Addition Formula for Trigonometric Functions: Various Proofs
📂Functions Addition Formula for Trigonometric Functions: Various Proofs Theorem sin ( α + β ) = sin α cos β + cos α sin β sin ( α − β ) = sin α cos β − cos α sin β cos ( α + β ) = cos α cos β − sin α sin β cos ( α − β ) = cos α cos β + sin α sin β tan ( α + β ) = tan α + tan β 1 − tan α tan β tan ( α − β ) = tan α − tan β 1 + tan α tan β
\sin\left( \alpha +\beta \right) =\sin\alpha \cos\beta +\cos\alpha \sin\beta
\\
\sin\left( \alpha -\beta \right) =\sin\alpha \cos\beta -\cos\alpha \sin\beta
\\
\cos\left( \alpha +\beta \right) =\cos\alpha \cos\beta -\sin\alpha \sin\beta
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\cos\left( \alpha -\beta \right) =\cos\alpha \cos\beta +\sin\alpha \sin\beta
\\
\tan\left( \alpha +\beta \right) =\frac { \tan\alpha +\tan\beta }{ 1-\tan\alpha \tan\beta }
\\
\tan\left( \alpha -\beta \right) =\frac { \tan\alpha -\tan\beta }{ 1+\tan\alpha \tan\beta }
sin ( α + β ) = sin α cos β + cos α sin β sin ( α − β ) = sin α cos β − cos α sin β cos ( α + β ) = cos α cos β − sin α sin β cos ( α − β ) = cos α cos β + sin α sin β tan ( α + β ) = 1 − tan α tan β tan α + tan β tan ( α − β ) = 1 + tan α tan β tan α − tan β
Proof Proof using the Law of Cosines
By Pythagoras’ theorem
A B ‾ 2 = ( cos α − cos β ) 2 + ( sin α − sin β ) 2 = 2 − 2 cos α cos β – 2 sin α sin β
\begin{align*}
{\overline { AB } } ^{ 2 } =& {( \cos \alpha -\cos \beta )}^{ 2 }+{(\sin\alpha -\sin\beta )}^{ 2 }
\\ =& 2-2 \cos \alpha \cos \beta –2 \sin \alpha \sin \beta
\end{align*}
A B 2 = = ( cos α − cos β ) 2 + ( sin α − sin β ) 2 2 − 2 cos α cos β –2 sin α sin β
By the Second Law of Cosines
A B ‾ 2 = 1 2 + 1 2 − 2 cos ( β − α ) = 2 − 2 cos ( β − α )
\begin{align*}
{ \overline { AB } } ^{ 2 } =& 1^{ 2 }+1^{ 2 }-2\cos(\beta -\alpha )
\\ =& 2-2\cos(\beta -\alpha )
\end{align*}
A B 2 = = 1 2 + 1 2 − 2 cos ( β − α ) 2 − 2 cos ( β − α )
Since the right-hand sides of both equations are equal
cos ( β − α ) = cos α cos β + sin α sin β
\cos(\beta -\alpha )=\cos\alpha \cos\beta +\sin\alpha \sin\beta
cos ( β − α ) = cos α cos β + sin α sin β
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This is the most basic method of proof, and although there are various methods, this is usually the first one encountered.
Proof using the Dot Product of Vectors cos ( β − α ) = O A ⃗ ⋅ O B ⃗ ∣ O A ⃗ ∣ ∣ O B ⃗ ∣ = cos α cos β + sin α sin β
\begin{align*}
\cos(\beta -\alpha ) =& \frac { \vec { OA }\cdot \vec { OB } }{ \left| \vec { OA } \right| \left| \vec { OB } \right| }
\\ =& \cos\alpha \cos\beta +\sin\alpha \sin\beta
\end{align*}
cos ( β − α ) = = O A OB O A ⋅ OB cos α cos β + sin α sin β
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Writing the dot product of vectors on paper is practically the same as writing a single line. The idea is simple and it’s the easiest method.
Proof using Triangles
(1) Let the area of the triangle be S S S
S = 1 2 a b sin ( α + β )
S=\frac { 1 }{ 2 }ab\sin(\alpha +\beta )
S = 2 1 ab sin ( α + β )
(2) Adding the areas of the two triangles bounded by the perpendicular
S = 1 2 b h sin α + 1 2 a h sin β
S=\frac { 1 }{ 2 }bh\sin\alpha +\frac { 1 }{ 2 }ah\sin\beta
S = 2 1 bh sin α + 2 1 ah sin β
Since h = b cos α = a cos β h=b\cos\alpha =a\cos\beta h = b cos α = a cos β
S = 1 2 a b cos β sin α + 1 2 a b cos α sin β
S=\frac { 1 }{ 2 }ab\cos\beta \sin\alpha +\frac { 1 }{ 2 }ab\cos\alpha \sin\beta
S = 2 1 ab cos β sin α + 2 1 ab cos α sin β
What is obtained in (1) and (2) are both S S S 1 2 a b \frac { 1 }{ 2 }ab 2 1 ab
sin ( α + β ) = cos β sin α + cos α sin β
\sin(\alpha +\beta )=\cos\beta \sin\alpha +\cos\alpha \sin\beta
sin ( α + β ) = cos β sin α + cos α sin β
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The proof using the area of a triangle is simple in idea, and dealing with
h h h is key.
A A A β \beta β
[ cos ( α + β ) sin ( α + β ) ] = [ cos β − sin β sin β cos β ] [ cos α sin α ] ⟹ { cos ( α + β ) = cos β cos α − sin β sin α sin ( α + β ) = sin β cos α + cos β sin α
\begin{bmatrix}
\cos(\alpha +\beta )
\\ \sin(\alpha +\beta )
\end{bmatrix}
= \begin{bmatrix}
{ \cos\beta }&{ -\sin\beta }
\\ { \sin\beta }&{ \cos\beta }
\end{bmatrix}
\begin{bmatrix} { \cos\alpha }
\\ { \sin\alpha }
\end{bmatrix}
\\ \implies \begin{cases}
\cos(\alpha +\beta )=\cos\beta \cos\alpha -\sin\beta \sin\alpha
\\ { \sin(\alpha +\beta )=\sin\beta \cos\alpha +\cos\beta \sin\alpha }
\end{cases}
[ cos ( α + β ) sin ( α + β ) ] = [ cos β sin β − sin β cos β ] [ cos α sin α ] ⟹ { cos ( α + β ) = cos β cos α − sin β sin α sin ( α + β ) = sin β cos α + cos β sin α
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This is a proof using rotational transformation. It requires setting the angle slightly differently but it’s great since it allows obtaining the cosines and sines at the same time.
Corollaries These cases are used more often than one might think, so it’s convenient to remember them.
sin ( π 4 + π 6 ) = cos ( π 4 − π 6 ) = 3 + 1 2 2 sin ( π 4 − π 6 ) = cos ( π 4 + π 6 ) = 3 − 1 2 2
\begin{align*}
\sin(\frac { \pi }{ 4 }+\frac { \pi }{ 6 })=\cos(\frac { \pi }{ 4 }-\frac { \pi }{ 6 })=\frac { \sqrt { 3 }+1 }{ 2\sqrt { 2 } }
\\
\sin(\frac { \pi }{ 4 }-\frac { \pi }{ 6 })=\cos(\frac { \pi }{ 4 }+\frac { \pi }{ 6 })=\frac { \sqrt { 3 }-1 }{ 2\sqrt { 2 } }
\end{align*}
sin ( 4 π + 6 π ) = cos ( 4 π − 6 π ) = 2 2 3 + 1 sin ( 4 π − 6 π ) = cos ( 4 π + 6 π ) = 2 2 3 − 1 Addition formula for tangent:
tan ( θ 1 ± θ 2 ) = tan θ 1 ± tan θ 2 1 ∓ tan θ 1 tan θ 2
\tan ( \theta_1 \pm \theta_2) = \dfrac{\tan\theta_1 \pm \tan\theta_2}{1 \mp \tan\theta_1\tan\theta_2}
tan ( θ 1 ± θ 2 ) = 1 ∓ tan θ 1 tan θ 2 tan θ 1 ± tan θ 2 tan ( θ 1 ± θ 2 ) = sin ( θ 1 ± θ 2 ) cos ( θ 1 ± θ 2 ) = sin θ 1 cos θ 2 ± sin θ 2 cos θ 2 cos θ 1 cos θ 2 ∓ sin θ 1 sin θ 2
\tan (\theta_1 \pm \theta2)=\dfrac{\sin ( \theta_1 \pm \theta_2)}{\cos ( \theta_1 \pm \theta_2)} =\dfrac{ \sin \theta_1 \cos \theta_2 \pm \sin \theta_2 \cos \theta_2}{\cos \theta_1 \cos\theta_2 \mp \sin\theta_1 \sin\theta_2}
tan ( θ 1 ± θ 2 ) = cos ( θ 1 ± θ 2 ) sin ( θ 1 ± θ 2 ) = cos θ 1 cos θ 2 ∓ sin θ 1 sin θ 2 sin θ 1 cos θ 2 ± sin θ 2 cos θ 2 cos θ 1 cos θ 2 \cos\theta_1\cos\theta_2 cos θ 1 cos θ 2 sin θ 1 cos θ 1 ± sin θ 2 cos θ 1 1 ∓ sin θ 1 sin θ 2 cos θ 1 cos θ 2 = tan θ 1 ± tan θ 2 1 ∓ tan θ 1 tan θ 2
\dfrac{ \dfrac{\sin \theta_1}{ \cos \theta_1} \pm \dfrac{\sin \theta_2}{ \cos \theta_1} } { 1 \mp \dfrac{\sin\theta_1 \sin\theta_2}{\cos\theta_1\cos\theta_2 }} = \dfrac{ \tan\theta_1 \pm \tan\theta_2}{1 \mp \tan\theta_1\tan\theta_2}
1 ∓ cos θ 1 cos θ 2 sin θ 1 sin θ 2 cos θ 1 sin θ 1 ± cos θ 1 sin θ 2 = 1 ∓ tan θ 1 tan θ 2 tan θ 1 ± tan θ 2
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Rotating point around the origin by