Sum and Difference of Angles Formulas

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, $\cos a$ is the horizontal component of point P and$\sin a$is its vertical component. While$\cos b$is the horizontal component of point Q and $\sin b$ is its vertical component. Thus, to find the distance PQ, we shall use the formula of the distance between two points.

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Where in point P, $(x_2,y_2)$ is$(\cos a,\sin a)$ and in point Q, $(x_1,y_1)$ is$(\cos b,\sin b)$. Thus

$$\begin{gathered} PQ=\sqrt{(cosa-cosb{)}^2+(sina-sinb{)}^2} \\ P{Q}^2=(cosa-cosb{)}^2+(sina-sinb{)}^2 \\ P{Q}^2=co{s}^{2}a-2cosacosb+co{s}^{2}b+si{n}^{2}a-2sinasinb+si{n}^{2}b \end{gathered}$$

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:

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

Then:

$$\begin{gathered} P{Q}^2=1+1-2cosacosb-2sinasinb \\ P{Q}^2=2-2cosacosb-2sinasinb \end{gathered}$$

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

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Where point S is $(x_2,y_2)$ and N is $(x_1,y_1)$, then

$$\begin{gathered} SN=\sqrt{(\cos {(a-b)-1)}^2+(\sin {(a-b)-0)}^2} \\ S{N}^2=(\cos {(a-b)-1)}^2+(\sin {(a-b)-0)}^2 \\ S{N}^2={\cos }^2(a-b)-2\cos (a-b)+1+{\sin }^2(a-b) \end{gathered}$$

Rearrange and bring like terms

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

Remember that

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

then;

$$\begin{gathered} S{N}^2=1-2\cos (a-b)+1 \\ S{N}^2=2-2\cos (a-b) \end{gathered}$$

Remember that

$PQ=SN$

then

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

Thus

$2-2\cos (a-b)=2-2\cos a\cos b-2\sin a\sin b$

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

$-2\cos (a-b)=-2\cos a\cos b-2\sin a\sin b$

Divide both sides by -2 on both sides

$\cos (a-b)=\cos a\cos b+\sin a\sin b$

Summing of cosine functions

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

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

Note that

$\cos (-b)=\cos b$

and

$\sin (-b)=-\sin b$

therefore

$$\begin{gathered} \cos (a+b)=\cos a\cos (-b)+\sin a\sin (-b) \\ \cos (a+b)=\cos a\cos b-\sin a\sin b \end{gathered}$$

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

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

Note that line EG = EH + HG, thus

$$\begin{gathered} \sin (\alpha +\beta )=\frac{EH+HG}{AE} \\ \sin (\alpha +\beta )=\frac{EH}{AE}+\frac{HG}{AE} \end{gathered}$$

Recall;

$HG=DC$

Thus

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

See that

$\angle DAC=\angle FDH$

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

$\angle HED+90°+90°-\alpha =180°$

$\angle HED+180°-180°=\alpha$

$\angle HED=\alpha$

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

$$\begin{gathered} \cos \alpha =\frac{EH}{ED} \\ EH=ED\cos \alpha \end{gathered}$$

Recall that

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

Substitute the value of EH

$$\begin{gathered} \sin (\alpha +\beta )=\frac{ED\cos \alpha }{AE}+\frac{DC}{AE} \\ \sin (\alpha +\beta )=(\frac{ED}{AE}\times \cos \alpha )+\frac{DC}{AE} \end{gathered}$$

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

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

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

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

From the right-angled triangle ADC, using SOHCAHTOA

$$\begin{gathered} \sin \alpha =\frac{DC}{AD} \\ DC=AD\sin \alpha \end{gathered}$$

Substitute the value of DC in the equation

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

Looking at the right-angled triangle AED and using SOHCAHTOA

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

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

$$\begin{gathered} \sin (\alpha +\beta )=\sin \beta \cos \alpha +\cos \beta \sin \alpha \\ \sin (\alpha +\beta )=\sin \alpha \cos \beta +\sin \beta \cos \alpha \end{gathered}$$

Difference of its functions

Knowing that

$\sin (\alpha +\beta )=\sin \alpha \cos \beta +\sin \beta \cos \alpha$

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

Therefore

$\sin (\alpha -\beta )=\sin \alpha \cos (-\beta )+\sin (-\beta )\cos \alpha$

Note that

$\cos (-\beta )=\cos \beta$

and

$\sin (-\beta )=-\sin \beta$

therefore

$\sin (\alpha -\beta )=\sin \alpha \cos \beta -\sin \beta \cos \alpha$

Proving Sum and Difference of Tangent Functions

Summing of tangent functions

Recall that

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

Therefore

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

Thus

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

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

$$\begin{gathered} \tan (A+B)=\frac{\frac{\sin A\cos B}{\cos A\cos B}+\frac{\sin B\cos A}{\cos A\cos B}}{\frac{\cos A\cos B}{\cos A\cos B}-\frac{\sin A\sin B}{\cos A\cos B}} \\ \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \end{gathered}$$

Difference of tangent functions

Recall that

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

Therefore

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

Thus

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

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

$$\begin{gathered} \tan (A-B)=\frac{\frac{\sin A\cos B}{\cos A\cos B}-\frac{\sin B\cos A}{\cos A\cos B}}{\frac{\cos A\cos B}{\cos A\cos B}+\frac{\sin A\sin B}{\cos A\cos B}} \\ \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B} \end{gathered}$$

Sum and difference of formulas application

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