Compound Angle Formulae

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°)