Compound Angle Formulae | Examples | A Level Maths Revision Notes

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1 Summary

1.1 Compound angle formulas are:

1.2 Half angle formulas are:

1.3 Function to trigonometric form:

1.4 All the compound angle formulas are listed below:

1.5 Double Angle formulae

1.5.1 Example #1

1.5.2 Example #2

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.