Trigonometric Ratios of Compound Angles - Practice Questions & MCQ

Trigonometric Ratio for Compound Angles (Part 1)

The sum or difference of two or more angles is called a compound angle. If $A, B$ and $C$ are any angle then $A+B, A-B, A+B+C, A+$ $B$ - C etc all are compound angles.
1. $\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta$
2. $\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta$
3. $\sin (\alpha-\beta)=\sin \alpha \cos \beta-\cos \alpha \sin \beta$
4. $\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$

Proof:

1. cos (α - β) 

Let's consider two points on the unit circle. Point $P$ is at an angle $\alpha$ from the positive $x$-axis with coordinates ( $\cos \alpha$, sin $\alpha$ ) and point $Q$ is at an angle of $\beta$ from the positive $x$-axis with coordinates $(\cos \beta, \sin \beta)$. Note the measure of angle POQ is $\alpha-\beta$. Label two more points: $A$ at an angle of $(\alpha-\beta)$ from the positive $x$-axis with coordinates $(\cos (\alpha-\beta), \sin (\alpha-\beta)$ ); and point $B$ with coordinates $(1,0)$. Triangle POQ is a rotation of triangle AOB and thus the distance from P to Q is the same as the distance from A to B .

We can find the distance from P to Q using the distance formula.

$\begin{aligned} & d_{P Q}=\sqrt{(\cos \alpha-\cos \beta)^2+(\sin \alpha-\sin \beta)^2} \\ & \quad=\sqrt{\cos ^2 \alpha-2 \cos \alpha \cos \beta+\cos ^2 \beta+\sin ^2 \alpha-2 \sin \alpha \sin \beta+\sin ^2 \beta} \\ & =\sqrt{\left(\cos ^2 \alpha+\sin ^2 \alpha\right)+\left(\cos ^2 \beta+\sin ^2 \beta\right)-2 \cos \alpha \cos \beta-2 \sin \alpha \sin \beta} \\ & =\sqrt{1+1-2 \cos \alpha \cos \beta-2 \sin \alpha \sin \beta} \\ & =\sqrt{2-2 \cos \alpha \cos \beta-2 \sin \alpha \sin \beta}\end{aligned}$

Similarly, using the distance formula we can find the distance from A to B.

$
\begin{aligned}
& d_{A B}=\sqrt{(\cos (\alpha-\beta)-1)^2+(\sin (\alpha-\beta)-0)^2} \\
& \quad=\sqrt{\cos ^2(\alpha-\beta)-2 \cos (\alpha-\beta)+1+\sin ^2(\alpha-\beta)} \\
& =\sqrt{\left(\cos ^2(\alpha-\beta)+\sin ^2(\alpha-\beta)\right)-2 \cos (\alpha-\beta)+1} \\
& =\sqrt{1-2 \cos (\alpha-\beta)+1} \\
& =\sqrt{2-2 \cos (\alpha-\beta)}
\end{aligned}
$
Because the two distances are the same, we set them equal to each other and simplify

$
\begin{gathered}
\sqrt{2-2 \cos \alpha \cos \beta-2 \sin \alpha \sin \beta}=\sqrt{2-2 \cos (\alpha-\beta)} \\
2-2 \cos \alpha \cos \beta-2 \sin \alpha \sin \beta=2-2 \cos (\alpha-\beta)
\end{gathered}
$
Finally, we subtract 2 from both sides and divide both sides by -2

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

$
\text { 2. } \begin{aligned}
\cos & (\alpha+\beta) \\
\quad & =\cos (\alpha-(-\beta)) \\
\quad= & \cos \alpha \cos (-\beta)+\sin \alpha \sin (-\beta) \\
& =\cos \alpha \cos \beta-\sin \alpha \sin \beta
\end{aligned}
$
Hence, $\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta$
3. Sine Compound Angle Formulae

$
\begin{aligned}
& \text { We have } \\
& \qquad \begin{aligned}
\sin (\alpha-\beta) & =\cos \left(90^{\circ}-(\alpha-\beta)\right) \\
& =\cos \left(\left(90^{\circ}-\alpha\right)+\beta\right) \\
& =\cos \left(90^{\circ}-\alpha\right) \cos \beta-\sin \left(90^{\circ}-\alpha\right) \sin \beta \\
& =\sin \alpha \cos \beta-\cos \alpha \sin \beta
\end{aligned} \\
& \begin{aligned}
\sin (\alpha+\beta) & =\sin (\alpha-(-\beta)) \\
& =\sin \alpha \cos (-\beta)-\cos \alpha \sin (-\beta) \\
& =\sin \alpha \cos \beta+\cos \alpha \sin \beta
\end{aligned}
\end{aligned}
$
 

Trigonometric Ratio for Compound Angles (Part 2)
$\begin{aligned} & \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ & \tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta} \\ & \cot (\alpha+\beta)=\frac{\cot \alpha \cot \beta-1}{\cot \alpha+\cot \beta} \\ & \cot (\alpha-\beta)=\frac{\cot \alpha \cot \beta+1}{\cot \beta-\cot \alpha}\end{aligned}$

Proof:

Finding the sum of two angles formula for tangent involves taking the quotient of the sum formulas for sine and cosine and simplifying,

$
\begin{aligned}
\tan (\alpha+\beta) & =\frac{\sin (\alpha+\beta)}{\cos (\alpha+\beta)} \\
& =\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\cos \alpha \cos \beta-\sin \alpha \sin \beta}
\end{aligned}
$

[Divide the numerator and denominator by $\cos \alpha \cos \beta$ ]

$
\begin{aligned}
& =\frac{\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\cos \alpha \cos \beta-\sin \alpha \sin \beta}{\cos \alpha \cos \beta}} \\
& =\frac{\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}}{1-\frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}} \\
& =\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}
\end{aligned}
$

For the difference of tangent, put - β in place of β in the above result.

Cot Formulae

$
\begin{aligned}
\cot (\alpha+\beta) & =\frac{\cos (\alpha+\beta)}{\sin (\alpha+\beta)} \\
& =\frac{\cos \alpha \cos \beta-\sin \alpha \sin \beta}{\sin \alpha \cos \beta+\cos \alpha \sin \beta} \\
& =\frac{\frac{\cos \alpha \cos \beta-\sin \alpha \sin \beta}{\sin \alpha \sin \beta}}{\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\sin \alpha \sin \beta}}
\end{aligned}
$

[Divide the numerator and denominator by $\sin \alpha \sin \beta$ ]

$
\begin{aligned}
& =\frac{\frac{\cos \alpha \cos \beta}{\sin \alpha \sin \beta}-\frac{\sin \alpha \sin \beta}{\sin \alpha \sin \beta}}{\frac{\sin \alpha \alpha \cos \beta}{\sin \alpha \sin \beta}+\frac{\cos \alpha \sin \beta}{\sin \alpha \sin \beta}} \\
& =\frac{\cot \alpha \cot \beta-1}{\cot \beta+\cot \alpha}
\end{aligned}
$
For the difference of cotangent, put - $\beta$ in place of $\beta$ in the above result

Trigonometric Ratio for Compound Angles (Part 3)
Proof Cotangent of the Sum and Difference of two Angles

$
\begin{aligned}
& \cot (\alpha+\beta)=\frac{\cot \alpha \cot \beta-1}{\cot \alpha+\cot \beta} \\
& \cot (\alpha-\beta)=\frac{\cot \alpha \cot \beta+1}{\cot \beta-\cot \alpha}
\end{aligned}
$
$
\begin{aligned}
\cot (\alpha+\beta) & =\frac{\cos (\alpha+\beta)}{\sin (\alpha+\beta)} \\
& =\frac{\cos \alpha \cos \beta-\sin \alpha \sin \beta}{\sin \alpha \cos \beta+\cos \alpha \sin \beta} \\
& =\frac{\frac{\cos \alpha \cos \beta-\sin \alpha \sin \beta}{\sin \alpha \sin \beta}}{\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\sin \alpha \sin \beta}}
\end{aligned}
$

[Divide the numerator and denominator by $\sin \alpha \sin \beta$ ]

$
\begin{aligned}
& =\frac{\frac{\cos \alpha \cos \beta}{\sin \alpha \sin \beta}-\frac{\sin \alpha \sin \beta}{\sin \alpha \sin \beta}}{\frac{\sin \alpha \cos \beta}{\sin \alpha \sin \beta}+\frac{\cos \alpha \sin \beta}{\sin \alpha \sin \beta}} \\
& =\frac{\cot \alpha \cot \beta-1}{\cot \beta+\cot \alpha}
\end{aligned}
$
For the difference of cotangent, put - $\beta$ in place of $\beta$ in the above derivation.

Trigonometric Ratio for Compound Angles (Some More Results)

$
\begin{aligned}
& \text { 1. } \sin (A+B) \sin (A-B)=\sin ^2 A-\sin ^2 B=\cos ^2 B-\cos ^2 A \\
& \text { 2. } \cos (A+B) \cos (A-B)=\cos ^2 A-\sin ^2 B=\cos ^2 B-\sin ^2 A \\
& \text { 3. } \tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A}
\end{aligned}
$
Proof:

$
\begin{aligned}
\sin (A+B) \sin (A-B) & =(\sin A \cos B+\cos A \sin B)(\sin A \cos B-\cos A \sin B) \\
& =\sin ^2 A \cos ^2 B-\cos ^2 A \sin ^2 B \\
& =\sin ^2 A\left(1-\sin ^2 B\right)-\left(1-\sin ^2 A\right) \sin ^2 B \\
& =\sin ^2 A-\sin ^2 A \sin ^2 B-\sin ^2 B+\sin ^2 A \sin ^2 B=\sin ^2 A-\sin ^2 B \\
& =\left(1-\cos ^2 A\right)-\left(1-\cos ^2 B\right)=\cos ^2 B-\cos ^2 A
\end{aligned}
$

1.
2.

$
\begin{aligned}
\cos (A+B) \cos (A-B) & =(\cos A \cos B-\sin A \sin B)(\cos A \cos B+\sin A \sin B) \\
& =\cos ^2 A \cos ^2 B-\sin ^2 A \sin ^2 B \\
& =\cos ^2 A\left(1-\sin ^2 B\right)-\left(1-\cos ^2 A\right) \sin ^2 B=\cos ^2 A-\sin ^2 B \\
& =\left(1-\sin ^2 A\right)-\left(1-\cos ^2 B\right)=\cos ^2 B-\sin ^2 A
\end{aligned}
$

3.

$
\begin{aligned}
& \tan (A+B+C)=\tan ((A+B)+C)=\frac{\tan (A+B)+\tan C}{1-\tan (A+B) \tan C} \\
& =\frac{\frac{\tan A+\tan B}{1-\tan A \tan B}+\tan C}{1-\left(\frac{\tan A \tan B}{1-\tan A \tan B}\right) \tan C}=\frac{\tan A+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A}
\end{aligned}
$