Multiple and Sub Multiple Angle Properties

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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.

Download SOLVED Practice Questions of Multiple and Sub Multiple Angle Properties for FREE
Trigonometry
grade 11 | Questions Set 1
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Download SOLVED Practice Questions of Multiple and Sub Multiple Angle Properties for FREE
Trigonometry
grade 11 | Questions Set 1
Trigonometry
grade 11 | Answers Set 1
Trigonometry
grade 11 | Questions Set 2
Trigonometry
grade 11 | Answers Set 2
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