# Compound Angle Formulae

Contents

## Summary

### Compound angle formulas are:

• $\sin { (A\quad +\quad B) } \quad =\quad \cos { \quad A\quad \sin { \quad B } } +\quad \sin { \quad A\quad \cos { \quad B } }$
• $\sin { (A\quad -\quad B) } \quad =\quad \sin { \quad A\quad \cos { \quad B\quad -\quad \cos { \quad A\quad \sin { \quad B } } } }$
• $\cos { (A\quad +\quad B) } \quad =\quad \cos { \quad A\quad } \cos { \quad B } \quad -\quad \sin { \quad A } \quad \sin { \quad B }$
• $\cos { (A\quad -\quad B) } \quad =\quad \cos { \quad A\quad } \cos { \quad B } \quad +\quad \sin { \quad A } \quad \sin { \quad B }$
• $\tan { (A\quad -\quad B) } \quad =\quad \frac { \tan { \quad A\quad -\quad \tan { \quad B } } }{ 1\quad +\quad \tan { A\tan { B } \quad \quad } } \quad$
• $\tan { (A\quad +\quad B) } \quad =\quad \frac { \tan { \quad A\quad +\quad \tan { \quad B } } }{ 1\quad -\quad \tan { A\tan { B } \quad \quad } } \quad$

### Half angle formulas are:

• $\sin { (2A) } \quad =\quad 2\quad \cos { A } \quad \sin { A }$
• $\cos { 2A\quad =\quad { cos }^{ 2 } } A\quad -\quad { sin }^{ 2 }A,\quad cos2A\quad =\quad 1\quad -\quad 2{ sin }^{ 2 }A,\quad cos2A\quad =\quad 2{ cos }^{ 2 }A\quad -\quad 1$
• $tan(2A)\quad =\quad \frac { 2tanA }{ 1\quad -\quad { tan }^{ 2 }A }$

### Function to trigonometric form:

• $\alpha \quad cos\theta \quad +\quad b\quad sin\theta \quad =\quad R\quad cos(\theta \quad +\quad \alpha )$

In Fig 1, $P\hat { O } { \underset { \sim }{ O } }\quad =\quad A$ and $L\hat { O } { \underset { \sim }{ O } }\quad =\quad B$ are acute angles and $L\hat { O } { N }\quad =\quad A\quad +\quad B$  $L\underset { \sim }{ O } \quad is\quad \bot \quad to\quad OY$ $LN\quad is\quad \bot \quad to\quad OP\quad \& \quad \underset { \sim }{ O } P\quad is\quad \bot \quad to\quad OP$ $\underset { \sim }{ O } M\quad is\quad \bot \quad to\quad LN$ $OP\quad is\quad parallel\quad to\quad M\underset { \sim }{ O }$ $Now\quad M\hat { L } \underset { \sim }{ O } \quad =\quad A\quad an\quad L\overset { \wedge }{ \underset { \sim }{ O } } M\quad =\quad 90\quad -\quad A$ $sin(A\quad +\quad B)\quad =\quad \frac { LN }{ OL } \quad =\quad \frac { LM\quad +\quad MN }{ OL } \quad =\quad \frac { LM }{ OL } \quad +\quad \frac { MN }{ OL }$

As $MN\quad =\quad \underset { \sim }{ O } P$ $=\quad \frac { LM }{ LQ } \quad *\quad \frac { LQ }{ OL } \quad +\quad \frac { QP }{ OQ } \quad *\quad \frac { OQ }{ OL }$ $=\quad cosA\quad sinB\quad +\quad sinA\quad cosB$

Hence, $sin(A\quad +\quad B)\quad =\quad cosA\quad sinB\quad +\quad sinA\quad cosB\quad \quad \quad =>\quad i)$

Replacing B by -B in i) $sin(A\quad -\quad B)\quad =\quad sinA\quad cosA\quad -\quad cosA\quad sinB\quad \quad \quad =>\quad ii)$

Also from the diagram: $cos(A\quad +\quad B)\quad =\quad \frac { ON }{ OL } \quad =\quad \frac { OP\quad -\quad PN }{ OL } \quad =\quad \frac { OP }{ OL } \quad -\quad \frac { PN }{ OL }$ $=\quad \frac { OP }{ OQ } \quad *\quad \frac { OQ }{ OL } \quad +\quad \frac { PN }{ LQ } \quad *\quad \frac { LQ }{ OL }$

Hence, $cos(A\quad+\quad B)\quad =\quad cosA\quad cosB\quad -\quad sinA\quad sinB\quad \quad \quad =>\quad iii)$ $cos(A\quad -\quad B)\quad =\quad cosA\quad cosB\quad +\quad sinA\quad sinB\quad \quad \quad =>\quad iv)$

Also: $tan(A\quad -\quad B)\quad =\quad \frac { tanA\quad -\quad tanB }{ 1\quad +\quad tanA\quad tanB } \quad \quad \quad =>\quad v)$ $tan(A\quad +\quad B)\quad =\quad \frac { tanA\quad +\quad tanB }{ 1\quad -\quad tanA\quad tanB } \quad \quad \quad =>\quad vi)$

We have derived the compound angle formulae above.

### All the compound angle formulas are listed below:

• $\sin { (A\quad +\quad B) } \quad =\quad \cos { \quad A\quad \sin { \quad B } } +\quad \sin { \quad A\quad \cos { \quad B } }$
• $\sin { (A\quad -\quad B) } \quad =\quad \sin { \quad A\quad \cos { \quad B\quad -\quad \cos { \quad A\quad \sin { \quad B } } } }$
• $\cos { (A\quad +\quad B) } \quad =\quad \cos { \quad A\quad } \cos { \quad B } \quad -\quad \sin { \quad A } \quad \sin { \quad B }$
• $\cos { (A\quad -\quad B) } \quad =\quad \cos { \quad A\quad } \cos { \quad B } \quad +\quad \sin { \quad A } \quad \sin { \quad B }$
• $\tan { (A\quad -\quad B) } \quad =\quad \frac { \tan { \quad A\quad -\quad \tan { \quad B } } }{ 1\quad +\quad \tan { A\tan { B } \quad \quad } } \quad$
• $\tan { (A\quad +\quad B) } \quad =\quad \frac { \tan { \quad A\quad +\quad \tan { \quad B } } }{ 1\quad -\quad \tan { A\tan { B } \quad \quad } } \quad$

### Double Angle formulae

We use compound angle formulas from above to find double angle formula i.e

If: $sin(A\quad +\quad B)\quad =\quad cosA\quad sinB\quad +\quad sinA\quad cosB$

Replacing B by A: $sin(A\quad +\quad A)\quad =\quad cosA\quad sinA\quad +\quad sinAcosA$ $sin(2A)\quad =\quad 2\quad cosA\quad sinA$

Similar replace B by A: $cos(A\quad +\quad B)\quad =\quad cosA\quad cosB\quad -\quad sinA\quad sinB$ $cos(A\quad +\quad A)\quad =\quad cosA\quad cosA\quad -\quad sinA\quad sinA$ $cos\quad 2A\quad =\quad { cos }^{ 2 }A\quad -\quad { sin }^{ 2 }A$ $cos\quad 2A\quad =\quad 1\quad -\quad { 2sin }^{ 2 }A$ $cos\quad 2A\quad =\quad 2{ cos }^{ 2 }A\quad -\quad 1$

Finally, $tan(A\quad +\quad A)\quad =\quad \frac { tanA\quad +\quad tanA }{ 1\quad -\quad tanAtanA }$ $tan(2A)\quad =\quad \frac { tanA\quad +\quad tanA }{ 1\quad -\quad tanAtanA }$ $tan(2A)\quad =\quad \frac { 2tanA }{ 1\quad -\quad { tan }^{ 2 }A }$

#### Example #1

Q. Given that $sin\theta \quad =\quad \frac { 5 }{ 13 }$, where $\theta$ is obtuse, find the exact value of

i) sin 2 $\theta$

ii) cos 2 $\theta$

iii)tan 2 $\theta$

Solution:

As $sin\theta \quad =\quad \frac { 5 }{ 13 }$ By pythagoras theorem ${ 5 }^{ 2 }\quad +\quad { x }^{ 2 }\quad =\quad { 13 }^{ 2 }$ $x\quad =\quad 12$

So, now we know $cos\theta \quad =\quad \frac { -12 }{ 13 }$ and $tan\theta \quad =\quad \frac { -5 }{ 12 }$ (minus sign because theta is in the second quadrant and cos & tan are negative in the second quadrant)

i) $sin\theta \quad =\quad 2sin\theta \quad cos\theta$ $=\quad 2(\frac { 5 }{ 13 } )(\frac { -12 }{ 13 } )$

Ans: $sin2\theta \quad =\quad \frac { -120 }{ 169 }$

ii) $cos2\theta \quad =\quad { cos }^{ 2 }\theta \quad -\quad { sin }^{ 2 }\theta$ $=\quad { (\frac { -12 }{ 13 } ) }^{ 2 }\quad -\quad { \left( \frac { 5 }{ 13 } \right) }^{ 2 }$

Ans: $cos2\theta \quad =\quad \frac { 119 }{ 169 }$

iii) $tan2\theta \quad =\quad \frac { 2tan\theta }{ 1\quad -\quad { tan }^{ 2 }\theta }$ $=\quad \frac { 2tan(-{ 5 }/{ 12 }) }{ 1\quad -\quad { tan }^{ 2 }(-{ 5 }/{ 12 }) }$

Ans: $tan2\theta \quad =\quad \frac { -120 }{ 119 }$

Function $\alpha \quad cos\quad \theta \quad +\quad b\quad sin\quad \theta$

Suppose a function $f\left( \theta \right) \quad =\quad \alpha \quad cos\quad \theta \quad +\quad b\quad sin\quad \theta$ , where a and b are constants, it can be easily converted into a single trigonometric function of the form $R\quad cos(\theta \quad +\quad \alpha )$ where R is a positive constant and $\alpha$ is an acute angle.

#### Example #2

If we have a function: $f\left( \theta \right) \quad =\quad 7\quad cos\quad \theta \quad +\quad 24\quad sin\quad \theta$

We equate it with its trigonometric form: $7\quad cos\quad \theta \quad +\quad 24\quad sin\quad \theta \quad =\quad R\quad cos(\theta \quad +\quad \alpha )$

Expand the right hand side: $7\quad cos\quad \theta \quad +\quad 24\quad sin\quad \theta \quad =\quad R\quad cos\theta cos\alpha \quad -\quad Rsin\theta sin\alpha \quad$

Compare the coefficients of $cos\theta$ and $sin\theta$: $7\quad =\quad R\quad cos\alpha$ $24\quad =\quad R\quad sin\alpha$

To find R we take the magnitude of 7 and 24 $R\quad =\quad \sqrt { { (7) }^{ 2 }\quad +\quad { (24) }^{ 2 } }$ $R\quad =\quad 25$

To find $\alpha$ we do: $tan(\alpha )\quad =\quad \frac { 27 }{ 7 }$ $\alpha \quad =\quad 73.7$°

Ans: Hence the trigonometric form is $25cos(\theta \quad +\quad 73.7$° $)$