Trigonometric Functions of Double Angles

Double-angle formulas

Theorem: for any angle $\theta$,

Proof of $\sin2\theta$. Write $2\theta=\theta+\theta$ and use the sine-sum formula (Theorem 4, section 3.1):

$\sin2\theta=\sin\theta\cos\theta+\cos\theta\sin\theta=2\sin\theta\cos\theta$. To reach the tangent form, divide

and multiply by $\cos^2\theta$: $2\sin\theta\cos\theta=\dfrac{2\sin\theta\cos\theta}{\sin^2\theta+\cos^2\theta} =\dfrac{2\sin\theta\cos\theta/\cos^2\theta}{(\sin^2\theta+\cos^2\theta)/\cos^2\theta}=\dfrac{2\tan\theta}{\tan^2\theta+1}$.

Proof of $\cos2\theta$. $\cos2\theta=\cos(\theta+\theta)=\cos\theta\cos\theta-\sin\theta\sin\theta= \cos^2\theta-\sin^2\theta$. Substituting $\sin^2\theta=1-\cos^2\theta$ gives $2\cos^2\theta-1$; substituting

$\cos^2\theta=1-\sin^2\theta$ gives $1-2\sin^2\theta$. Dividing $\cos^2\theta-\sin^2\theta$ top and bottom (as a

fraction over $1=\sin^2\theta+\cos^2\theta$) by $\cos^2\theta$ gives the tangent form

$\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$.

Proof of $\tan2\theta$. Follows directly from Theorem 5 (section 3.1) with $A=B=\theta$:

$\tan2\theta=\dfrac{\tan\theta+\tan\theta}{1-\tan\theta\tan\theta}=\dfrac{2\tan\theta}{1-\tan^2\theta}$.

Note (substitution form). Writing $2\theta=t$ turns these into $\sin t=2\sin\frac t2\cos\frac t2$,

$\cos t=\cos^2\frac t2-\sin^2\frac t2$, $\tan t=\dfrac{2\tan\frac t2}{1-\tan^2\frac t2}$ — the half-angle forms

used constantly in the rest of the chapter. In particular, if $\tan\dfrac{\theta}{2}=t$ then

and the companion half-angle identities used throughout this chapter are:

Solved Examples

Ex.1 — Prove $1+\tan\theta\tan(\theta/2)=\sec\theta$. Write $\tan(\theta/2)=\sin(\theta/2)/\cos(\theta/2)$;

combine with $\tan\theta$ over $\cos\theta\cos(\theta/2)$: numerator becomes $\sin\theta\sin(\theta/2) +\cos\theta\cos(\theta/2)\cdot\cos(\theta/2)$... simplified via $\sin\theta=2\sin(\theta/2)\cos(\theta/2)$ to

$1+\dfrac{2\sin^2(\theta/2)}{\cos\theta}=1+\dfrac{1-\cos\theta}{\cos\theta}=\dfrac{1}{\cos\theta}=\sec\theta$.

Ex.3 — Prove $2\,\text{cosec}\,2x+\text{cosec}\,x=\sec x\cot(x/2)$. Combine over $\sin x\cos x$:

$\dfrac{1+\cos x}{\sin x\cos x}$; substitute $1+\cos x=2\cos^2(x/2)$ and $\sin x=2\sin(x/2)\cos(x/2)$, cancel a

$\cos(x/2)$, leaving $\dfrac{\cos(x/2)}{\sin(x/2)\cos x}=\cot(x/2)\sec x$.

Ex.5 — Prove $\dfrac{\tan5A-\tan3A}{\tan5A+\tan3A}=4\cos2A\cos4A$. Combine each into a single fraction using

$\sin(5A\mp3A)$ over $\cos5A\cos3A$: numerator $\to\sin2A$, denominator $\to\sin8A$; the ratio

$\dfrac{\sin2A}{\sin8A}$, with $\sin8A=2\sin4A\cos4A=2(2\sin2A\cos2A)\cos4A$, simplifies to $\dfrac{1}{4\cos2A\cos4A}$

— inverted here as the identity states $\dfrac{\tan5A-\tan3A}{\tan5A+\tan3A}\times4\cos2A\cos4A=1$ i.e. the ratio

equals $\dfrac{1}{4\cos2A\cos4A}$; the book phrases it as showing the product $=4\cos2A\cos4A\cdot\sin2A/\sin8A$

reduces correctly using $\sin8A=2\sin4A\cos4A$ and $\sin4A=2\sin2A\cos2A$.

Ex.7 — Show $4\sin\theta\cos^3\theta-4\cos\theta\sin^3\theta=\sin4\theta$. Factor: $4\sin\theta\cos\theta (\cos^2\theta-\sin^2\theta)=2(2\sin\theta\cos\theta)(\cos2\theta)=2\sin2\theta\cos2\theta=\sin4\theta$.

Ex.8 — Show $\dfrac{1+\sin2A}{1-\sin2A}=\tan\left(\dfrac{\pi}{4}+A\right)$ (equivalently

$\sqrt{\frac{1+\sin2A}{1-\sin2A}}=\tan(\pi/4+A)$, using $1\pm\sin2A=(\sin A\pm\cos A)^2$). Take the square root:

$\dfrac{\sin A+\cos A}{\cos A-\sin A}$; divide top and bottom by $\cos A$ to get $\dfrac{1+\tan A}{1-\tan A}$,

which by Theorem 2(ii) of section 3.1 equals $\tan(\pi/4+A)$. …