Compound Angles
The Compound Angle Formulae or Sine, Cosine and Tangent in Solving Trigonometric Problems
The aim is to express sin ( α ± β ) \sin(\alpha \pm \beta) sin ( α ± β ) cos ( α ± β ) \cos(\alpha \pm \beta) cos ( α ± β ) sin α \sin\alpha sin α sin β \sin\beta sin β cos α \cos\alpha cos α cos β \cos\beta cos β
Consider the following diagram:
From the figure above
sin ( α + β ) = B D B C ..................(1) \sin(\alpha + \beta) = \frac{BD}{BC} \quad \text{..................(1)} sin ( α + β ) = B C B D ..................(1)
From the same figure sin α = B D A B = E C A C \sin\alpha = \frac{BD}{AB} = \frac{EC}{AC} sin α = A B B D = A C E C cos α = A E A C = A D A B \cos\alpha = \frac{AE}{AC} = \frac{AD}{AB} cos α = A C A E = A B A D sin β = E C B C \sin\beta = \frac{EC}{BC} sin β = B C E C cos β = E B B C \cos\beta = \frac{EB}{BC} cos β = B C E B
Now sin ( α + β ) = B D B C = A B sin α B C \sin(\alpha + \beta) = \frac{BD}{BC} = \frac{AB\sin\alpha}{BC} sin ( α + β ) = B C B D = B C A B s i n α
sin ( α + β ) = ( A E + E B ) sin α B C \sin(\alpha + \beta) = \frac{(AE + EB)\sin\alpha}{BC} sin ( α + β ) = B C ( A E + E B ) s i n α
= A E B C sin α + E B B C sin α = \frac{AE}{BC}\sin\alpha + \frac{EB}{BC}\sin\alpha = B C A E sin α + B C E B sin α
= ( A E B C ) × ( A C A C ) + ( E B B C ) sin α = \left(\frac{AE}{BC}\right) \times \left(\frac{AC}{AC}\right) + \left(\frac{EB}{BC}\right)\sin\alpha = ( B C A E ) × ( A C A C ) + ( B C E B ) sin α
But A E A C = cos α \frac{AE}{AC} = \cos\alpha A C A E = cos α E C B C = sin β \frac{EC}{BC} = \sin\beta B C E C = sin β E B B C = cos β \frac{EB}{BC} = \cos\beta B C E B = cos β
It follows that sin ( α + β ) = cos α sin β + cos β sin α \sin(\alpha + \beta) = \cos\alpha\sin\beta + \cos\beta\sin\alpha sin ( α + β ) = cos α sin β + cos β sin α
∴ sin ( α + β ) = cos α sin β + cos β sin α \therefore \boxed{\sin(\alpha + \beta) = \cos\alpha\sin\beta + \cos\beta\sin\alpha} ∴ sin ( α + β ) = cos α sin β + cos β sin α
From ΔBCD
For cos ( α ± β ) \cos(\alpha \pm \beta) cos ( α ± β ) α \alpha α β \beta β
d 2 = 2 − 2 cos ( α − β ) d^2 = 2 - 2\cos(\alpha - \beta) d 2 = 2 − 2 cos ( α − β )
Therefore 2 − 2 ( cos α cos β + sin α sin β ) = 2 − 2 cos ( α − β ) 2 - 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = 2 - 2\cos(\alpha - \beta) 2 − 2 ( cos α cos β + sin α sin β ) = 2 − 2 cos ( α − β )
Or cos ( α − β ) = cos α cos β + sin α sin β \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta cos ( α − β ) = cos α cos β + sin α sin β
∴ cos ( α − β ) = cos α cos β + sin α sin β \therefore \boxed{\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta} ∴ cos ( α − β ) = cos α cos β + sin α sin β
Because cosine and sine are even and odd functions respectively, then we can find cos ( α + β ) \cos(\alpha + \beta) cos ( α + β ) sin ( α − β ) \sin(\alpha - \beta) sin ( α − β )
α + β = α − ( − β ) \alpha + \beta = \alpha - (-\beta) α + β = α − ( − β )
So cos ( α + β ) = cos α cos ( − β ) + sin α sin ( − β ) \cos(\alpha + \beta) = \cos\alpha\cos(-\beta) + \sin\alpha\sin(-\beta) cos ( α + β ) = cos α cos ( − β ) + sin α sin ( − β )
But cos ( − β ) = cos β \cos(-\beta) = \cos\beta cos ( − β ) = cos β sin ( − β ) = − sin β \sin(-\beta) = -\sin\beta sin ( − β ) = − sin β
Then cos ( α + β ) = cos α cos β − sin α sin β \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta cos ( α + β ) = cos α cos β − sin α sin β
∴ cos ( α + β ) = cos α cos β − sin α sin β \therefore \boxed{\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta} ∴ cos ( α + β ) = cos α cos β − sin α sin β
cos ( α + β ) = cos α cos β − sin α sin β \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta cos ( α + β ) = cos α cos β − sin α sin β
cos ( α − β ) = cos α cos β + sin α sin β \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta cos ( α − β ) = cos α cos β + sin α sin β
sin ( α + β ) = cos α sin β + cos β sin α \sin(\alpha + \beta) = \cos\alpha\sin\beta + \cos\beta\sin\alpha sin ( α + β ) = cos α sin β + cos β sin α
sin ( α − β ) = sin α cos β − cos α sin β \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta sin ( α − β ) = sin α cos β − cos α sin β
Example 1
Without using tables find the value of each of the following:
sin 75 ∘ \sin 75^\circ sin 7 5 ∘ cos 105 ∘ \cos 105^\circ cos 10 5 ∘
Solution:
(a) sin 75 ∘ \sin 75^\circ sin 7 5 ∘
75 ∘ = 45 ∘ + 30 ∘ 75^\circ = 45^\circ + 30^\circ 7 5 ∘ = 4 5 ∘ + 3 0 ∘
sin 75 ∘ = sin ( 45 ∘ + 30 ∘ ) \sin 75^\circ = \sin(45^\circ + 30^\circ) sin 7 5 ∘ = sin ( 4 5 ∘ + 3 0 ∘ ) = sin 45 ∘ cos 30 ∘ + sin 30 ∘ cos 45 ∘ = \sin 45^\circ \cos 30^\circ + \sin 30^\circ \cos 45^\circ = sin 4 5 ∘ cos 3 0 ∘ + sin 3 0 ∘ cos 4 5 ∘
sin 75 ∘ = 2 2 × 3 2 + 1 2 × 2 2 = 1 4 ( 6 + 2 ) \sin 75^\circ = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{\sqrt{2}}{2} = \frac{1}{4}(\sqrt{6} + \sqrt{2}) sin 7 5 ∘ = 2 2 × 2 3 + 2 1 × 2 2 = 4 1 ( 6 + 2 )
∴ sin 75 ∘ = 1 4 ( 6 + 2 ) \therefore \sin 75^\circ = \frac{1}{4}(\sqrt{6} + \sqrt{2}) ∴ sin 7 5 ∘ = 4 1 ( 6 + 2 )
(b) cos 105 ∘ \cos 105^\circ cos 10 5 ∘
105 ∘ = 60 ∘ + 45 ∘ 105^\circ = 60^\circ + 45^\circ 10 5 ∘ = 6 0 ∘ + 4 5 ∘
cos 105 ∘ = cos ( 60 ∘ + 45 ∘ ) \cos 105^\circ = \cos(60^\circ + 45^\circ) cos 10 5 ∘ = cos ( 6 0 ∘ + 4 5 ∘ )
cos 105 ∘ = cos 60 ∘ cos 45 ∘ − sin 60 ∘ sin 45 ∘ \cos 105^\circ = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ cos 10 5 ∘ = cos 6 0 ∘ cos 4 5 ∘ − sin 6 0 ∘ sin 4 5 ∘ = 1 2 × 2 2 − 3 2 × 2 2 = \frac{1}{2} \times \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} = 2 1 × 2 2 − 2 3 × 2 2 = 1 4 ( 2 − 6 ) = \frac{1}{4}(\sqrt{2} - \sqrt{6}) = 4 1 ( 2 − 6 )
cos 105 ∘ = 2 − 6 4 \boxed{\cos 105^\circ = \frac{\sqrt{2} - \sqrt{6}}{4}} cos 10 5 ∘ = 4 2 − 6
Example 2
Find:
sin 150 ∘ \sin 150^\circ sin 15 0 ∘ cos 15 ∘ \cos 15^\circ cos 1 5 ∘
150 ∘ = 90 ∘ + 60 ∘ 150^\circ = 90^\circ + 60^\circ 15 0 ∘ = 9 0 ∘ + 6 0 ∘
sin 150 ∘ = sin ( 90 ∘ + 60 ∘ ) \sin 150^\circ = \sin(90^\circ + 60^\circ) sin 15 0 ∘ = sin ( 9 0 ∘ + 6 0 ∘ ) = sin 90 ∘ ⋅ cos 60 ∘ + sin 60 ∘ ⋅ cos 90 ∘ = \sin 90^\circ \cdot \cos 60^\circ + \sin 60^\circ \cdot \cos 90^\circ = sin 9 0 ∘ ⋅ cos 6 0 ∘ + sin 6 0 ∘ ⋅ cos 9 0 ∘ = 1 × 1 2 + 3 2 × 0 = 1 2 = 1 \times \frac{1}{2} + \frac{\sqrt{3}}{2} \times 0 = \frac{1}{2} = 1 × 2 1 + 2 3 × 0 = 2 1
∴ sin 150 ∘ = 1 2 \therefore \sin 150^\circ = \frac{1}{2} ∴ sin 15 0 ∘ = 2 1
cos 15 ∘ = 45 ∘ − 30 ∘ \cos 15^\circ = 45^\circ - 30^\circ cos 1 5 ∘ = 4 5 ∘ − 3 0 ∘ 60 ∘ − 45 ∘ 60^\circ - 45^\circ 6 0 ∘ − 4 5 ∘
cos 15 ∘ = cos 45 ∘ ⋅ cos 30 ∘ + sin 45 ∘ ⋅ sin 30 ∘ \cos 15^\circ = \cos 45^\circ \cdot \cos 30^\circ + \sin 45^\circ \cdot \sin 30^\circ cos 1 5 ∘ = cos 4 5 ∘ ⋅ cos 3 0 ∘ + sin 4 5 ∘ ⋅ sin 3 0 ∘ = 1 2 × 2 2 + 3 2 × 2 2 = \frac{1}{2} \times \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} = 2 1 × 2 2 + 2 3 × 2 2 = 1 4 ( 2 + 6 ) = \frac{1}{4}(\sqrt{2} + \sqrt{6}) = 4 1 ( 2 + 6 )
∴ cos 15 ∘ = 1 4 ( 2 + 6 ) \therefore \cos 15^\circ = \frac{1}{4}(\sqrt{2} + \sqrt{6}) ∴ cos 1 5 ∘ = 4 1 ( 2 + 6 )
