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

 Page 3 / 8

## 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:

$\begin{array}{ccc}\hfill {\mathrm{sin}}^{2}\theta & =& \frac{1-\mathrm{cos}\left(2\theta \right)}{2}\hfill \\ \hfill {\mathrm{sin}}^{2}\left(\frac{\alpha }{2}\right)& =& \frac{1-\left(\mathrm{cos}2\cdot \frac{\alpha }{2}\right)}{2}\hfill \\ & =& \frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}\hfill \\ \hfill \mathrm{sin}\left(\frac{\alpha }{2}\right)& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \end{array}$

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

$\begin{array}{ccc}\hfill {\mathrm{cos}}^{2}\theta & =& \frac{1+\mathrm{cos}\left(2\theta \right)}{2}\hfill \\ \hfill {\mathrm{cos}}^{2}\left(\frac{\alpha }{2}\right)& =& \frac{1+\mathrm{cos}\left(2\cdot \frac{\alpha }{2}\right)}{2}\hfill \\ & =& \frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}\hfill \\ \hfill \mathrm{cos}\left(\frac{\alpha }{2}\right)& =& ±\sqrt{\frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \end{array}$

For the tangent identity, we have

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

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

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

$\begin{array}{ccc}\hfill \mathrm{sin}\text{\hspace{0.17em}}\frac{30°}{2}& =& \sqrt{\frac{1-\mathrm{cos}30°}{2}}\hfill \\ & =& \sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\hfill \\ & =& \sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}}\hfill \\ & =& \sqrt{\frac{2-\sqrt{3}}{4}}\hfill \\ & =& \frac{\sqrt{2-\sqrt{3}}}{2}\hfill \end{array}$

Remember that we can check the answer with a graphing calculator.

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.

$\begin{array}{ccc}\hfill \mathrm{sin}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \\ & =& ±\sqrt{\frac{1-\left(-\frac{15}{17}\right)}{2}}\hfill \\ & =& ±\sqrt{\frac{\frac{32}{17}}{2}}\hfill \\ & =& ±\sqrt{\frac{32}{17}\cdot \frac{1}{2}}\hfill \\ & =& ±\sqrt{\frac{16}{17}}\hfill \\ & =& ±\frac{4}{\sqrt{17}}\hfill \\ & =& \frac{4\sqrt{17}}{17}\hfill \end{array}$

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.
$\begin{array}{ccc}\hfill \mathrm{cos}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \\ & =& ±\sqrt{\frac{1+\left(-\frac{15}{17}\right)}{2}}\hfill \\ & =& ±\sqrt{\frac{\frac{2}{17}}{2}}\hfill \\ & =& ±\sqrt{\frac{2}{17}\cdot \frac{1}{2}}\hfill \\ & =& ±\sqrt{\frac{1}{17}}\hfill \\ & =& -\frac{\sqrt{17}}{17}\hfill \end{array}$

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.
$\begin{array}{ccc}\hfill \mathrm{tan}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }}\hfill \\ & =& ±\sqrt{\frac{1-\left(-\frac{15}{17}\right)}{1+\left(-\frac{15}{17}\right)}}\hfill \\ & =& ±\sqrt{\frac{\frac{32}{17}}{\frac{2}{17}}}\hfill \\ & =& ±\sqrt{\frac{32}{2}}\hfill \\ & =& -\sqrt{16}\hfill \\ & =& -4\hfill \end{array}$

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.

x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
x exposent4+4x exposent3+8x exposent2+4x+1=0
HERVE
How can I solve for a domain and a codomains in a given function?
ranges
EDWIN
Thank you I mean range sir.
Oliver
proof for set theory
don't you know?
Inkoom
find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad
factoring polynomial
find general solution of the Tanx=-1/root3,secx=2/root3
find general solution of the following equation
Nani
the value of 2 sin square 60 Cos 60
0.75
Lynne
0.75
Inkoom
when can I use sin, cos tan in a giving question
depending on the question
Nicholas
I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks.
John
I want to learn the calculations
where can I get indices
I need matrices
Nasasira
hi
Raihany
Hi
Solomon
need help
Raihany
maybe provide us videos
Nasasira
Raihany
Hello
Cromwell
a
Amie
What do you mean by a
Cromwell
nothing. I accidentally press it
Amie
you guys know any app with matrices?
Khay
Ok
Cromwell
Solve the x? x=18+(24-3)=72
x-39=72 x=111
Suraj
Solve the formula for the indicated variable P=b+4a+2c, for b
Need help with this question please
b=-4ac-2c+P
Denisse
b=p-4a-2c
Suddhen
b= p - 4a - 2c
Snr
p=2(2a+C)+b
Suraj
b=p-2(2a+c)
Tapiwa
P=4a+b+2C
COLEMAN
b=P-4a-2c
COLEMAN
like Deadra, show me the step by step order of operation to alive for b
John
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has