Sine and Cosine of Sum

Theorem

$\cos \left({a + b}\right) = \cos a \cos b - \sin a \sin b$

$\sin \left({a + b}\right) = \sin a \cos b + \cos a \sin b$

where $\sin$ and $\cos$ are sine and cosine.

Corollary

$\cos \left({a - b}\right) = \cos a \cos b + \sin a \sin b$

$\sin \left({a - b}\right) = \sin a \cos b - \cos a \sin b$

Proof from Euler's Formula

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \cos \left({a + b}\right) + i \sin \left({a + b}\right)\)	\(=\)	\(\displaystyle e^{i \left({a + b}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Euler's Formula
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^{i a} e^{i b}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Exponent of Sum
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\cos a + i \sin a}\right) \left({\cos b + i \sin b}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Euler's Formula
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\cos a \cos b - \sin a \sin b}\right) + i \left({\sin a \cos b + \cos a \sin b}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Distribution

By equating real and imaginary parts, we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \cos \left({a + b}\right)\)	\(=\)	\(\displaystyle \cos a \cos b - \sin a \sin b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sin \left({a + b}\right)\)	\(=\)	\(\displaystyle \sin a \cos b + \cos a \sin b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Proof from Algebraic Definitions

We have:

• From the definition of sine:

$\displaystyle \sin x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$

• From the definition of cosine:

$\displaystyle \cos x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$.

Let:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle g \left({a}\right)\)	\(=\)	\(\displaystyle \sin \left({a + b}\right) - \sin a \cos b - \cos a \sin b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle h \left({a}\right)\)	\(=\)	\(\displaystyle \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Let us derive these with respect to $a$, keeping $b$ constant.

Then from Derivative of Sine Function and Derivative of Cosine Function, we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle g^{\prime} \left({a}\right)\)	\(=\)	\(\displaystyle \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b = h \left({a}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle h^{\prime} \left({a}\right)\)	\(=\)	\(\displaystyle - \sin \left({a + b}\right) + \sin a \cos b + \cos a \sin b = - g \left({a}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Hence:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle D_a \left({g \left({a}\right)^2 + h \left({a}\right)^2}\right)\)	\(=\)	\(\displaystyle 2 g \left({a}\right) g^{\prime} \left({a}\right) + 2 h \left({a}\right) h^{\prime} \left({a}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Thus from Differentiation of a Constant, $\forall a: g \left({a}\right)^2 + h \left({a}\right)^2 = c$.

But this is true for $a = 0$, and $g \left({0}\right)^2 + h \left({0}\right)^2 = 0$.

So $g \left({a}\right)^2 + h \left({a}\right)^2 = 0$

But $g \left({a}\right)^2 \ge 0$ and $g \left({a}\right)^2 \ge 0$ from Even Powers are Positive.

So it follows that $g \left({a}\right) = 0$ and $h \left({a}\right) = 0$.

Hence the result.

$\blacksquare$

Geometric Proof

$AB$, $AC$, $AE$, and $AD$ are radii of the circle centered at $A$.

Let $\angle BAC = a$ and $\angle DAC = \angle BAE = b$.

By Euclid's First Postulate, we can construct line segments $BD$ and $CE$.

By Euclid's second common notion, $\angle DAB = \angle CAE$.

Thus by Triangle Side-Angle-Side Equality, $\triangle DAB \cong \triangle CAE$.

Therefore, $DB = CE$.

We now assign Cartesian coordinates to the points $B$, $C$, $D$, and $E$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle B\)	\(=\)	\(\displaystyle \left({1, 0}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle C\)	\(=\)	\(\displaystyle \left({\cos a, \sin a}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle D\)	\(=\)	\(\displaystyle \left({\cos \left({a + b}\right), \sin \left({a + b}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle E\)	\(=\)	\(\displaystyle \left({\cos b, -\sin b}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from Cosine Function is Even and Sine Function is Odd: $\cos \left({-x}\right) = \cos x$ and $\sin \left({-x}\right) = - \sin x$

We use the definition of the distance function on the Euclidean space $\left({\R^2, d}\right)$ as defined by the Euclidean metric:

$\forall x, y \in \R^2: d \left({x, y}\right) = \sqrt {\left({x_1 - y_1}\right)^2 + \left({x_2 - y_2}\right)^2}$

where $x = \left({x_1, y_1}\right), y = \left({x_2, y_2}\right)$.

Thus $DB \cong CE \iff d \left({D, B}\right) = d \left({C, E}\right)$.

So, plugging in the coordinates of $B, C, D, E$, we get:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle (\cos(a + b)-1)^2 + \sin^2(a + b)\)	\(=\)	\(\displaystyle (\cos a - \cos b)^2 + (\sin a + \sin b)^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \underbrace{\cos^2(a + b) + \sin^2(a + b)}_1 - 2 \cos(a + b) + 1\)	\(=\)	\(\displaystyle \cos^2 a - 2 \cos a \cos b + \cos^2 b + \sin^2 a + 2\sin a \sin b + \sin^2 b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 2-2\cos(a + b)\)	\(=\)	\(\displaystyle \underbrace{\cos^2 a + \sin^2 a}_1 + \underbrace{\cos^2 b+\sin^2 b}_{1}+ -2\cos a \cos b+2\sin a \sin b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \cos a \cos b - \sin a \sin b\)	\(=\)	\(\displaystyle \cos(a + b)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

In the above, repeated use is made of the identity $\cos^2 \theta + \sin^2 \theta \equiv 1$ from Sum of Squares of Sine and Cosine.

Now, using the identity $\cos \left({\frac \pi 2 - a}\right) = \sin a$ from Sine equals Cosine of Complement, we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sin \left({a + b}\right)\)	\(=\)	\(\displaystyle \cos \left({\frac \pi 2 - \left({a + b}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cos \left({\left({\frac \pi 2 - a}\right) - b}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cos \left({\frac \pi 2 - a}\right) \left({\cos \left({-b}\right)}\right) - \sin \left({\frac \pi 2 - a}\right) \left({\sin \left({-b}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	using identity demonstrated above
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cos \left({\frac \pi 2 - a}\right) \cos b + \sin \left({\frac \pi 2 - a}\right) \sin b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	($\cos \left({-x}\right) = \cos x$ and $\sin \left({-x}\right) = - \sin x$ as above)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sin a \cos b + \cos a \sin b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sine equals Cosine of Complement

$\blacksquare$

Proof of Corollary

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \cos \left({a - b}\right)\)	\(=\)	\(\displaystyle \cos a \cos \left({- b}\right) - \sin a \sin \left({- b}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Main result
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cos a \cos b + \sin a \sin b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Cosine Function is Even; Sine Function is Odd

Similarly, we obtain:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sin \left({a - b}\right)\)	\(=\)	\(\displaystyle \sin a \cos \left({- b}\right) + \cos a \sin \left({- b}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Main result
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sin a \cos b - \cos a \sin b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Cosine Function is Even; Sine Function is Odd

$\blacksquare$

Historical Note

These formulas were proved by François Viète in about 1579.

See Also

• Tangent of Sum, a closely related result.

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $5.34, \ 5.35, \ 5.36$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 16.3 \ (2)$