Double Angle and Half Angle Formulas

Did you know that

$\sin 2\theta \ne 2\sin \theta$

and

$\cos \frac{\theta }{2}\ne \frac{\cos \theta }{2}$?

In this article, you would understand what happens when trigonometric identities are either doubled or halved.

Double-angle formulas

Trigonometric functions can be doubled but not in the same way as normal numbers are doubled.

If you have the expression 3y and you are to double it, it is easy to multiply 3y by 2 to get 6y. Note that sin30° is 0.5 and doubling the angle gives 60°, but sin60° would not give you 1. As normal mathematical operations would multiply 0.5 by 2 to give 1, trigonometric identities would require their own formula to double their function.

Deriving the double-angle formula for Sine function

We intend on finding the formula for sin2θ. Note that

$\sin 2\theta =\sin (\theta +\theta )$

Recall that,

$\sin (A+B)=\sin A\cos B+\sin B\cos A$

Now take$A=B=\theta ,$ we get

$$\begin{gathered} \sin 2\theta =\sin \theta \cos \theta +\sin \theta \cos \theta \\ =2\sin \theta \cos \theta \end{gathered}$$

Deriving the double-angle formula for Cosine function

We will now develop a formula for a double-angle for the cosine function. We derive three equal formulas.

We first note that

$\cos 2\theta =\cos (\theta +\theta )$

Now, recall that

$\cos (A+B)=\cos A\cos B-\sin A\sin B$

By taking$A=B=\theta ,$we get

$$\begin{gathered} \cos 2\theta =\cos (\theta +\theta ) \\ =\cos \theta \cos \theta -\sin \theta \sin \theta \\ ={\cos }^2\theta -{\sin }^2\theta \end{gathered}$$

Thus, we derive the first formula for $\cos 2\theta$,

$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$

Now we replace this with the obtained formula for$\cos 2\theta ,$to get

$$\begin{gathered} \cos 2\theta =1-{\sin }^2\theta -{\sin }^2\theta \\ =1-2{\sin }^2\theta \end{gathered}$$

Thus, the second formula for $\cos 2\theta$ is

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

In a similar way, we have ${\sin }^2\theta =1-{\cos }^2\theta$.

Substituting the value of ${\sin }^2\theta$ into the formula of $\cos 2\theta$, we have

$$\begin{gathered} \cos 2\theta ={\cos }^2\theta -(1-{\cos }^2\theta ) \\ ={\cos }^2\theta -1+{\cos }^2\theta \\ =2{\cos }^2\theta -1 \end{gathered}$$

Thus, the third formula for $\cos 2\theta$ is

$\cos 2\theta =2{\cos }^2\theta -1$

Deriving the double-angle formula for Tangent function

We will develop a formula for a double-angle of the tangent function.

We recall that

$\tan \theta =\frac{\sin \theta }{\cos \theta }$

and,

$\sin 2\theta =2\sin \theta \cos \theta$

$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$

Thus,

$\tan 2\theta =\frac{\sin 2\theta }{\cos 2\theta }$

Substituting $\sin 2\theta$ and $\cos 2\theta$ by their expressions, we get

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

To simplify this further, we divide both the numerator and the denominator of the right-hand side of the equation by ${\cos }^2\theta$, to get

$$\begin{gathered} \tan 2\theta =\frac{\frac{2\sin \theta \cos \theta }{{\cos }^2\theta }}{\frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta }} \\ =\frac{\frac{2\sin \theta }{\cos \theta }}{\frac{{\cos }^2\theta }{{\cos }^2\theta }-\frac{{\sin }^2\theta }{{\cos }^2\theta }} \\ =\frac{2\tan \theta }{1-{\tan }^2\theta } \end{gathered}$$

Deriving the double-angle formulas for Secant, Cosecant and Cotangent functions

Secant, cosecant, cotangent functions are the reciprocals of cosine, sine and tangent respectively. In order to derive their double-angle formulas, you just need to find the multiplicative inverse of the corresponding double-angle formulas.

Double angle formula for Secant

We recall by the definition of the secant function that

$sec\theta =\frac{1}{\cos \theta }$ so $sec2\theta =\frac{1}{\cos 2\theta }$

but from the double angle formula for cosine we have $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$ , thus

$sec2\theta =\frac{1}{{\cos }^2\theta -{\sin }^2\theta }$

Now let us express $sec2\theta$in terms of $sec\theta$ and $csc\theta$.

In fact, $\cos \theta =\frac{1}{sec}$and $csc\theta =\frac{1}{\sin \theta }$, thus we have

$$\begin{gathered} sec2\theta =\frac{1}{{\left(\frac{1}{sec\theta }\right)}^2-{\left(\frac{1}{csc\theta }\right)}^2} \\ =\frac{1}{\frac{1}{se{c}^2\theta }-\frac{1}{cs{c}^2\theta }} \\ =\frac{1}{\frac{cs{c}^2\theta -se{c}^2\theta }{se{c}^2\theta cs{c}^2\theta }} \\ =\frac{se{c}^2\theta cs{c}^2\theta }{cs{c}^2\theta -se{c}^2\theta }. \end{gathered}$$

Double angle formula for Cosecant

We recall by the definition of the secant function that

$csc\theta =\frac{1}{\sin \theta }$so $csc2\theta =\frac{1}{\sin 2\theta }$

but from the double angle formula for sine function, we have$\sin 2\theta =2\sin \theta \cos \theta$, thus

$$\begin{gathered} csc2\theta =\frac{1}{2\sin \theta \cos \theta } \\ =\frac{1}{2}\times \frac{1}{\sin \theta }\times \frac{1}{\cos \theta } \\ =\frac{1}{2}\times csc\theta \times sec\theta \\ csc2\theta =\frac{1}{2}csc\theta sec\theta \end{gathered}$$

Double angle formula for Cotangent

We recall by the definition of the secant function that

$cot\theta =\frac{1}{\tan \theta }$so $cot2\theta =\frac{1}{\tan 2\theta }$

We recall from the double angle formula for the tangent function that $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$, we have

$$\begin{gathered} cot2\theta =\frac{1}{\frac{2\tan \theta }{1-{\tan }^2\theta }} \\ =\frac{1-{\tan }^2\theta }{2\tan \theta } \end{gathered}$$

Half-angle formulas

Trigonometric functions can be halved but not in the same manner normal numbers are halved. If you have the expression 6y and you are to half it, it is easy to multiply 6y by 0.5 to get 3y. Note that sin30° is 0.5 and halving the angle gives 15 degrees, but sin15° would not give you 0.25. As normal mathematical operations would multiply 0.5 by 0.5 (half) to give 0.25, trigonometric identities would require their own formula to half its function.

Deriving the half-angle formula for Sine

To find $\sin \frac{\theta }{2}$, we recall first that

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

Let $\theta =\frac{\varnothing }{2}$, thus

$$\begin{gathered} \cos \left(2\times \frac{\varphi }{2}\right)=1-2{\sin }^2\frac{\varphi }{2} \\ \cos \varphi =1-2{\sin }^2\frac{\varphi }{2} \end{gathered}$$

In order to isolate ${\sin }^2\frac{\varphi }{2}$, subtract 1 from both sides, to get

$$\begin{gathered} \mathrm{co}\varphi -1=1-2{\sin }^2\frac{\varphi }{2}-1 \\ -2{\sin }^2\frac{\varphi }{2}=\cos \varphi -1 \end{gathered}$$

${\sin }^2\frac{\varphi }{2}=\frac{1-\cos \varphi }{2}$

Taking the square root of both sides of the equation, we get

$\sin \frac{\varphi }{2}=\pm \sqrt{\frac{1-\cos \varphi }{2}}$

Deriving the half-angle formula for cosine

Recall that

$\cos 2\theta =2{\cos }^2\theta -1$

Where,

$\theta =\frac{\varnothing }{2}$

Therefore

$$\begin{gathered} \cos (2\times \frac{\varnothing }{2})=(2\times {\cos }^2\frac{\varnothing }{2})-1 \\ \cos \varnothing =(2\times {\cos }^2\frac{\varnothing }{2})-1 \end{gathered}$$

$$\begin{gathered} \cos \varnothing +1=(2\times {\cos }^2\frac{\varnothing }{2})-1+1 \\ \cos \varnothing +1=2\times {\cos }^2\frac{\varnothing }{2} \end{gathered}$$

Divide both sides by 2

$\frac{\cos \varnothing +1}{2}={\cos }^2\frac{\varnothing }{2}$

Find the square root of both sides of the equation

$\sqrt{\frac{\cos \varnothing +1}{2}}=\cos \frac{\varnothing }{2}$

Thus

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

Deriving the half-angle formula for tangents

Knowing that

$$\begin{gathered} \tan \theta =\frac{\sin \theta }{\cos \theta } \\ \sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}} \\ \cos \frac{\theta }{2}=\pm \sqrt{\frac{\cos \theta +1}{2}} \end{gathered}$$

Then

$$\begin{gathered} \tan \frac{\theta }{2}=\frac{\sqrt{\frac{1-\cos \theta }{2}}}{\sqrt{\frac{\cos \theta +1}{2}}} \\ \tan \frac{\theta }{2}=\frac{\frac{\sqrt{1-\cos \theta }}{\sqrt{2}}}{\frac{\sqrt{\cos \theta +1}}{\sqrt{2}}} \\ \tan \frac{\theta }{2}=\frac{\sqrt{1-\cos \theta }}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{\cos \theta +1}} \\ \tan \frac{\theta }{2}=\frac{\sqrt{1-\cos \theta }}{\sqrt{\cos \theta +1}} \end{gathered}$$

Multiply the right-hand side of the equation by $\frac{\sqrt{\cos \theta +1}}{\sqrt{\cos \theta +1}}$ and you would have

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

Recall that

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

Then

$$\begin{gathered} \tan \frac{\theta }{2}=\frac{\sqrt{{\sin }^2\theta }}{\cos \theta +1} \\ t\mathrm{an}\frac{\theta }{2}=\frac{\sin \theta }{\cos \theta +1} \end{gathered}$$

Deriving the half-angle formula for secant, cosecant and cotangent

As mentioned earlier, secant, cosecant, cotangent are the inverse of cosine, sine and tangent respectively. So as to derive their half-angle formulas, you just need to find the multiplicative inverse of the corresponding half-angle formulas. Thus the half-angle formula of secant becomes:

$$\begin{gathered} sec\theta =\frac{1}{\cos \theta } \\ \cos \frac{\theta }{2}=\pm \sqrt{\frac{\cos \theta +1}{2}} \\ sec\frac{\theta }{2}=\pm \sqrt{\frac{2}{\cos \theta +1}} \end{gathered}$$

the half-angle formula of cosecant becomes:

$$\begin{gathered} \cos ec\theta =\frac{1}{\sin \theta } \\ \sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}} \\ \cos ec\frac{\theta }{2}=\pm \sqrt{\frac{2}{1-\cos \theta }} \end{gathered}$$

and the half-angle formula of cotangent becomes:

$$\begin{gathered} cot\theta =\frac{1}{\tan \theta } \\ \tan \frac{\theta }{2}=\frac{\sin \theta }{\cos \theta +1} \\ cot\frac{\theta }{2}=\frac{\cos \theta +1}{\sin \theta } \end{gathered}$$

This is the same as

$$\begin{gathered} cot\frac{\theta }{2}=\frac{\cos \theta +1}{\sin \theta } \\ cot\frac{\theta }{2}=\frac{\cos \theta }{\sin \theta }+\frac{1}{\sin \theta } \\ cot\frac{\theta }{2}=cot\theta +\cos ec\theta \end{gathered}$$

Applications of double-angle and half-angle formulas

Here are a few examples that show the application of double-angle and half-angle formulas.