there is only one side length that we have yet to determine, CD. We can find CD using similar triangles.
We will temporarily simplify the figure to make these similar triangles more apparent (they are AEB and ADC).
Using similarity, we find CD = sin(2theta).
At this point, there is only one side length that we have yet to determine, CD. We can find CD using similar triangles.
We will temporarily simplify the figure to make these similar triangles more apparent (they are AEB and ADC).
Using similarity, we find CD = sin(2theta).
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Now, we have calculated the lengths of all of the sides in our figure. Whew! That was a lot of work.
Let's take a look at the elegant, completed diagram.
Now, we have calculated the lengths of all of the sides in our figure. Whew! That was a lot of work.
Let's take a look at the elegant, completed diagram.
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Now we will prove the double angle identity for sine.
The key will be to calculate the area of the huge triangle, BDC. But there are two ways to do this: we can either express this as (BD * DC)/2 or (BC * DF)/2!
See below for justification.
Now we will prove the double angle identity for sine.
The key will be to calculate the area of the huge triangle, BDC. But there are two ways to do this: we can either express this as (BD * DC)/2 or (BC * DF)/2!
See below for justification.
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However, we already know the lengths of BD, CD, DF, and CB!
Plugging these expressions into our equations, we get a very interesting expression involving sin(2theta) ...
(If you are confused by any of these expressions, please revisit Tweet 14).
However, we already know the lengths of BD, CD, DF, and CB!
Plugging these expressions into our equations, we get a very interesting expression involving sin(2theta) ...
(If you are confused by any of these expressions, please revisit Tweet 14).
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Now, we only have to do algebra!
If we simply cancel the twos in both of the expressions to get rid of the fractions, we obtain:
sin(2theta) = 2 * sin(theta) * cos(theta).
This is the double angle formula for sine!
Now, we only have to do algebra!
If we simply cancel the twos in both of the expressions to get rid of the fractions, we obtain:
sin(2theta) = 2 * sin(theta) * cos(theta).
This is the double angle formula for sine!
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We now move to the double angle formula for cosine.
Let's begin by "angle chasing" - finding expressions for lots of angles. In particular, we will determine a few angles (BCD, CDF) in terms of theta.
We now move to the double angle formula for cosine.
Let's begin by "angle chasing" - finding expressions for lots of angles. In particular, we will determine a few angles (BCD, CDF) in terms of theta.
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Observe that triangles BDF, FDC, and BCD have angles with measures of:
theta, 90˚ - theta, and 90˚.
Thus, they must be similar - all of their angles are congruent.
FDB and DCB will be particularly interesting, so let's take a look at the side lengths of these triangles.
Observe that triangles BDF, FDC, and BCD have angles with measures of:
theta, 90˚ - theta, and 90˚.
Thus, they must be similar - all of their angles are congruent.
FDB and DCB will be particularly interesting, so let's take a look at the side lengths of these triangles.
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Now, we know that BF/BD = BD/BC by the similarity of the two triangles.
This is the key step, because we will get a proportion involving cos(2theta).
This proportion should be fairly easy to solve!
Now, we know that BF/BD = BD/BC by the similarity of the two triangles.
This is the key step, because we will get a proportion involving cos(2theta).
This proportion should be fairly easy to solve!
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We will cross-multiply both sides to solve for cos(2theta).
Brief algebra reveals that
cos(2theta) = 2[cos(theta)]^2 - 1.
This is *one* version of the double angle identity for cosine (there are other equivalent identities but they are easy to derive!).
We will cross-multiply both sides to solve for cos(2theta).
Brief algebra reveals that
cos(2theta) = 2[cos(theta)]^2 - 1.
This is *one* version of the double angle identity for cosine (there are other equivalent identities but they are easy to derive!).
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There are two other ways to express the double angle formula for cosine, which are shown below.
See if you can prove them from the first identity!
Hint: use the Pythagorean Identity from Tweet 7. The proofs shouldn't be very long!
There are two other ways to express the double angle formula for cosine, which are shown below.
See if you can prove them from the first identity!
Hint: use the Pythagorean Identity from Tweet 7. The proofs shouldn't be very long!
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The double angle formulas are extremely useful when solving trigonometric identities.
They can also be considered special cases of the sine/cosine sum formulas (shown below), which have their own proofs without words - although these are not quite as elegant.
The double angle formulas are extremely useful when solving trigonometric identities.
They can also be considered special cases of the sine/cosine sum formulas (shown below), which have their own proofs without words - although these are not quite as elegant.
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Congrats! You've made it through an involved proof of the double angle identities.
Keep in mind that this proof only relied on the basic definitions of sine and cosine with some tools from geometry. Neat!
/End
Congrats! You've made it through an involved proof of the double angle identities.
Keep in mind that this proof only relied on the basic definitions of sine and cosine with some tools from geometry. Neat!
/End
CORRECTION on this tweet:
The figure has a typo when I explain the scale factor.
The scale factor of similarity is still 2, but the proper ratios are BC/AC = DB/EB = 2 (the original figure had CB instead of DB).
The figure has a typo when I explain the scale factor.
The scale factor of similarity is still 2, but the proper ratios are BC/AC = DB/EB = 2 (the original figure had CB instead of DB).
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