Conversion of sum or difference into product
Theorem: for any angles C C C D D D
sin C + sin D = 2 sin C + D 2 cos C − D 2 ( 1 ) \sin C+\sin D=2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}\qquad(1) sin C + sin D = 2 sin 2 C + D cos 2 C − D ( 1 )
sin C − sin D = 2 cos C + D 2 sin C − D 2 ( 2 ) \sin C-\sin D=2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}\qquad(2) sin C − sin D = 2 cos 2 C + D sin 2 C − D ( 2 )
cos C + cos D = 2 cos C + D 2 cos C − D 2 ( 3 ) \cos C+\cos D=2\cos\dfrac{C+D}{2}\cos\dfrac{C-D}{2}\qquad(3) cos C + cos D = 2 cos 2 C + D cos 2 C − D ( 3 )
cos C − cos D = − 2 sin C + D 2 sin C − D 2 = 2 sin C + D 2 sin D − C 2 ( 4 ) \cos C-\cos D=-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}=2\sin\dfrac{C+D}{2}\sin\dfrac{D-C}{2}\qquad(4) cos C − cos D = − 2 sin 2 C + D sin 2 C − D = 2 sin 2 C + D sin 2 D − C ( 4 )
Proof. Let A = C + D 2 A=\dfrac{C+D}{2} A = 2 C + D B = C − D 2 B=\dfrac{C-D}{2} B = 2 C − D A + B = C A+B=C A + B = C A − B = D A-B=D A − B = D
compound-angle identities sin ( A + B ) = sin A cos B + cos A sin B \sin(A+B)=\sin A\cos B+\cos A\sin B sin ( A + B ) = sin A cos B + cos A sin B sin ( A − B ) = sin A cos B − cos A sin B \sin(A-B)=\sin A\cos B-\cos A\sin B sin ( A − B ) = sin A cos B − cos A sin B
sin ( A + B ) + sin ( A − B ) = 2 sin A cos B , sin ( A + B ) − sin ( A − B ) = 2 cos A sin B \sin(A+B)+\sin(A-B)=2\sin A\cos B,\qquad \sin(A+B)-\sin(A-B)=2\cos A\sin B sin ( A + B ) + sin ( A − B ) = 2 sin A cos B , sin ( A + B ) − sin ( A − B ) = 2 cos A sin B
Substituting A + B = C , A − B = D , A = C + D 2 , B = C − D 2 A+B=C,A-B=D,A=\frac{C+D}2,B=\frac{C-D}2 A + B = C , A − B = D , A = 2 C + D , B = 2 C − D
Similarly, adding and subtracting cos ( A + B ) = cos A cos B − sin A sin B \cos(A+B)=\cos A\cos B-\sin A\sin B cos ( A + B ) = cos A cos B − sin A sin B cos ( A − B ) = cos A cos B + sin A sin B \cos(A-B)=\cos A\cos B+\sin A\sin B cos ( A − B ) = cos A cos B + sin A sin B
gives cos ( A + B ) + cos ( A − B ) = 2 cos A cos B \cos(A+B)+\cos(A-B)=2\cos A\cos B cos ( A + B ) + cos ( A − B ) = 2 cos A cos B cos ( A − B ) − cos ( A + B ) = 2 sin A sin B \cos(A-B)-\cos(A+B)=2\sin A\sin B cos ( A − B ) − cos ( A + B ) = 2 sin A sin B
gives formula (3), and formula (4) after applying sin ( − θ ) = − sin θ \sin(-\theta)=-\sin\theta sin ( − θ ) = − sin θ − sin C − D 2 -\sin\dfrac{C-D}2 − sin 2 C − D
sin D − C 2 \sin\dfrac{D-C}2 sin 2 D − C
Solved Examples
Ex.1(i) — Prove sin 40 ° − cos 70 ° = 3 cos 80 ° \sin40°-\cos70°=\sqrt3\cos80° sin 40° − cos 70° = 3 cos 80° Rewrite sin 40 ° = cos ( 90 ° − 40 ° ) = cos 50 ° \sin40°=\cos(90°-40°)=\cos50° sin 40° = cos ( 90° − 40° ) = cos 50°
cos 50 ° − cos 70 ° = − 2 sin 60 ° sin ( − 10 ° ) = 2 sin 60 ° sin 10 ° = 2 ⋅ 3 2 sin 10 ° = 3 sin 10 ° = 3 cos 80 ° \cos50°-\cos70°=-2\sin60°\sin(-10°)=2\sin60°\sin10°=2\cdot\dfrac{\sqrt3}2\sin10°=\sqrt3\sin10°=\sqrt3\cos80° cos 50° − cos 70° = − 2 sin 60° sin ( − 10° ) = 2 sin 60° sin 10° = 2 ⋅ 2 3 sin 10° = 3 sin 10° = 3 cos 80°
(using sin 10 ° = cos 80 ° \sin10°=\cos80° sin 10° = cos 80°
Ex.1(ii) — Prove cos 40 ° + cos 50 ° + cos 70 ° + cos 80 ° = cos 20 ° + cos 10 ° \cos40°+\cos50°+\cos70°+\cos80°=\cos20°+\cos10° cos 40° + cos 50° + cos 70° + cos 80° = cos 20° + cos 10° Pair ( cos 80 ° + cos 40 ° ) (\cos80°+\cos40°) ( cos 80° + cos 40° )
( cos 70 ° + cos 50 ° ) (\cos70°+\cos50°) ( cos 70° + cos 50° ) 2 cos 60 ° cos 20 ° 2\cos60°\cos20° 2 cos 60° cos 20° 2 cos 60 ° cos 10 ° 2\cos60°\cos10° 2 cos 60° cos 10°
= 2 cos 60 ° ( cos 20 ° + cos 10 ° ) = 2 ⋅ 1 2 ( cos 20 ° + cos 10 ° ) = cos 20 ° + cos 10 ° =2\cos60°(\cos20°+\cos10°)=2\cdot\dfrac12(\cos20°+\cos10°)=\cos20°+\cos10° = 2 cos 60° ( cos 20° + cos 10° ) = 2 ⋅ 2 1 ( cos 20° + cos 10° ) = cos 20° + cos 10°
Ex.3(i) — Show sin 8 x + sin 2 x cos 2 x − cos 8 x = cot 3 x \dfrac{\sin8x+\sin2x}{\cos2x-\cos8x}=\cot3x cos 2 x − cos 8 x sin 8 x + sin 2 x = cot 3 x Numerator = 2 sin 5 x cos 3 x =2\sin5x\cos3x = 2 sin 5 x cos 3 x
C = 8 x , D = 2 x C=8x,D=2x C = 8 x , D = 2 x = 2 sin 5 x sin 3 x =2\sin5x\sin3x = 2 sin 5 x sin 3 x = cos 3 x sin 3 x = cot 3 x =\dfrac{\cos3x}{\sin3x}=\cot3x = sin 3 x cos 3 x = cot 3 x
Ex.3(ii) — Show sin 2 α − sin 2 β sin 2 α + sin 2 β = tan ( α + β ) cot ( α − β ) \dfrac{\sin^2\alpha-\sin^2\beta}{\sin^2\alpha+\sin^2\beta}=\tan(\alpha+\beta)\cot(\alpha-\beta) sin 2 α + sin 2 β sin 2 α − sin 2 β = tan ( α + β ) cot ( α − β )
Each sine-square is written using power reduction sin 2 θ = 1 − cos 2 θ 2 \sin^2\theta=\frac{1-\cos2\theta}2 sin 2 θ = 2 1 − c o s 2 θ
cos 2 α − cos 2 β \cos2\alpha-\cos2\beta cos 2 α − cos 2 β
tan ( α + β ) cot ( α − β ) \tan(\alpha+\beta)\cot(\alpha-\beta) tan ( α + β ) cot ( α − β )
Ex.4(i) — Prove cos ( 7 x − 5 y ) − cos ( 7 y − 5 x ) sin ( 7 x − 5 y ) + sin ( 7 y − 5 x ) = cot ( x + y ) \dfrac{\cos(7x-5y)-\cos(7y-5x)}{\sin(7x-5y)+\sin(7y-5x)}=\cot(x+y) sin ( 7 x − 5 y ) + sin ( 7 y − 5 x ) cos ( 7 x − 5 y ) − cos ( 7 y − 5 x ) = cot ( x + y ) Apply formula (4) to the
numerator and formula (1) to the denominator; both share the factor 6 ( x + y ) 6(x+y) 6 ( x + y ) cos 6 ( x + y ) \cos6(x+y) cos 6 ( x + y )
sin 6 ( x + y ) \sin6(x+y) sin 6 ( x + y ) ( 7 x − 5 y ) + ( 7 y − 5 x ) 2 = x + y \frac{(7x-5y)+(7y-5x)}2=x+y 2 ( 7 x − 5 y ) + ( 7 y − 5 x ) = x + y
collapses to cot ( x + y ) \cot(x+y) cot ( x + y )
Ex.4(ii) — Show sin 6 θ + sin 4 θ − sin 2 θ = 4 cos θ sin 2 θ cos 3 θ \sin6\theta+\sin4\theta-\sin2\theta=4\cos\theta\sin2\theta\cos3\theta sin 6 θ + sin 4 θ − sin 2 θ = 4 cos θ sin 2 θ cos 3 θ Group as
2 sin 5 θ cos θ − 2 sin θ cos θ 2\sin5\theta\cos\theta-2\sin\theta\cos\theta 2 sin 5 θ cos θ − 2 sin θ cos θ sin 2 θ = 2 sin θ cos θ \sin2\theta=2\sin\theta\cos\theta sin 2 θ = 2 sin θ cos θ
factor 2 cos θ 2\cos\theta 2 cos θ = 2 cos θ ( sin 5 θ − sin θ ) = 2 cos θ ⋅ 2 cos 3 θ sin 2 θ =2\cos\theta(\sin5\theta-\sin\theta)=2\cos\theta\cdot2\cos3\theta\sin2\theta = 2 cos θ ( sin 5 θ − sin θ ) = 2 cos θ ⋅ 2 cos 3 θ sin 2 θ
= 4 cos θ sin 2 θ cos 3 θ =4\cos\theta\sin2\theta\cos3\theta = 4 cos θ sin 2 θ cos 3 θ