# Power Reducing Calculator

Welcome to Omni's **power reducing calculator**, where we'll study **the power reducing formulas** that connect the squares of the trigonometric function (`sin²(x)`

, `cos²(x)`

, and `tan²(x)`

) to **the cosine of the angle doubled** (i.e., using the `cos(2x)`

identity). It is a quick and easy way to go between different powers of trigonometric functions, **either from smaller to larger, or vice versa**.

So sit back, grab a cup of coffee, and **let's get to it**!

## Trigonometric functions

Geometrically speaking, **the world is made of triangles**. The best proof of this is found in video games: all graphics, especially those in three-dimensions, are done by combining a lot of triangles together to form **a pattern that resembles a person, animal, or object**. Starting from a square, you can just slice it in two along the diagonal to **get two triangles**. The more complex the object, the more triangles it needs. But in the end, **it's just more of the same three-edged shape**. Even circles can be approximated by a sufficient number of triangles.

In fact, **this is a good thing** because triangles are easy to work with. They only have three sides and, what is more, once we know their lengths, **we also know their angles**. The opposite is also true, but can't know the scale of the triangle.

The above is a very non-trivial fact. What is says is that basically **no weird things can happen**. The correspondence between the angles and the sides **must satisfy several properties**. This is precisely the motivation behind and trigonometric functions. It greatly simplifies triangle studies since instead of worrying about measuring three sides and three angles, **we can work only with the functions**.

The introduction to trigonometric functions is usually done on right triangles. Below you can find the formulas for almost all the trigonometric functions. Let us mention here that, in reality, we mostly limit ourselves to the sine, cosine, and tangent functions (`sin`

, `cos`

, and `tan`

, respectively).

Note that as the name suggests, **these are functions**, so they associate a number to an argument, the latter being an angle. Also, since we're dealing with right triangles, **the argument cannot be larger than** `90°`

(or `π/2`

in radians).

However, **we can extend all the trigonometric functions to any angle**, even a negative one. To do that, we look at the two-dimensional Euclidean space. For a given point `A = (x, y)`

in that space, denote by `α`

**the angle between the positive half of the horizontal axis to the interval from** `(0,0)`

**to** `A`

**going counterclockwise**. Clearly, this angle can be larger than `90°`

.

Now, we translate all the trigonometric function formulas from above to this new setting by exchanging `b`

for `x`

, `a`

for `y`

, and `c`

for the distance from `A`

to `(0,0)`

(which is equal to `√(x² + y²)`

.

We've already mentioned that trigonometric functions describe **the correspondence between the sides and the angles of a triangle**. What we'd like to do next is find some **correspondences between the correspondences**, or, if you prefer, formulas using the above functions. We call such equations **trigonometric identities**, and we can't stress enough how *important* they are.

Sometimes, like in the case of the law of sines or the law of cosines, they can provide you with additional info about the triangle. Other times, they simplify longer, tedious trigonometric calculations. **The power reducing identities are an example of the latter.**

## Power reducing identities

Before we see the power reducing formulas themselves, let's introduce **two other trigonometric properties**, which will, in fact, imply the power reducing identities.

**The (fundamental) Pythagorean trigonometric identity** states that

`sin²(x) + cos²(x) = 1`

for any angle `x`

. If you're not convinced that **such a pretty formula is true**, we encourage you to take a minute to prove it. It's enough to substitute for `sin(x)`

and `cos(x)`

the quotients from and use the Pythagorean theorem.

Next, we have a bunch of double angle formulas. However, **we'll only need the one for cosine**, i.e., the `cos(2x)`

identity. It states that

`cos(2x) = cos²(x) - sin²(x)`

for any angle `x`

.

Now, if we use the first formula to rewrite the second, on the one hand, we'll get

`cos(2x) = cos²(x) - sin²(x) =`

`cos²(x) - sin²(x) + 1 - 1 =`

`cos²(x) - sin²(x) + 1 - cos²(x) - sin²(x) =`

`-2sin²(x) + 1`

,

and, on the other hand,

`cos(2x) = cos²(x) - sin²(x) =`

`cos²(x) - sin²(x) + 1 - 1 =`

`cos²(x) - sin²(x) + cos²(x) + sin²(x) - 1 =`

`2cos²(x) - 1`

.

Next, in the first one, we move the `1`

to the left and divide both sides by `-2`

. This way, we obtain"

`cos(2x) = -2sin²(x) + 1`

`(cos(2x) - 1) / (-2) = sin²(x)`

`sin²(x) = (1 - cos(2x)) / 2`

.

Similarly, the second formula gives

`cos(2x) = 2cos²(x) - 1`

`(cos(2x) + 1) / 2 = cos²(x)`

`cos²(x) = (1 + cos(2x)) / 2`

.

These are our **first two power reducing identities**. To find the last one, that with tangent, recall **one more basic trigonometric formula**:

`tan(x) = sin(x) / cos(x)`

for `x`

not equal to `90° + k*180°`

(or `π/2 + kπ`

in radians), where `k`

is any integer. This allows us to find **the third power reducing formula** by simply applying the two we got above:

`tan²(x) = (sin(x) / cos(x))² = sin²(x) / cos²(x) =`

`= [(1 - cos(2x)) / 2] / [(1 + cos(2x)) / 2] =`

`= (1 - cos(2x)) / (1 + cos(2x))`

.

**Now, those don't seem too bad, do they?** Certainly, the proofs didn't take too long. Just to be on the safe side, let's conclude the section by listing the three of them **in a nice, concise way**:

And with that, **we conclude the theoretic part** of the article. It was fun, wasn't it? Still, how much time can you spend looking at symbols instead of the good ol' numbers? Well, next up is a section to address just that and show you **the possibilities of the power reducing calculator**!

## Example: using the power reducing calculator

Suppose that you have a test **in trigonometric functions** coming later this week. From what the teacher says, it will focus on calculating the values of the functions at specific angles and using some trigonometric identities (including **the power reducing formulas**) to help you along the way.

You decide to grab the exercise book and **go through at least a few examples** to prepare yourself. The first one tells you to calculate

`sin²(x) - sin⁴(x)`

for `x = 15°`

(or `x = π/12`

in radians).

**Now, this looks like a job for the power reducing calculator!**

First of all, let's see **how easy it is to find the answer using our tool**. For that, we'll divide the task into **three parts**:

- Finding
`sin²(x)`

; - Finding
`sin⁴(x)`

; and - Calculating the difference.

Take a look at the power reducing calculator. At the top, we have **a field with the variable** "*angle*". This is the `x`

in the formulas you can hopefully see above the calculator. Oh, just like in the exercise! **What a coincidence!**

We input the number:

`x = 15°`

.

The moment we do that, **the power reducing calculator spits out the answer** for `sin²(x)`

(and other functions, which we don't need at the moment). Yay, **we have the first part**!

Now how do we find `sin⁴(x)`

? Well, it's crucial to observe that

`(sin²(x))² = sin⁴(x)`

.

This means that we can either just **take the square of** `sin²(x)`

that we got above, or... **use the power reducing calculator again**!

The best thing about **they work in multiple ways**. This means that instead of inputting the angle, we can now **input the value of the sine**, and that's precisely what we'll do here. If we write the value of `sin²(x)`

under `sin(x)`

in the power reducing calculator, then the tool will **spit out the value of** `sin⁴(x)`

(i.e., the square of what we input) in the `sin²(x)`

window.

**Magical, isn't it?** Note, however, that the calculator will also show the angle that the value corresponds to, which is different from the `15°`

from the exercise.

And once we have the two values, we can subtract them and **get the answer** to our problem. However, **let's not spoiler it for ourselves** just yet! It's time to grab a piece of paper and **calculate the whole thing ourselves**.

## Power reducing formula in an example by hand

First of all, note that `15°`

is a *weird* angle, in the sense that **it's not so clear what the trigonometric functions of it are**. Fortunately, however, we don't need it! "*How so?*" you ask? Well, **let's see**.

We need to find `sin²(x)`

and `sin⁴(x)`

. But we've learned a thing or two about `sin²(x)`

in , didn't we? That's right - **in the power reducing identities**!

According to the power reducing formula for the sine, we have

`sin²(x) = (1 - cos(2x)) / 2`

.

In our case, this translates to

`sin²(15°) = (1 - cos(2 * 15°)) / 2 = (1 - cos(30°)) / 2`

.

Wait, **doesn't it look familiar** - the cosine of `30°`

? Sure it does, it's the function corresponding to a right triangle with angles `90°`

, `60°`

, and `30°`

. There's no need for the `cos(2x)`

identity or anything.

What is more, we know its sides quite well, or rather the ratios between them. If we recall the formula of the cosine, this tells us that

`cos(30°) = √3 / 2 ≈ 0.865`

.

We substitute it into the power reducing formula:

`sin²(15°) = (1 - cos(30°)) / 2 ≈ (1 - 0.865) / 2 ≈ 0.068`

.

Lastly, we use a similar trick as before to find `sin⁴(15°)`

:

`sin⁴(15°) = (sin²(15°))² ≈ 0.068² ≈ 0.005`

,

which tells us that **the final answer is**

`sin²(15°) - sin⁴(15°) ≈ 0.068 - 0.005 = 0.063`

.

Great, **we've finished the exercise**! If anything like that appears on the test, **we'll be more than ready**!

Hm, should we continue with the next one, or take a short break and a nap?