7.2 Sum and difference identities  (Page 5/6)

 Page 5 / 6

Verify the identity: $\text{\hspace{0.17em}}\mathrm{tan}\left(\pi -\theta \right)=-\mathrm{tan}\text{\hspace{0.17em}}\theta .$

Using sum and difference formulas to solve an application problem

Let $\text{\hspace{0.17em}}{L}_{1}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{L}_{2}\text{\hspace{0.17em}}$ denote two non-vertical intersecting lines, and let $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ denote the acute angle between $\text{\hspace{0.17em}}{L}_{1}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{L}_{2}.\text{\hspace{0.17em}}$ See [link] . Show that

$\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}$

where $\text{\hspace{0.17em}}{m}_{1}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{m}_{2}\text{\hspace{0.17em}}$ are the slopes of $\text{\hspace{0.17em}}{L}_{1}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{L}_{2}\text{\hspace{0.17em}}$ respectively. ( Hint: Use the fact that $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}{\theta }_{1}={m}_{1}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}{\theta }_{2}={m}_{2}.$ )

Using the difference formula for tangent, this problem does not seem as daunting as it might.

Investigating a guy-wire problem

For a climbing wall, a guy-wire $\text{\hspace{0.17em}}R\text{\hspace{0.17em}}$ is attached 47 feet high on a vertical pole. Added support is provided by another guy-wire $\text{\hspace{0.17em}}S\text{\hspace{0.17em}}$ attached 40 feet above ground on the same pole. If the wires are attached to the ground 50 feet from the pole, find the angle $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ between the wires. See [link] .

Let’s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta =\frac{47}{50},$ and $\text{\hspace{0.17em}}\mathrm{tan}\left(\beta -\alpha \right)=\frac{40}{50}=\frac{4}{5}.\text{\hspace{0.17em}}$ We can then use difference formula for tangent.

$\mathrm{tan}\left(\beta -\alpha \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\beta -\mathrm{tan}\text{\hspace{0.17em}}\alpha }{1+\mathrm{tan}\text{\hspace{0.17em}}\beta \mathrm{tan}\text{\hspace{0.17em}}\alpha }$

Now, substituting the values we know into the formula, we have

Use the distributive property, and then simplify the functions.

$\begin{array}{l}\text{\hspace{0.17em}}4\left(1\right)+4\left(\frac{47}{50}\right)\mathrm{tan}\text{\hspace{0.17em}}\alpha =5\left(\frac{47}{50}\right)-5\text{\hspace{0.17em}}\mathrm{tan}\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}}4+3.76\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha =4.7-5\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha \hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}5\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha +3.76\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha =0.7\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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8.76\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha =0.7\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}}\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}}\mathrm{tan}\text{\hspace{0.17em}}\alpha \approx 0.07991\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}}{\mathrm{tan}}^{-1}\left(0.07991\right)\approx .079741\hfill \end{array}$

Now we can calculate the angle in degrees.

$\alpha \approx 0.079741\left(\frac{180}{\pi }\right)\approx {4.57}^{\circ }$

Access these online resources for additional instruction and practice with sum and difference identities.

Key equations

 Sum Formula for Cosine $\mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta -\mathrm{sin}\text{\hspace{0.17em}}\alpha \mathrm{sin}\text{\hspace{0.17em}}\beta$ Difference Formula for Cosine $\mathrm{cos}\left(\alpha -\beta \right)=\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta +\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta$ Sum Formula for Sine $\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta +\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta$ Difference Formula for Sine $\mathrm{sin}\left(\alpha -\beta \right)=\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta -\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta$ Sum Formula for Tangent $\mathrm{tan}\left(\alpha +\beta \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha +\mathrm{tan}\text{\hspace{0.17em}}\beta }{1-\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }$ Difference Formula for Tangent $\mathrm{tan}\left(\alpha -\beta \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha -\mathrm{tan}\text{\hspace{0.17em}}\beta }{1+\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }$ Cofunction identities $\begin{array}{l}\mathrm{sin}\text{\hspace{0.17em}}\theta =\mathrm{cos}\left(\frac{\pi }{2}-\theta \right)\\ \mathrm{cos}\text{\hspace{0.17em}}\theta =\mathrm{sin}\left(\frac{\pi }{2}-\theta \right)\\ \mathrm{tan}\text{\hspace{0.17em}}\theta =\mathrm{cot}\left(\frac{\pi }{2}-\theta \right)\\ \mathrm{cot}\text{\hspace{0.17em}}\theta =\mathrm{tan}\left(\frac{\pi }{2}-\theta \right)\\ \mathrm{sec}\text{\hspace{0.17em}}\theta =\mathrm{csc}\left(\frac{\pi }{2}-\theta \right)\\ \mathrm{csc}\text{\hspace{0.17em}}\theta =\mathrm{sec}\left(\frac{\pi }{2}-\theta \right)\end{array}$

Key concepts

• The sum formula for cosines states that the cosine of the sum of two angles equals the product of the cosines of the angles minus the product of the sines of the angles. The difference formula for cosines states that the cosine of the difference of two angles equals the product of the cosines of the angles plus the product of the sines of the angles.
• The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle. See [link] and [link] .
• The sum formula for sines states that the sine of the sum of two angles equals the product of the sine of the first angle and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. The difference formula for sines states that the sine of the difference of two angles equals the product of the sine of the first angle and cosine of the second angle minus the product of the cosine of the first angle and the sine of the second angle. See [link] .
• The sum and difference formulas for sine and cosine can also be used for inverse trigonometric functions. See [link] .
• The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. The difference formula for tangent states that the tangent of the difference of two angles equals the difference of the tangents of the angles divided by 1 plus the product of the tangents of the angles. See [link] .
• The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles. See [link] .
• The cofunction identities apply to complementary angles and pairs of reciprocal functions. See [link] .
• Sum and difference formulas are useful in verifying identities. See [link] and [link] .
• Application problems are often easier to solve by using sum and difference formulas. See [link] and [link] .

Questions & Answers

what is f(x)=
Karim Reply
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
GREAT ANSWER THOUGH!!!
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?
unknown Reply
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
Ef Reply
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
KARMEL Reply
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
Rima Reply
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.
Brittany Reply
how do you find the period of a sine graph
Imani Reply
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
Jhon Reply
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
Baptiste Reply
the sum of any two linear polynomial is what
Esther Reply
divide simplify each answer 3/2÷5/4
Momo Reply
divide simplify each answer 25/3÷5/12
Momo
how can are find the domain and range of a relations
austin Reply
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?
Diddy Reply
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
can I see the picture
Zairen Reply
How would you find if a radical function is one to one?
Peighton Reply

Read also:

Get the best Precalculus course in your pocket!

Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Precalculus' conversation and receive update notifications?

 By By By By By By By