Simple trigonometric proofs, part II

Ola! It’s time to do some more trigonometry! I already gave you proofs of the first seven identities, which you can find here. There’s still a lot more in what my friend Tomáš sent me and so here we go, seven more formulas, seven more proofs. And today we’re starting with…

(8) Cosine sum-to-product identities

Statement: Let $\alpha, \beta$ be any real numbers. Then

$\cos \alpha + \cos \beta = 2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right) ,$
$\cos \alpha - \cos \beta = -2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right) .$

Proof: We will use the cosine angle sum formula to easily derive both identities at once. Before that, however, observe the following in greater detail:

$\alpha = \frac{\alpha+\alpha}{2} = \frac{\alpha+\beta-\beta+\alpha}{2} = \frac{\alpha+\beta}{2} + \frac{\alpha-\beta}{2} .$

Similarly, it is true that $\beta = \frac{\alpha+\beta}{2}-\frac{\alpha-\beta}{2}$ . Substituting these expressions for $\alpha$ and $\beta$ respectively, we get

$\cos \alpha \pm \cos \beta = \cos \left( \frac{\alpha+\beta}{2} + \frac{\alpha-\beta}{2} \right) \pm \cos \left( \frac{\alpha + \beta}{2} - \frac{\alpha-\beta}{2} \right) .$

Now we can use the cosine angle sum formulas where we put $\alpha = \frac{\alpha+\beta}{2}$ and $\beta = \frac{\alpha-\beta}{2}$ to obtain

$\cos \alpha \pm \cos \beta =$

$= \cos \left( \frac{\alpha+\beta}{2} \right) \cos \left( \frac{\alpha-\beta}{2} \right) - \sin \left( \frac{\alpha+\beta}{2} \right) \sin \left( \frac{\alpha-\beta}{2} \right) \pm$

$= \left( \cos \left( \frac{\alpha+\beta}{2} \right) \cos \left( \frac{\alpha-\beta}{2} \right) + \sin \left( \frac{\alpha+\beta}{2} \right) \sin \left( \frac{\alpha-\beta}{2} \right) \right) .$

At last, we get any of the desired formulas by substituting respective signs into the last equation.

■

(9) Angle sum identities for tangent and contangent

Statement: Let $\alpha, \beta$ be any real numbers. Then

$\tan (\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta} ,$
$\cot (\alpha \pm \beta) = \frac{\cot \alpha \cot \beta \mp 1}{\cot \beta \pm \cot \alpha} .$

Proof: To prove the tangent identity, we will start with the right-hand side (RHS) of the equation. Using the definition of tangent: $\tan x = \frac{\sin x}{\cos x}$ and angle sum identities for cosine and sine, we get

$\frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta} = \frac{\frac{\sin \alpha}{\cos \alpha} \pm \frac{\sin \beta}{\cos \beta}}{1 \mp \frac{\sin \alpha}{\cos \alpha} \frac{\sin \beta}{\cos \beta}} = \frac{\frac{\sin \alpha \cos \beta \pm \cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\cos \alpha \cos \beta \mp \sin \alpha \sin \beta}{\cos \alpha \cos \beta}} =$

$\frac{\sin(\alpha \pm \beta)}{\cos (\alpha \pm \beta)} = \tan(\alpha \pm \beta)$

To obtain the cotangent identity we proceed similarly – starting with RHS of the equation and using the definition of contangent and angle sum formulas for sine and consine, we write

$\frac{\cot \alpha \cot \beta \mp 1}{\cot \beta \pm \cot \alpha} = \frac{\frac{\cos \alpha \cos \beta}{\sin \alpha \sin \beta} \mp 1}{\frac{\cos \beta}{\sin \beta} \pm \frac{\cos \alpha}{\sin \alpha}} = \frac{\frac{\cos \alpha \cos \beta \mp \sin \alpha \sin \beta}{\sin \alpha \sin \beta}}{\frac{\sin \alpha \cos \beta \pm \cos \alpha \sin \beta}{\sin \alpha \sin \beta}}=$

$\frac{\cos (\alpha \pm \beta)}{\sin(\alpha \pm \beta)} = \cot(\alpha \pm \beta)$

Finally, as always, substituting the respective signs into the equalities will give us any of the desired formulas and so our proof is concluded.

■

(10) Tangent of double angle

Statement: Let $\theta \in \mathbb{R} \setminus \{(2k+1)\frac{\pi}{4} : k \in \mathbb{Z} \}$ . Then

$\tan 2 \theta = \frac{2\tan \theta}{1-\tan^2 \theta}$ .

Proof: Once again, we start with the RHS of the equality and apply the definition of tangent to obtain

$\frac{2\tan \theta}{1-\tan^2 \theta} = \frac{\frac{2\sin \theta}{\cos \theta}}{\frac{\cos^2\theta - \sin^2 \theta}{\cos^2 \theta}} = \frac{2\sin \theta \cos \theta}{\cos^2 \theta - \sin^2 \theta}.$

At this moment we apply the formula for sine of double angle to the numerator and the formula for cosine of double angle to the denominator and we get the result:

$\frac{2\sin \theta \cos \theta}{\cos^2 \theta - \sin^2 \theta} = \frac{\sin 2\theta}{\cos 2\theta} = \tan 2 \theta .$

■

(11) Cotangent of double angle

Statement: Let $\theta \in \mathbb{R} \setminus \{ k\frac{\pi}{2}:k \in \mathbb{Z}\}$ . Then

$\cot 2 \theta = \frac{\cot^2 \theta-1}{2\cot \theta}.$

Proof: The proof is by all means analogous to the previous one – we get the result by starting with RHS, using the definition of cotangent and applying the double angle formulas for sine and cosine:

$\frac{\cot^2 \theta -1}{2\cot \theta} = \frac{\frac{\cos^2 \theta - \sin^2 \theta}{\sin^2 \theta}}{\frac{2\cos \theta}{\sin \theta}} = \frac{\cos^2 \theta - \sin^2 \theta}{2\sin \theta \cos \theta} = \frac{\cos 2\theta}{\sin 2\theta} = \cot 2\theta.$

■

(12) Tangent of half-angle

Statement: Let $\theta \in \mathbb{R} \setminus \{(2k+1)\pi : k \in \mathbb{Z}\}$ . Then

$\tan \frac{\theta}{2} = \pm \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}.$

Proof: We will use the sine and cosine of half-angle identities to compute $\left|\tan \frac{\theta}{2} \right|$ :

$\left|\tan \frac{\theta}{2} \right| = \left| \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right| = \frac{\sqrt{\frac{1-\cos \theta}{2}}}{\sqrt{\frac{1+\cos \theta}{2}}} = \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}.$

From the properties of absolute value it now follows, that indeed

$\tan \frac{\theta}{2} = \pm \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}.$

■

(13) Cotangent of half-angle

Statement: Let $\theta \in \mathbb{R} \setminus \{2k\pi : k \in \mathbb{Z} \}$ . Then

$\cot \frac{\theta}{2} = \pm \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}.$

Proof: One way of proving this would be analogous to the previous proof. However, we can use the previous formula to obtain a slightly simpler proof: We know from the definition of tangent and cotangent, that $\cot x = \frac{1}{\tan x}$ for every admittable value of $x$ . Therefore

$\cot \frac{\theta}{2} = \frac{1}{\tan \frac{\theta}{2}} = \frac{1}{\pm \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}} = \pm \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}.$

■

(14) Sum-to-product identities for tangent and cotangent

Statement: Let $\alpha, \beta$ be real numbers such that $\tan \alpha, \tan \beta$ or $\cot \alpha, \cot \beta$ are well-defined, respectively. Then

$\tan \alpha \pm \tan \beta = \frac{\sin(\alpha \pm \beta)}{\cos \alpha \cos \beta}$ ,
$\cot \alpha \pm \cot \beta = \pm \frac{\sin(\alpha+\beta)}{\sin \alpha \sin \beta}.$

Proof: The only thing we will use in these simple proofs are definitions of tangent (cotangent) and angle-sum formulas for sine. The tangent formula can be proved as follows:

$\tan \alpha \pm \tan \beta = \frac{\sin \alpha \cos \beta \pm \cos \alpha \sin \beta}{\cos \alpha \cos \beta} = \frac{\sin(\alpha \pm \beta)}{\cos \alpha \cos \beta} ,$

whereas the cotangent formula

$\cos \alpha \pm \cot \beta = \frac{\cos \alpha \sin \beta \pm \sin \alpha \cos \beta}{\sin \alpha \sin \beta} = \frac{\pm \sin(\alpha \pm \beta)}{\sin \alpha \sin \beta}.$

■

That concludes the second part of simple trigonometric proofs. I’m not sure whether there will be more of them, but if so, they will be much more interesting and fun. These two parts covered most of very basic high-school trigonometry and I sincerely hope that someone found them at least a bit helpful.

At this moment, I am starting my fall semester and hopefully I will be able to gather enough free time to keep this blog alive. Until next time! 🙂

Did you find this post helpful? Please share it with others!