Trigonometric ratios of compound Angles

Definition :

The algebraic sum of two or more angles is called a ' compound angle ' .

If A , B , C are angles , then A + B , A - B , A + B - C  , A - B + C , A + B + C , etc., are compound angles .

Formulas

cos(A + B)  =  cosA cosB  -  sinA sinB  for all A , B in  R .
cos(A - B)   =  cosA cosB  + sinA sinB  for all A , B  in  R .
sin(A + B)   =  sinA cosB  +  cosA sinB  for all  A , B in  R.
sin(A - B)    =  sinA cosB   -  cosA sinB  for all A , B  in  R.
tan(A + B)   =  (tanA+tanB)/(1-tanAtanB)   if  none of A , B , A + B is an odd multiple of (pi)/2  .

cot(A + B)   =  (cotBcotA-1)/(cotB+cotA)   if  none of A , B , A + B is an odd multiple of   (pi)/2 .
tan(A - B)    =   (tanA-tanB)/(1+tanAtanB)    if  none of A , B , A - B is an odd multiple of (pi)/2.
cot(A - B)    =   (cotBcotA+1)/(cotB-cotA)    if  none of A , B , A + B is an odd multiple of  (pi)/2.
sin(A + B) sin(A - B)   =  sin2A - sin2B    =    cos2B  -  cos2A .
cos(A + B) cos(A - B)   =  cos2A - sin2B   =   cos2B  -  sin2A


Solved Problems on Trigonometric Ratios of Compound Angles

1) Find the value of sin 105o and cos 165o .

Sol: sin 105o   =   sin (45o + 60o)

=    sin45o  cos60o   +  cos45o sin60o

=   (1/sqrt(2)) (1/2)   +  (1/sqrt(2)) (sqrt(3)/2)

sin105o   =   (sqrt(3)+1)/(2sqrt(2))

Now , cos165o   =   cos (180o - 15o)    =   - cos15o

=   - cos(45o - 30o)

=   - [ cos45o cos30o + sin 45o sin 30o]

=   -[(1/sqrt(2))(sqrt(3)/2)+(1/sqrt(2))(1/2)]

cos 165o   =   -(sqrt(3)+1)/(2sqrt(2))

2) Show that   cos42o + cos78o + cos162o = 0 .

Solution :  cos42o + cos78o + cos162o

=   cos (60o - 18o)  +  cos(60o + 18o)  +  cos(180o - 18o)

=   (cos60o cos18o + sin60o sin18o)  +  (cos60ocos18o - sin60o sin18o)  -  cos18o

=   2 cos60ocos18o  -  cos18o

=   2(1/2) cos18o  -  cos18o  =  0


More Problems on Trigonometric Ratios of Compound Angles:

1)Find the values of sin 15o , cos 15o , tan15o

Sol:

sin15o   =  sin (60o - 45o)   =  sin 60o cos45o  -  cos60o sin45o

=  (sqrt(3)/2)(1/sqrt(2)) -  (1/2)(1/sqrt(2))

sin 15o  =  (sqrt(3)-1)/(2sqrt(2))

cos 15o   =   cos(60o - 45o)   =   cos60o cos45o  +  sin60o sin45o

=   (1/2)(1/(sqrt(2)))   +  (sqrt(3)/2)(1/sqrt(2))

cos15o     =   (sqrt(3)+1)/(2sqrt(2))

tan15o   =   tan(60o - 45o)   =   (tan60^o-tan45^o)/(1+tan60^otan45^o)

=    (sqrt(3)-1)/(2sqrt(2))

tan15o   =   2 -  sqrt(3)

2) Show that  cos100o cos40o  +  sin100o sin40o   =  1/2

Solution :  cos100o cos40o  +  sin100o sin40o

The above equation is in the form of  cosA cosB  +  sinA sinB  which is equal to cos(A - B)

Here A = 100o  and  B = 40o

cos(A - B)  =  cos(100o - 40o)

=  cos(60o)

=  1/2

Hence  cos100o cos40o  +  sin100o sin40o  =  1/2