We know that:
sin
2
θ
=
1
−
cos
2
θ
{\displaystyle \sin ^{2}\theta =1-\cos ^{2}\theta \,}
So do we really need the
sin
{\displaystyle \sin \,}
Or put another way, could we have worked out all our interesting formulas for things like
cos
(
a
+
b
)
{\displaystyle \cos(a+b)\,}
cos
{\displaystyle \cos \,}
sin
{\displaystyle \sin \,}
The answer is yes.
We don't need to have one geometric argument for
cos
(
a
+
b
)
{\displaystyle \cos(a+b)\,}
sin
(
a
+
b
)
{\displaystyle \sin(a+b)\,}
sin
{\displaystyle \sin \,}
cos
{\displaystyle \cos \,}
To find a formula for
cos
(
θ
1
+
θ
2
)
{\displaystyle \cos(\theta _{1}+\theta _{2})\,}
cos
(
θ
1
)
{\displaystyle \cos \left(\theta _{1}\right)}
cos
(
θ
2
)
{\displaystyle \cos \left(\theta _{2}\right)}
c
{\displaystyle c\,}
θ
1
≠
θ
2
{\displaystyle \theta _{1}\neq \theta _{2}}
ψ
1
≠
ψ
2
{\displaystyle \psi _{1}\neq \psi _{2}}
a
2
{\displaystyle a_{2}\,}
c
1
{\displaystyle c_{1}\,}
c
1
{\displaystyle c_{1}\,}
a
2
{\displaystyle a_{2}\,}
θ
1
{\displaystyle \theta _{1}\,}
θ
2
{\displaystyle \theta _{2}\,}
θ
3
=
θ
1
+
θ
2
{\displaystyle \theta _{3}=\theta _{1}+\theta _{2}\,}
a
1
{\displaystyle a_{1}\,}
g
{\displaystyle g\,}
φ
2
{\displaystyle \varphi _{2}}
g
{\displaystyle g\,}
c
2
{\displaystyle c_{2}\,}
a
2
{\displaystyle a_{2}\,}
c
1
{\displaystyle c_{1}\,}
cos
(
θ
1
)
{\displaystyle \cos \left(\theta _{1}\right)}
a
1
{\displaystyle a_{1}\,}
cos
(
θ
1
)
⋅
cos
(
θ
2
)
{\displaystyle \cos \left(\theta _{1}\right)\cdot \cos \left(\theta _{2}\right)}
g
{\displaystyle g\,}
a
1
{\displaystyle a_{1}\,}
(1) it is right angled, and
(2)
θ
4
+
(
π
−
π
2
−
θ
1
−
θ
2
)
=
φ
2
=
π
−
π
2
−
θ
2
⇒
θ
4
=
θ
1
{\displaystyle \theta _{4}+(\pi -{\frac {\pi }{2}}-\theta _{1}-\theta _{2})=\varphi _{2}=\pi -{\frac {\pi }{2}}-\theta _{2}\Rightarrow \theta _{4}=\theta _{1}}
The length of side
b
4
{\displaystyle b_{4}\,}
c
o
s
(
φ
2
)
c
o
s
(
φ
4
)
=
c
o
s
(
φ
2
)
c
o
s
(
φ
1
)
{\displaystyle cos(\varphi _{2})cos(\varphi _{4})=cos(\varphi _{2})cos(\varphi _{1})}
θ
4
=
θ
1
{\displaystyle \theta _{4}\,=\theta _{1}\,}
g
{\displaystyle g\,}
c
o
s
(
θ
1
+
θ
2
)
=
c
o
s
(
θ
1
)
c
o
s
(
θ
2
)
−
c
o
s
(
φ
1
)
c
o
s
(
φ
2
)
,
{\displaystyle cos(\theta _{1}+\theta _{2})=cos(\theta _{1})cos(\theta _{2})-cos(\varphi _{1})cos(\varphi _{2}),}
θ
1
+
φ
1
=
θ
2
+
φ
2
=
π
2
{\displaystyle \theta _{1}+\varphi _{1}=\theta _{2}+\varphi _{2}={\frac {\pi }{2}}}
giving us the "Cosine Angle Sum Formula".
We can apply this formula immediately to sum two equal angles:
c
o
s
(
2
θ
)
=
c
o
s
(
θ
+
θ
)
=
c
o
s
(
θ
)
c
o
s
(
θ
)
−
c
o
s
(
φ
)
c
o
s
(
φ
)
=
c
o
s
(
θ
)
2
−
c
o
s
(
φ
)
2
{\displaystyle cos(2\theta )=cos(\theta +\theta )=cos(\theta )cos(\theta )-cos(\varphi )cos(\varphi )=cos(\theta )^{2}-cos(\varphi )^{2}}
θ
+
φ
=
π
2
{\displaystyle \theta +\varphi ={\frac {\pi }{2}}}
From the theorem of Pythagoras we know that:
a
2
+
b
2
=
c
2
{\displaystyle a^{2}+b^{2}=c^{2}\,}
in this case:
c
o
s
(
θ
)
2
+
c
o
s
(
φ
)
2
=
1
2
{\displaystyle cos(\theta )^{2}+cos(\varphi )^{2}=1^{2}}
⇒
c
o
s
(
φ
)
2
=
1
−
c
o
s
(
θ
)
2
{\displaystyle \Rightarrow cos(\varphi )^{2}=1-cos(\theta )^{2}}
θ
+
φ
=
π
2
{\displaystyle \theta +\varphi ={\frac {\pi }{2}}}
Substituting into (I) gives:
c
o
s
(
2
θ
)
=
c
o
s
(
θ
)
2
−
c
o
s
(
φ
)
2
{\displaystyle cos(2\theta )=cos(\theta )^{2}-cos(\varphi )^{2}}
=
c
o
s
(
θ
)
2
−
(
1
−
c
o
s
(
θ
)
2
)
{\displaystyle =cos(\theta )^{2}-(1-cos(\theta )^{2})\,}
=
2
c
o
s
(
θ
)
2
−
1
{\displaystyle =2cos(\theta )^{2}-1\,}
θ
+
φ
=
π
4
{\displaystyle \theta +\varphi ={\frac {\pi }{4}}}
which is identical to the "Cosine Double Angle Sum Formula":
2
c
o
s
(
δ
)
2
−
1
=
c
o
s
(
2
δ
)
{\displaystyle 2cos(\delta )^{2}-1=cos(2\delta )\,}
Pythagorean identity
edit
The price we have to pay for the notational convenience of this new function
s
i
n
(
)
{\displaystyle sin()\,}
c
o
s
(
)
{\displaystyle cos()\,}
c
o
s
(
θ
)
2
{\displaystyle cos(\theta )^{2}\,}
1
−
s
i
n
(
θ
)
2
{\displaystyle 1-sin(\theta )^{2}\,}
c
o
s
(
θ
1
+
θ
2
)
=
c
o
s
(
θ
1
)
c
o
s
(
θ
2
)
−
s
i
n
(
θ
1
)
s
i
n
(
θ
2
)
{\displaystyle cos(\theta _{1}+\theta _{2})=cos(\theta _{1})cos(\theta _{2})-sin(\theta _{1})sin(\theta _{2})\,}
⇒
c
o
s
(
θ
1
+
θ
2
)
2
=
(
c
o
s
(
θ
1
)
c
o
s
(
θ
2
)
−
s
i
n
(
θ
1
)
s
i
n
(
θ
2
)
)
2
{\displaystyle \Rightarrow cos(\theta _{1}+\theta _{2})^{2}=(cos(\theta _{1})cos(\theta _{2})-sin(\theta _{1})sin(\theta _{2}))^{2}\,}
⇒
1
−
c
o
s
(
θ
1
+
θ
2
)
2
=
1
−
(
c
o
s
(
θ
1
)
c
o
s
(
θ
2
)
−
s
i
n
(
θ
1
)
s
i
n
(
θ
2
)
)
2
{\displaystyle \Rightarrow 1-cos(\theta _{1}+\theta _{2})^{2}=1-(cos(\theta _{1})cos(\theta _{2})-sin(\theta _{1})sin(\theta _{2}))^{2}\,}
⇒
s
i
n
(
θ
1
+
θ
2
)
2
=
1
−
(
c
o
s
(
θ
1
)
2
c
o
s
(
θ
2
)
2
+
s
i
n
(
θ
1
)
2
s
i
n
(
θ
2
)
2
−
2
c
o
s
(
θ
1
)
c
o
s
(
θ
2
)
s
i
n
(
θ
1
)
s
i
n
(
θ
2
)
)
{\displaystyle \Rightarrow sin(\theta _{1}+\theta _{2})^{2}=1-(cos(\theta _{1})^{2}cos(\theta _{2})^{2}+sin(\theta _{1})^{2}sin(\theta _{2})^{2}-2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2}))\,}
=
1
−
(
c
o
s
(
θ
1
)
2
(
1
−
s
i
n
(
θ
2
)
2
)
+
s
i
n
(
θ
1
)
2
(
1
−
c
o
s
(
θ
2
)
2
)
−
2
c
o
s
(
θ
1
)
c
o
s
(
θ
2
)
s
i
n
(
θ
1
)
s
i
n
(
θ
2
)
)
{\displaystyle =1-(cos(\theta _{1})^{2}(1-sin(\theta _{2})^{2})+sin(\theta _{1})^{2}(1-cos(\theta _{2})^{2})-2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2}))\,}
=
1
−
(
c
o
s
(
θ
1
)
2
−
c
o
s
(
θ
1
)
2
s
i
n
(
θ
2
)
2
+
s
i
n
(
θ
1
)
2
−
s
i
n
(
θ
1
)
2
c
o
s
(
θ
2
)
2
−
2
c
o
s
(
θ
1
)
c
o
s
(
θ
2
)
s
i
n
(
θ
1
)
s
i
n
(
θ
2
)
)
{\displaystyle =1-(cos(\theta _{1})^{2}-cos(\theta _{1})^{2}sin(\theta _{2})^{2}+sin(\theta _{1})^{2}-sin(\theta _{1})^{2}cos(\theta _{2})^{2}-2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2}))\,}
=
1
−
(
1
−
c
o
s
(
θ
1
)
2
s
i
n
(
θ
2
)
2
−
s
i
n
(
θ
1
)
2
c
o
s
(
θ
2
)
2
−
2
c
o
s
(
θ
1
)
c
o
s
(
θ
2
)
s
i
n
(
θ
1
)
s
i
n
(
θ
2
)
)
{\displaystyle =1-(1-cos(\theta _{1})^{2}sin(\theta _{2})^{2}-sin(\theta _{1})^{2}cos(\theta _{2})^{2}-2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2}))\,}
=
c
o
s
(
θ
1
)
2
s
i
n
(
θ
2
)
2
+
s
i
n
(
θ
1
)
2
c
o
s
(
θ
2
)
2
+
2
c
o
s
(
θ
1
)
c
o
s
(
θ
2
)
s
i
n
(
θ
1
)
s
i
n
(
θ
2
)
{\displaystyle =cos(\theta _{1})^{2}sin(\theta _{2})^{2}+sin(\theta _{1})^{2}cos(\theta _{2})^{2}+2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2})\,}
=
(
c
o
s
(
θ
1
)
s
i
n
(
θ
2
)
+
s
i
n
(
θ
1
)
c
o
s
(
θ
2
)
)
2
{\displaystyle =(cos(\theta _{1})sin(\theta _{2})+sin(\theta _{1})cos(\theta _{2}))^{2}\,}
taking the square root of both sides produces the "Sine Angle Sum Formula"
⇒
s
i
n
(
θ
1
+
θ
2
)
=
c
o
s
(
θ
1
)
s
i
n
(
θ
2
)
+
s
i
n
(
θ
1
)
c
o
s
(
θ
2
)
{\displaystyle \Rightarrow sin(\theta _{1}+\theta _{2})=cos(\theta _{1})sin(\theta _{2})+sin(\theta _{1})cos(\theta _{2})}
We can use a similar technique to find the "Sine Half Angle Formula" from the "Cosine Half Angle Formula":
c
o
s
(
θ
2
)
=
1
+
c
o
s
(
θ
)
2
{\displaystyle cos\left({\frac {\theta }{2}}\right)={\sqrt {\frac {1+cos(\theta )}{2}}}}
We know that
1
−
c
o
s
(
θ
2
)
2
=
s
i
n
(
θ
2
)
2
{\displaystyle 1-cos\left({\frac {\theta }{2}}\right)^{2}=sin\left({\frac {\theta }{2}}\right)^{2}}
⇒
1
−
c
o
s
(
θ
2
)
2
=
1
−
1
+
c
o
s
(
θ
)
2
{\displaystyle \Rightarrow 1-cos\left({\frac {\theta }{2}}\right)^{2}=1-{\frac {1+cos(\theta )}{2}}}
⇒
s
i
n
(
θ
2
)
2
=
1
−
c
o
s
(
θ
)
2
{\displaystyle \Rightarrow sin\left({\frac {\theta }{2}}\right)^{2}={\frac {1-cos(\theta )}{2}}}
⇒
s
i
n
(
θ
2
)
=
1
−
c
o
s
(
θ
)
2
{\displaystyle \Rightarrow sin\left({\frac {\theta }{2}}\right)={\sqrt {\frac {1-cos(\theta )}{2}}}}
So far so good, but we still have a
c
o
s
(
θ
)
2
{\displaystyle {\frac {cos(\theta )}{2}}}
s
i
n
(
θ
2
)
=
1
−
1
−
s
i
n
(
θ
)
2
2
{\displaystyle sin\left({\frac {\theta }{2}}\right)={\sqrt {\frac {1-{\sqrt {1-sin(\theta )^{2}}}}{2}}}}
or perhaps a little more legibly as:
s
i
n
(
θ
2
)
=
1
2
1
−
1
−
s
i
n
(
θ
)
2
{\displaystyle sin\left({\frac {\theta }{2}}\right)={\sqrt {\frac {1}{2}}}{\sqrt {1-{\sqrt {1-sin(\theta )^{2}}}}}}
