# 7.3 Double-angle, half-angle, and reduction formulas  (Page 9/8)

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In this section, you will:
• Use double-angle formulas to find exact values.
• Use double-angle formulas to verify identities.
• Use reduction formulas to simplify an expression.
• Use half-angle formulas to find exact values.

Bicycle ramps made for competition (see [link] ) must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{5}{3}.\text{\hspace{0.17em}}$ The angle is divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one.

## Using double-angle formulas to find exact values

In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas    are a special case of the sum formulas, where $\text{\hspace{0.17em}}\alpha =\beta .\text{\hspace{0.17em}}$ Deriving the double-angle formula for sine begins with the sum formula,

$\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta +\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta$

If we let $\text{\hspace{0.17em}}\alpha =\beta =\theta ,$ then we have

Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, $\text{\hspace{0.17em}}\mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta -\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta ,$ and letting $\text{\hspace{0.17em}}\alpha =\beta =\theta ,$ we have

Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:

The second interpretation is:

Similarly, to derive the double-angle formula for tangent, replacing $\text{\hspace{0.17em}}\alpha =\beta =\theta \text{\hspace{0.17em}}$ in the sum formula gives

$\begin{array}{c}\mathrm{tan}\left(\alpha +\beta \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha +\mathrm{tan}\text{\hspace{0.17em}}\beta }{1-\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }\\ \mathrm{tan}\left(\theta +\theta \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\theta +\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-\mathrm{tan}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta }\\ \mathrm{tan}\left(2\theta \right)=\frac{2\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-{\mathrm{tan}}^{2}\theta }\end{array}$

## Double-angle formulas

The double-angle formulas    are summarized as follows:

$\mathrm{sin}\left(2\theta \right)=2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta$

$\mathrm{tan}\left(2\theta \right)=\frac{2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-{\mathrm{tan}}^{2}\theta }$

Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.

1. Draw a triangle to reflect the given information.
2. Determine the correct double-angle formula.
3. Substitute values into the formula based on the triangle.
4. Simplify.

## Using a double-angle formula to find the exact value involving tangent

Given that $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =-\frac{3}{4}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in quadrant II, find the following:

1. $\mathrm{sin}\left(2\theta \right)$
2. $\mathrm{cos}\left(2\theta \right)$
3. $\mathrm{tan}\left(2\theta \right)$

If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =-\frac{3}{4},$ such that $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in the second quadrant, the adjacent side is on the x -axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse:

Now we can draw a triangle similar to the one shown in [link] .

1. Let’s begin by writing the double-angle formula for sine.
$\mathrm{sin}\left(2\theta \right)=2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta$

We see that we to need to find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta .\text{\hspace{0.17em}}$ Based on [link] , we see that the hypotenuse equals 5, so $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{3}{5},$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =-\frac{4}{5}.\text{\hspace{0.17em}}$ Substitute these values into the equation, and simplify.

Thus,

2. Write the double-angle formula for cosine.
$\mathrm{cos}\left(2\theta \right)={\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta$

Again, substitute the values of the sine and cosine into the equation, and simplify.

3. Write the double-angle formula for tangent.
$\mathrm{tan}\left(2\theta \right)=\frac{2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-{\mathrm{tan}}^{2}\theta }$

In this formula, we need the tangent, which we were given as $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =-\frac{3}{4}.\text{\hspace{0.17em}}$ Substitute this value into the equation, and simplify.

I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
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Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert