# 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] .

how can are find the domain and range of a relations
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
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
i want to sure my answer of the exercise
what is the diameter of(x-2)²+(y-3)²=25
how to solve the Identity ?
what type of identity
Jeffrey
Confunction Identity
Barcenas
how to solve the sums
meena
hello guys
meena
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
what is a complex number used for?
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
They are used in any profession where the phase of a waveform has to be accounted for in the calculations. Imagine two electrical signals in a wire that are out of phase by 90°. At some times they will interfere constructively, others destructively. Complex numbers simplify those equations
Tim