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# Sum and Difference of Angles Formulas

In a class of trigonometry, our Maths teacher said  that the sum of 30° and 40° would give 70° but the sum of sin30° and sin40° would not give sin70° and that caused some commotion in the class. How then do you add and subtract sines or cosines of angles? Hereafter, all you need to know about such operations will be explained.

## What are the sum and difference of angles formulas in trigonometry?

The sum and difference of angle formulas are equations used in carrying out the addition and subtraction of trigonometric identities.

Unlike normal arithmetic operations, addition and subtraction of trigonometric functions have a different approach. For example, cos (45° -15°) is not the same as cos45° - cos15°. It becomes more challenging when trigonometric functions are involved in such arithmetic operations. So, formulas have to be derived to carry out to solve this problem.

Having the knowledge of the trigonometric functions of special angles such as sines, cosines, and tangents of 30, 45, 60, and 90 degrees, means that the addition or subtraction of these angles can give other angles. For instance, sin15° can be derived, since sin15° is the same as sin(45-30)° . Afterwards, we shall be deriving formulas to solve these operations.

## Proving Sum and Difference of Cosine Functions

### Difference of cosine functions

Consider the figure below:

Figure 1: An image showing the use of the standard position of a unit circle to prove the difference of cosine functions, - StudySmarter Originals

The figure above is taken from the standard position of a unit circle. If a is the angle PON and b is the angle ∠QON, then the angle POQ is (a - b) . Therefore, $\mathrm{cos}a$ is the horizontal component of point P and$\mathrm{sin}a$is its vertical component. While$\mathrm{cos}b$is the horizontal component of point Q and $\mathrm{sin}b$ is its vertical component. Thus, to find the distance PQ, we shall use the formula of the distance between two points.

$d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

Where in point P, $\left({x}_{2},{y}_{2}\right)$ is$\left(\mathrm{cos}a,\mathrm{sin}a\right)$ and in point Q, $\left({x}_{1},{y}_{1}\right)$ is$\left(\mathrm{cos}b,\mathrm{sin}b\right)$. Thus

$PQ=\sqrt{\left(cosa-cosb{\right)}^{2}+\left(sina-sinb{\right)}^{2}}\phantom{\rule{0ex}{0ex}}P{Q}^{2}=\left(cosa-cosb{\right)}^{2}+\left(sina-sinb{\right)}^{2}\phantom{\rule{0ex}{0ex}}P{Q}^{2}=co{s}^{2}a-2cosacosb+co{s}^{2}b+si{n}^{2}a-2sinasinb+si{n}^{2}b$

Rearrange the equation

$P{Q}^{2}=co{s}^{2}a+si{n}^{2}a+co{s}^{2}b+si{n}^{2}b-2cosacosb-2sinasinb$

Remember:

$co{s}^{2}\theta +si{n}^{2}\theta =1;so,co{s}^{2}a+{\mathrm{sin}}^{2}a=1andsi{n}^{2}b+{\mathrm{cos}}^{2}b=1$

Then:

$P{Q}^{2}=1+1-2cosacosb-2sinasinb\phantom{\rule{0ex}{0ex}}P{Q}^{2}=2-2cosacosb-2sinasinb$

If the angle (a-b) were to be replotted into the standard position of a unit circle from the origin O to the point S in the figure below

Figure 2: An image of the angle (a-b) being replotted, - StudySmarter Originals

Then, the distance SN in figure 2 (which is equal to the distance PQ in figure 1) can be derived with respect to the angle (a-b) and the corresponding points in S (cos (a-b), sin(a-b) ) and N (1 , 0).

Using

$d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

Where point S is $\left({x}_{2},{y}_{2}\right)$ and N is $\left({x}_{1},{y}_{1}\right)$, then

$SN=\sqrt{\left(\mathrm{cos}{\left(a-b\right)-1\right)}^{2}+\left(\mathrm{sin}{\left(a-b\right)-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}S{N}^{2}=\left(\mathrm{cos}{\left(a-b\right)-1\right)}^{2}+\left(\mathrm{sin}{\left(a-b\right)-0\right)}^{2}\phantom{\rule{0ex}{0ex}}S{N}^{2}={\mathrm{cos}}^{2}\left(a-b\right)-2\mathrm{cos}\left(a-b\right)+1+{\mathrm{sin}}^{2}\left(a-b\right)$

Rearrange and bring like terms

$S{N}^{2}={\mathrm{cos}}^{2}\left(a-b\right)+{\mathrm{sin}}^{2}\left(a-b\right)-2\mathrm{cos}\left(a-b\right)+1$

Remember that

$co{s}^{2}\theta +si{n}^{2}\theta =1;so,co{s}^{2}\left(a-b\right)+{\mathrm{sin}}^{2}\left(a-b\right)=1$

then;

$S{N}^{2}=1-2\mathrm{cos}\left(a-b\right)+1\phantom{\rule{0ex}{0ex}}S{N}^{2}=2-2\mathrm{cos}\left(a-b\right)$

Remember that

$PQ=SN$

then

$P{Q}^{2}=S{N}^{2}$

Thus

$2-2\mathrm{cos}\left(a-b\right)=2-2\mathrm{cos}a\mathrm{cos}b-2\mathrm{sin}a\mathrm{sin}b$

Solve the algebra by subtracting 2 from both sides of the equation

$-2\mathrm{cos}\left(a-b\right)=-2\mathrm{cos}a\mathrm{cos}b-2\mathrm{sin}a\mathrm{sin}b$

Divide both sides by -2 on both sides

$\mathrm{cos}\left(a-b\right)=\mathrm{cos}a\mathrm{cos}b+\mathrm{sin}a\mathrm{sin}b$

### Summing of cosine functions

$cos\left(a+b\right)=\mathrm{cos}\left(a-\left(-b\right)\right)$

Thus, substitute the value of b as -b in the equation.

Note that

$\mathrm{cos}\left(-b\right)=\mathrm{cos}b$

and

$\mathrm{sin}\left(-b\right)=-\mathrm{sin}b$

therefore

$\mathrm{cos}\left(a+b\right)=\mathrm{cos}a\mathrm{cos}\left(-b\right)+\mathrm{sin}a\mathrm{sin}\left(-b\right)\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(a+b\right)=\mathrm{cos}a\mathrm{cos}b-\mathrm{sin}a\mathrm{sin}b$

## Proving Sum and Difference of Sine Functions

### Summing of sine functions

Draw a right-angled triangle ABC as shown below.

An image of a right triangle, - StudySmarter Originals

Draw another line intersecting A and touching line BC at D, such that angle BAD is β and angle DAC is α as seen below.

Draw a line perpendicular to point D which touches line AB at E as seen below.

Draw a line from point E which is perpendicular to line AC cuts through line AD at F and meets line AC at G as shown below.

Draw a line from point D to point H on the line EG which is perpendicular to line EG as seen below.

Note that for each step hereafter, you should refer to the figure above.

Therefore

Using SOHCAHTOA

$\mathrm{sin}\left(\alpha +\beta \right)=\frac{EG}{AE}$

Note that line EG = EH + HG, thus

$\mathrm{sin}\left(\alpha +\beta \right)=\frac{EH+HG}{AE}\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(\alpha +\beta \right)=\frac{EH}{AE}+\frac{HG}{AE}$

Recall;

$HG=DC$

the lines HG and DC are parallel and equal.

Thus

$\mathrm{sin}\left(\alpha +\beta \right)=\frac{EH}{AE}+\frac{DC}{AE}$

See that

$\angle DAC=\angle FDH$

They are alternate angles because of lines HD and AC are parallel and is being cut through by line AD.

Note below

$\angle DAC=\angle FDH=\alpha$

Recall that line AD is perpendicular to line ED. Therefore

$\angle HDE=90°-\alpha$

Knowing that

$\angle EHD=90°$

thus

$\angle HED+90°+90°-\alpha =180°$

sum of angles in a triangle is equal to 180°

$\angle HED+180°-180°=\alpha$

$\angle HED=\alpha$

Looking at their angles, it means that triangle ADC and EDH are similar. see below

An image that proves the summation of sine of angles, StudySmarter Originals

From the right-angled triangle EDH

$\mathrm{cos}\alpha =\frac{EH}{ED}\phantom{\rule{0ex}{0ex}}EH=ED\mathrm{cos}\alpha$

Recall that

$\mathrm{sin}\left(\alpha +\beta \right)=\frac{EH}{AE}+\frac{DC}{AE}$

Substitute the value of EH

$\mathrm{sin}\left(\alpha +\beta \right)=\frac{ED\mathrm{cos}\alpha }{AE}+\frac{DC}{AE}\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(\alpha +\beta \right)=\left(\frac{ED}{AE}×\mathrm{cos}\alpha \right)+\frac{DC}{AE}$

Meanwhile, from the right-angled triangle AED, using SOHCAHTOA

$\mathrm{sin}\beta =\frac{ED}{AE}$

Substitute the value of $\frac{ED}{AE}$ in the equation

$\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\beta \mathrm{cos}\alpha +\frac{DC}{AE}$

From the right-angled triangle ADC, using SOHCAHTOA

$\mathrm{sin}\alpha =\frac{DC}{AD}\phantom{\rule{0ex}{0ex}}DC=AD\mathrm{sin}\alpha$

Substitute the value of DC in the equation

$\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\beta \mathrm{cos}\alpha +\frac{AD\mathrm{sin}\alpha }{AE}$

Looking at the right-angled triangle AED and using SOHCAHTOA

$\mathrm{cos}\beta =\frac{AD}{AE}$

Substitute the value of$\frac{AD}{AE}$ in the equation

$\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\beta \mathrm{cos}\alpha +\mathrm{cos}\beta \mathrm{sin}\alpha \phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\alpha \mathrm{cos}\beta +\mathrm{sin}\beta \mathrm{cos}\alpha$

### Difference of its functions

Knowing that

$\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\alpha \mathrm{cos}\beta +\mathrm{sin}\beta \mathrm{cos}\alpha$

Thus $\mathrm{sin}\left(\alpha -\beta \right)$ can be derived by exchanging β with -β throughout the equation.

Therefore

$\mathrm{sin}\left(\alpha -\beta \right)=\mathrm{sin}\alpha \mathrm{cos}\left(-\beta \right)+\mathrm{sin}\left(-\beta \right)\mathrm{cos}\alpha$

Note that

$\mathrm{cos}\left(-\beta \right)=\mathrm{cos}\beta$

and

$\mathrm{sin}\left(-\beta \right)=-\mathrm{sin}\beta$

therefore

$\mathrm{sin}\left(\alpha -\beta \right)=\mathrm{sin}\alpha \mathrm{cos}\beta -\mathrm{sin}\beta \mathrm{cos}\alpha$

## Proving Sum and Difference of Tangent Functions

### Summing of tangent functions

Recall that

$\mathrm{tan}\varnothing =\frac{\mathrm{sin}\varnothing }{\mathrm{cos}\varnothing }$

Therefore

$\mathrm{tan}\left(A+B\right)=\frac{\mathrm{sin}\left(A+B\right)}{\mathrm{cos}\left(A+B\right)}$

Thus

$\mathrm{tan}\left(A+B\right)=\frac{\mathrm{sin}A\mathrm{cos}B+\mathrm{sin}B\mathrm{cos}A}{\mathrm{cos}A\mathrm{cos}B-\mathrm{sin}A\mathrm{sin}B}$

Divide every entity of the right-hand side of the equation by cosAcosB

### Difference of tangent functions

Recall that

$\mathrm{tan}\varnothing =\frac{\mathrm{sin}\varnothing }{\mathrm{cos}\varnothing }$

Therefore

$\mathrm{tan}\left(A-B\right)=\frac{\mathrm{sin}\left(A-B\right)}{\mathrm{cos}\left(A-B\right)}$

Thus

$\mathrm{tan}\left(A-B\right)=\frac{\mathrm{sin}A\mathrm{cos}B-\mathrm{sin}B\mathrm{cos}A}{\mathrm{cos}A\mathrm{cos}B+\mathrm{sin}A\mathrm{sin}B}$

Divide every entity of the right-hand side of the equation by cosAcosB

## Sum and difference of formulas application

Below you shall see how to apply the sum and difference formulas.

Find the value of cos15°

Solution:

The first step is to find the best possible combination of special angles that will yield that angle. In this case, 15° can be gotten by subtracting 30° from 45°.

Therefore

$\mathrm{cos}15°=\mathrm{cos}\left(45°-30°\right)\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(45°-30°\right)=\mathrm{cos}45°\mathrm{cos}30°+\mathrm{sin}45°\mathrm{sin}30°$

recall

$\mathrm{cos}30°=\frac{\sqrt{3}}{2},\phantom{\rule{0ex}{0ex}}\mathrm{sin}30°=\frac{1}{2},\phantom{\rule{0ex}{0ex}}\mathrm{cos}45°=\mathrm{sin}45°=\frac{\sqrt{2}}{2}$

Therefore;

$\mathrm{cos}\left(45°-30°\right)=\left(\frac{\sqrt{2}}{2}×\frac{\sqrt{3}}{2}\right)+\left(\frac{\sqrt{2}}{2}×\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(45°-30°\right)=\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(45°-30°\right)=\frac{\sqrt{6}+\sqrt{2}}{4}$

Factorize further

id="2970782" role="math" $\mathrm{cos}\left(45°-30°\right)=\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{4}$

Thus

id="2970783" role="math" $\mathrm{cos}15°=\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{4}$

Prove that:

$\mathrm{sin}210°=-\frac{1}{2}$

Solution:

$\mathrm{sin}210°=\mathrm{sin}\left(180°+30°\right)$

knowing that

$\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\alpha \mathrm{cos}\beta +\mathrm{sin}\beta \mathrm{cos}\alpha$

Therefore

$\mathrm{sin}\left(180°+30°\right)=\mathrm{sin}180°\mathrm{cos}30°+\mathrm{sin}30°\mathrm{cos}180°$

Note that

$\mathrm{sin}180°=0,\mathrm{cos}180°=-1,\mathrm{sin}30°=\frac{1}{2},\mathrm{cos}30°=\frac{\sqrt{3}}{2}$:

Thus,

$\mathrm{sin}\left(180°+30°\right)=\left(0×\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{2}×-1\right)\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(180°+30°\right)=-\frac{1}{2}$

Hence;

$\mathrm{sin}210°=\mathrm{sin}\left(180°+30°\right)=-\frac{1}{2}$

If a man leaves a point P to a point R which is 20 km due east of P, then, he walks to a point S due North of R. Find the distance from R to S if S is 75 degrees Northeast of P without using calculators or mathematical tables.

Solution:

We are asked to calculate the distance RS. Using SOHCAHTOA

$\mathrm{tan}15°=\frac{RS}{20}\phantom{\rule{0ex}{0ex}}RS=20\mathrm{tan}15°\phantom{\rule{0ex}{0ex}}\mathrm{tan}15°=\mathrm{tan}\left(45°-30°\right)$

Note that

$\mathrm{tan}\left(A-B\right)=\frac{\mathrm{tan}A-\mathrm{tan}B}{1+\mathrm{tan}A\mathrm{tan}B}$

Therefore

$\mathrm{tan}\left(45°-30°\right)=\frac{\mathrm{tan}45°-\mathrm{tan}30°}{1+\mathrm{tan}45°\mathrm{tan}30°}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Where

$\mathrm{tan}45°=1$

and

$\mathrm{tan}30°=\frac{\sqrt{3}}{3}$

Then

Multiply the numerator and denominator by $1-\frac{\sqrt{3}}{3}$

Thus

$RS=20\mathrm{tan}15°\phantom{\rule{0ex}{0ex}}RS=20×\left(2-\sqrt{3}\right)km$

## Sum and Difference of Angles Formulas - Key takeaways

• The sum and difference of trigonometric functions are not calculated using a direct arithmetic approach.
• The formula for the sum and difference of sine is$\mathrm{sin}\left(\alpha ±\beta \right)=\mathrm{sin}\alpha \mathrm{cos}\beta ±\mathrm{sin}\beta \mathrm{cos}\alpha$
• The formula for the sum and difference of cosine is$\mathrm{cos}\left(a±b\right)=\mathrm{cos}a\mathrm{cos}b\mp \mathrm{sin}a\mathrm{sin}b$
• The formula for the sum and difference of tangent is$\mathrm{tan}\left(A±B\right)=\frac{\mathrm{tan}A±\mathrm{tan}B}{1\mp \mathrm{tan}A\mathrm{tan}B}$

#### Flashcards in Sum and Difference of Angles Formulas 2

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What is the angle difference formula?

This is the formula which calculates the difference between angles in trigonometry. It varies depending on the trigonometric function involved.

What are sum and difference formulas?

These are the formulas which calculate the sum of angles in trigonometry. They vary depending on the trigonometric function involved.

Why is the sum and difference of angles formula useful?

The sum and difference of angles formula is useful because angles of trigonometric functions cannot be calculated in a direct arithmetic manner.

How to do the sum and difference of angles formulas?

The sum and difference formulas is done by using the standard position of a unit circle.

What is an example of sum and difference of angles formulas?

An example of the sum and difference formula is the addition of cosine functions which is; cos(A+B) = cosAcosB - sinAsinB

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