# 7.3 Double-angle, half-angle, and reduction formulas  (Page 3/8)

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## Using the power-reducing formulas to prove an identity

Use the power-reducing formulas to prove

${\mathrm{sin}}^{3}\left(2x\right)=\left[\frac{1}{2}\text{\hspace{0.17em}}\mathrm{sin}\left(2x\right)\right]\text{\hspace{0.17em}}\left[1-\mathrm{cos}\left(4x\right)\right]$

We will work on simplifying the left side of the equation:

Use the power-reducing formulas to prove that $\text{\hspace{0.17em}}10\text{\hspace{0.17em}}{\mathrm{cos}}^{4}x=\frac{15}{4}+5\text{\hspace{0.17em}}\mathrm{cos}\left(2x\right)+\frac{5}{4}\text{\hspace{0.17em}}\mathrm{cos}\left(4x\right).$

## Using half-angle formulas to find exact values

The next set of identities is the set of half-angle formulas    , which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}\frac{\alpha }{2},$ the half-angle formula for sine is found by simplifying the equation and solving for $\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\alpha }{2}\right).\text{\hspace{0.17em}}$ Note that the half-angle formulas are preceded by a $\text{\hspace{0.17em}}±\text{\hspace{0.17em}}$ sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which $\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ terminates.

The half-angle formula for sine is derived as follows:

To derive the half-angle formula for cosine, we have

For the tangent identity, we have

## Half-angle formulas

The half-angle formulas    are as follows:

$\mathrm{sin}\left(\frac{\alpha }{2}\right)=±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}$
$\mathrm{cos}\left(\frac{\alpha }{2}\right)=±\sqrt{\frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}$
$\begin{array}{l}\mathrm{tan}\left(\frac{\alpha }{2}\right)=±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{\mathrm{sin}\text{\hspace{0.17em}}\alpha }\hfill \end{array}$

## Using a half-angle formula to find the exact value of a sine function

Find $\text{\hspace{0.17em}}\mathrm{sin}\left({15}^{\circ }\right)\text{\hspace{0.17em}}$ using a half-angle formula.

Since $\text{\hspace{0.17em}}{15}^{\circ }=\frac{{30}^{\circ }}{2},$ we use the half-angle formula for sine:

Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.

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

## Finding exact values using half-angle identities

Given that $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha =\frac{8}{15}$ and $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ lies in quadrant III, find the exact value of the following:

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

Using the given information, we can draw the triangle shown in [link] . Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =-\frac{8}{17}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\alpha =-\frac{15}{17}.$

1. Before we start, we must remember that, if $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ is in quadrant III, then $\text{\hspace{0.17em}}180°<\alpha <270°,$ so $\text{\hspace{0.17em}}\frac{180°}{2}<\frac{\alpha }{2}<\frac{270°}{2}.\text{\hspace{0.17em}}$ This means that the terminal side of $\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ is in quadrant II, since $\text{\hspace{0.17em}}90°<\frac{\alpha }{2}<135°.$

To find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\frac{\alpha }{2},$ we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in [link] and simplify.

We choose the positive value of $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ because the angle terminates in quadrant II and sine is positive in quadrant II.

2. To find $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\frac{\alpha }{2},$ we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in [link] , and simplify.

We choose the negative value of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ because the angle is in quadrant II because cosine is negative in quadrant II.

3. To find $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\frac{\alpha }{2},$ we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in [link] and simplify.

We choose the negative value of $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ lies in quadrant II, and tangent is negative in quadrant II.

#### Questions & Answers

"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
x=-b+_Гb2-(4ac) ______________ 2a
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
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
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
Â
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
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
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
divide simplify each answer 3/2÷5/4
divide simplify each answer 25/3÷5/12
Momo