# Multiple and Sub Multiple Angle Properties

Now, let us list down the multiple and sub-multiple angle properties followed by the trigonometric functions. We’ll justify a few of these properties and the rest can be inferred analogously.

• $$\sin \left( {A \pm B} \right) = \sin A\cos B \pm \cos A\sin B$$

Consider a unit circle with angles A and B depicted as shown:

Similarly, we can prove the relation for $$\sin \left( {A - B} \right).$$

• $$\cos (A \pm B) = \cos A\cos B \mp \sin A\sin B$$

Referring to Fig - 8, we have

\begin{align}&\cos (A + B) = OU = OT - TU \\ \,\,\,\,\, &\qquad\qquad\qquad\quad\;\;\;= OR\cos A - RS \\ \,\,\,\,\,&\qquad\qquad\qquad\quad\;\;\;= OR\cos A - PQ\cos \theta\\ \,\,\,\,\,&\qquad\qquad\qquad\quad\;\;\;= \cos A\cos B - \sin A\sin B \\ \end{align}

• \begin{align}tan (A \pm B) = \frac{{\tan A \pm \tan B}}{{1 \mp \tan A\tan B}}\end{align}

We can prove this both geometrically and algebraically, as we do here:

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

Dividing the numerator and denominator on the RHS by cos A cos B gives us the desired relation.
We can similarly arrive at such relations for the other three trigonometric rations

• \begin{align}\sin 2A = 2\sin A\cos A = \frac{{2\tan A}}{{1 + {{\tan }^2}A}}\end{align}

Proof:

\begin{align}& \sin 2A = \sin (A + A) = \sin A\cos A + \cos A\sin A \\ \;\;\;\;\;\;\, &\qquad\qquad\qquad\quad\;\;\;\quad= 2\sin A\cos A \\ \;\;\;\;\;\;\, &\qquad\qquad\qquad\quad\;\;\;\quad= 2\frac{{\sin A}}{{\cos A}}{\cos ^2}A \\ \;\;\;\;\;\;\, &\qquad\qquad\qquad\quad\;\;\;\quad= \frac{{2\tan A}}{{{{\sec }^2}A}} = \frac{{2\tan A}}{{1 + {{\tan }^2}A}} \end{align}

• \begin{align}\cos 2A = {\cos ^2}A - {\sin ^2}A = \frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = 2{\cos ^2}A - 1 = 1 - 2{\sin ^2}A\end{align}
• $$\sin 3A = 3\sin A - 4{\sin ^3}A$$

Proof:

\begin{align}&\sin 3A = \sin (2A + A) = \sin 2A\cos A + \cos 2A\sin A \\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;= 2\sin A{\cos ^2}A + (1 - 2{\sin ^2}A)\sin A\\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;= 2\sin A(1 - {\sin ^2}A) + (1 - 2{\sin ^2}A)\sin A \\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\quad\;= 3\sin A - 4{\sin ^3}A \end{align}

• $$\cos 3A = 4{\cos ^3}A - 3\cos A$$
• \begin{align}\tan 3A = \frac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}\end{align}
• \begin{align}\sin \frac{A}{2} = \sqrt {\frac{{1 - \cos A}}{2}}\end{align}

Proof:  \begin{align}\cos A = \cos \left( {2\left( {\frac{A}{2}} \right)} \right) = 1 - 2{\sin ^2}\frac{A}{2}\end{align}

The result follows from rearranging this relation.

• \begin{align}\cos \frac{A}{2} = \sqrt {\frac{{1 + \cos A}}{2}} \end{align}
• \begin{align}\tan \frac{A}{2} = \sqrt {\frac{{1 - \cos A}}{{1 + \cos A}}} = \frac{{1 - \cos A}}{{\sin A}} = \frac{{\sin A}}{{1 + \cos A}}\end{align}
• \begin{align}\sin C + \sin D = 2\sin \frac{{C + D}}{2}\cos \frac{{C - D}}{2}\end{align}

Proof: Let  \begin{align}\frac{{C + D}}{2} = A\;{\text{and}}\;\frac{{C - D}}{2} = B \Rightarrow A + B = C, A - B = D\end{align}
The RHS is:
\begin{align}&2\sin A\cos B = \sin (A + B) + \sin (A - B) ({\text{Verify}}) \\ \,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\quad\;= \sin C + \sin D \\ \,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\quad\;= {\text{LHS}} \\ \end{align}

\begin{align}&\bullet \qquad \sin C - \sin D = 2\cos \frac{{C + D}}{2}\;\sin \frac{{C - D}}{2} \\&\bullet \qquad\cos C + \cos D = 2\cos \frac{{C + D}}{2}\;\cos \frac{{C - D}}{2} \\&\bullet \qquad\cos C - \cos D = 2\sin \frac{{C + D}}{2}\;\sin \frac{{D - C}}{2}\end{align}

Observe that the last angle on the RHS is \begin{align}\frac{{D - C}}{2}\end{align}

\begin{align}&\bullet \qquad\tan C + \tan D = \frac{{\sin (C + D)}}{{\cos C\cos D}}\\&\bullet \qquad \sin (A + B)\sin (A - B) = {\sin ^2}A - {\sin ^2}B\\&\bullet \qquad \cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B\\&\bullet \qquad {\sin ^2}A = \frac{1}{2}(1 - \cos 2A) {\cos ^2}A = \frac{1}{2}(1 + \cos 2A)\\&\bullet \qquad {\tan ^2}A = \frac{{1 - \cos 2A}}{{1 + \cos 2A}}\\&\bullet \qquad {\sin ^3}A = \frac{{3\sin A - \sin 3A}}{4} {\cos ^3}A = \frac{{3\cos A + \cos 3A}}{4}\end{align}

These relations are very frequently used and it is advisable to commit these to memory as soon as possible.