# 9.2 Sum and difference identities  (Page 3/6)

 Page 3 / 6

## Finding the exact value of an expression involving an inverse trigonometric function

Find the exact value of $\text{\hspace{0.17em}}\mathrm{sin}\left({\mathrm{cos}}^{-1}\frac{1}{2}+{\mathrm{sin}}^{-1}\frac{3}{5}\right).\text{\hspace{0.17em}}$ Then check the answer with a graphing calculator.

The pattern displayed in this problem is $\text{\hspace{0.17em}}\mathrm{sin}\left(\alpha +\beta \right).\text{\hspace{0.17em}}$ Let $\text{\hspace{0.17em}}\alpha ={\mathrm{cos}}^{-1}\frac{1}{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\beta ={\mathrm{sin}}^{-1}\frac{3}{5}.\text{\hspace{0.17em}}$ Then we can write

$\begin{array}{ccc}\hfill \mathrm{cos}\text{\hspace{0.17em}}\alpha & =& \frac{1}{2},0\le \alpha \le \pi \hfill \\ \hfill \mathrm{sin}\text{\hspace{0.17em}}\beta & =& \frac{3}{5},-\frac{\pi }{2}\le \beta \le \frac{\pi }{2}\hfill \end{array}$

We will use the Pythagorean identities to find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta .$

$\begin{array}{ccc}\hfill \mathrm{sin}\text{\hspace{0.17em}}\alpha & =& \sqrt{1-{\mathrm{cos}}^{2}\alpha }\hfill \\ & =& \sqrt{1-\frac{1}{4}}\hfill \\ & =& \sqrt{\frac{3}{4}}\hfill \\ & =& \frac{\sqrt{3}}{2}\hfill \\ \hfill \mathrm{cos}\text{\hspace{0.17em}}\beta & =& \sqrt{1-{\mathrm{sin}}^{2}\beta }\hfill \\ & =& \sqrt{1-\frac{9}{25}}\hfill \\ & =& \sqrt{\frac{16}{25}}\hfill \\ & =& \frac{4}{5}\hfill \end{array}$

Using the sum formula for sine,

$\begin{array}{ccc}\hfill \mathrm{sin}\left({\mathrm{cos}}^{-1}\frac{1}{2}+{\mathrm{sin}}^{-1}\frac{3}{5}\right)& =& \mathrm{sin}\left(\alpha +\beta \right)\hfill \\ & =& \mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta +\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta \hfill \\ & =& \frac{\sqrt{3}}{2}\cdot \frac{4}{5}+\frac{1}{2}\cdot \frac{3}{5}\hfill \\ & =& \frac{4\sqrt{3}+3}{10}\hfill \end{array}$

## Using the sum and difference formulas for tangent

Finding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern.

Finding the sum of two angles formula for tangent involves taking quotient of the sum formulas for sine and cosine and simplifying. Recall, $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x=\frac{\mathrm{sin}\text{\hspace{0.17em}}x}{\mathrm{cos}\text{\hspace{0.17em}}x},\mathrm{cos}\text{\hspace{0.17em}}x\ne 0.$

Let’s derive the sum formula for tangent.

We can derive the difference formula for tangent in a similar way.

## Sum and difference formulas for tangent

The sum and difference formulas for tangent are:

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

Given two angles, find the tangent of the sum of the angles.

1. Write the sum formula for tangent.
2. Substitute the given angles into the formula.
3. Simplify.

## Finding the exact value of an expression involving tangent

Find the exact value of $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{\pi }{6}+\frac{\pi }{4}\right).$

Let’s first write the sum formula for tangent and then substitute the given angles into the formula.

$\begin{array}{ccc}\hfill \mathrm{tan}\left(\alpha +\beta \right)& =& \frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha +\mathrm{tan}\text{\hspace{0.17em}}\beta }{1-\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }\hfill \\ \hfill \mathrm{tan}\left(\frac{\pi }{6}+\frac{\pi }{4}\right)& =& \frac{\mathrm{tan}\left(\frac{\pi }{6}\right)+\mathrm{tan}\left(\frac{\pi }{4}\right)}{1-\left(\mathrm{tan}\left(\frac{\pi }{6}\right)\right)\left(\mathrm{tan}\left(\frac{\pi }{4}\right)\right)}\hfill \end{array}$

Next, we determine the individual function values within the formula:

$\mathrm{tan}\left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}},\mathrm{tan}\left(\frac{\pi }{4}\right)=1$

So we have

$\begin{array}{ccc}\hfill \mathrm{tan}\left(\frac{\pi }{6}+\frac{\pi }{4}\right)& =& \frac{\frac{1}{\sqrt{3}}+1}{1-\left(\frac{1}{\sqrt{3}}\right)\left(1\right)}\hfill \\ & =& \frac{\frac{1+\sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}\hfill \\ & =& \frac{1+\sqrt{3}}{\sqrt{3}}\left(\frac{\sqrt{3}}{\sqrt{3}-1}\right)\hfill \\ & =& \frac{\sqrt{3}+1}{\sqrt{3}-1}\hfill \end{array}$

Find the exact value of $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{2\pi }{3}+\frac{\pi }{4}\right).$

$\frac{1-\sqrt{3}}{1+\sqrt{3}}$

## Finding multiple sums and differences of angles

Given find

1. $\mathrm{sin}\left(\alpha +\beta \right)$
2. $\mathrm{cos}\left(\alpha +\beta \right)$
3. $\mathrm{tan}\left(\alpha +\beta \right)$
4. $\mathrm{tan}\left(\alpha -\beta \right)$

We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.

1. To find $\text{\hspace{0.17em}}\mathrm{sin}\left(\alpha +\beta \right),$ we begin with $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{3}{5}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}0<\alpha <\frac{\pi }{2}.\text{\hspace{0.17em}}$ The side opposite $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ has length 3, the hypotenuse has length 5, and $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ is in the first quadrant. See [link] . Using the Pythagorean Theorem, we can find the length of side $\text{\hspace{0.17em}}a\text{:}$
$\begin{array}{ccc}\hfill {a}^{2}+{3}^{2}& =& {5}^{2}\hfill \\ \hfill {a}^{2}& =& 16\hfill \\ \hfill a& =& 4\hfill \end{array}$

Since $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta =-\frac{5}{13}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\pi <\beta <\frac{3\pi }{2},$ the side adjacent to $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}-5,$ the hypotenuse is 13, and $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ is in the third quadrant. See [link] . Again, using the Pythagorean Theorem, we have

$\begin{array}{ccc}\hfill {\left(-5\right)}^{2}+{a}^{2}& =& {13}^{2}\hfill \\ \hfill 25+{a}^{2}& =& 169\hfill \\ \hfill {a}^{2}& =& 144\hfill \\ \hfill a& =& ±12\hfill \end{array}$

Since $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ is in the third quadrant, $\text{\hspace{0.17em}}a=–12.$

The next step is finding the cosine of $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ and the sine of $\text{\hspace{0.17em}}\beta .\text{\hspace{0.17em}}$ The cosine of $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ is the adjacent side over the hypotenuse. We can find it from the triangle in [link] : $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\alpha =\frac{4}{5}.\text{\hspace{0.17em}}$ We can also find the sine of $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ from the triangle in [link] , as opposite side over the hypotenuse: $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta =-\frac{12}{13}.\text{\hspace{0.17em}}$ Now we are ready to evaluate $\text{\hspace{0.17em}}\mathrm{sin}\left(\alpha +\beta \right).$

$\begin{array}{ccc}\hfill \mathrm{sin}\left(\alpha +\beta \right)& =& \mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta +\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta \hfill \\ & =& \left(\frac{3}{5}\right)\left(-\frac{5}{13}\right)+\left(\frac{4}{5}\right)\left(-\frac{12}{13}\right)\hfill \\ & =& -\frac{15}{65}-\frac{48}{65}\hfill \\ & =& -\frac{63}{65}\hfill \end{array}$
2. We can find $\text{\hspace{0.17em}}\mathrm{cos}\left(\alpha +\beta \right)\text{\hspace{0.17em}}$ in a similar manner. We substitute the values according to the formula.
$\begin{array}{ccc}\hfill \mathrm{cos}\left(\alpha +\beta \right)& =& \mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta -\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta \hfill \\ & =& \left(\frac{4}{5}\right)\left(-\frac{5}{13}\right)-\left(\frac{3}{5}\right)\left(-\frac{12}{13}\right)\hfill \\ & =& -\frac{20}{65}+\frac{36}{65}\hfill \\ & =& \frac{16}{65}\hfill \end{array}$
3. For $\text{\hspace{0.17em}}\mathrm{tan}\left(\alpha +\beta \right),$ if $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{3}{5}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\alpha =\frac{4}{5},$ then
$\mathrm{tan}\text{\hspace{0.17em}}\alpha =\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$

If $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta =-\frac{12}{13}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta =-\frac{5}{13},$ then

$\mathrm{tan}\text{\hspace{0.17em}}\beta =\frac{\frac{-12}{13}}{\frac{-5}{13}}=\frac{12}{5}$

Then,

4. To find $\text{\hspace{0.17em}}\mathrm{tan}\left(\alpha -\beta \right),$ we have the values we need. We can substitute them in and evaluate.
$\begin{array}{ccc}\hfill \mathrm{tan}\left(\alpha -\beta \right)& =& \frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha -\mathrm{tan}\text{\hspace{0.17em}}\beta }{1+\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }\hfill \\ & =& \frac{\frac{3}{4}-\frac{12}{5}}{1+\frac{3}{4}\left(\frac{12}{5}\right)}\hfill \\ & =& \frac{-\frac{33}{20}}{\frac{56}{20}}\hfill \\ & =& -\frac{33}{56}\hfill \end{array}$

answer and questions in exercise 11.2 sums
how do u calculate inequality of irrational number?
Alaba
give me an example
Chris
and I will walk you through it
Chris
cos (-z)= cos z .
what is a algebra
(x+x)3=?
what is the identity of 1-cos²5x equal to?
__john __05
Kishu
Hi
Abdel
hi
Ye
hi
Nokwanda
C'est comment
Abdel
Hi
Amanda
hello
SORIE
Hiiii
Chinni
hello
Ranjay
hi
ANSHU
hiiii
Chinni
h r u friends
Chinni
yes
Hassan
so is their any Genius in mathematics here let chat guys and get to know each other's
SORIE
I speak French
Abdel
okay no problem since we gather here and get to know each other
SORIE
hi im stupid at math and just wanna join here
Yaona
lol nahhh none of us here are stupid it's just that we have Fast, Medium, and slow learner bro but we all going to work things out together
SORIE
it's 12
what is the function of sine with respect of cosine , graphically
tangent bruh
Steve
cosx.cos2x.cos4x.cos8x
sinx sin2x is linearly dependent
what is a reciprocal
The reciprocal of a number is 1 divided by a number. eg the reciprocal of 10 is 1/10 which is 0.1
Shemmy
Reciprocal is a pair of numbers that, when multiplied together, equal to 1. Example; the reciprocal of 3 is ⅓, because 3 multiplied by ⅓ is equal to 1
Jeza
each term in a sequence below is five times the previous term what is the eighth term in the sequence
I don't understand how radicals works pls
How look for the general solution of a trig function
stock therom F=(x2+y2) i-2xy J jaha x=a y=o y=b
sinx sin2x is linearly dependent
cr
root under 3-root under 2 by 5 y square
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
cosA\1+sinA=secA-tanA
Wrong question
why two x + seven is equal to nineteen.
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI