The Trigonometry Formula for Double Angles is a continuation of the Sum and Difference of Trigonometry Angles Formula

After we previously studied Formulas for the Sum and Difference of Trigonometry Angles↝ , we continued the material to Trigonometry Formulas for Double Angles. The double angle in question is $2\alpha$ and also the angle $\dfrac12 \alpha$.

Like the formula for the sum and difference of two angles, the double angle formula is used to determine the trigonometric value for an angle that is not a special angle ($0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ $) without calculating tools such as calculators or tables.

How to use this double angle formula? see the following review.

1. Double Angle Trigonometry Formula

We can obtain or prove this formula by using the trigonometry formula for the sum of two angles.

Proving the double angle trigonometry formula:

$\bigstar$ Rumus Sinus sukta rangkap $ \sin 2\alpha = 2 \sin \alpha \cos \alpha $

Do not multiply sine roots: $ \sin (A+B) = \sin A \cos B + \cos A \sin B $

$$\begin{align*}\sin 2\alpha &= \sin ( \alpha + \alpha ) \\ &= \sin \alpha \cos \alpha + \cos \alpha \sin \alpha \\ &= 2 \sin \alpha \cos \alpha \end{align*}$$

So it is proved: $ \sin 2\alpha = 2 \sin \alpha \cos \alpha $

$\bigstar$ Rumus Cosinus sugat rangkap : $ \cos 2\alpha = \cos ^2 \alpha - \sin ^2 \alpha $

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

$$\begin{align*} \cos 2\alpha &= \cos (\alpha + \alpha ) \\ &= \cos \alpha \cos \alpha - \sin \alpha \sin \alpha \\ &= \cos ^2 \alpha - \sin ^2 \alpha \end{align*} $$

So it is proven: $ \cos 2\alpha = \cos ^2 \alpha - \sin ^2 \alpha $

Using the identity formula: $ \sin ^2 A + \cos ^2 A = 1 \rightarrow \sin ^2 A = 1 - \cos ^2 A $ we obtain

$ \bigstar$ Rumus : $ \cos 2\alpha = 2\cos ^2 \alpha - 1 $

$$\begin{align*}\cos 2\alpha &= \cos ^2 \alpha - \sin ^2 \alpha \\ &= \cos ^2 \alpha - (1 - \cos ^2 \alpha ) \\ &= 2\cos ^2 \alpha - 1 \end{align*} $$

Terbukti : $ \cos 2\alpha = 2\cos ^2 \alpha - 1 $

$ \bigstar$ Rumus : $ \cos 2\alpha = 1 - 2\sin ^2 \alpha $

$$\begin{align*}\cos 2\alpha &= \cos ^2 \alpha - \sin ^2 \alpha \\ &= ( 1 - \sin ^2 \alpha ) - \sin ^2 \alpha \\ &= 1 - 2\sin ^2 \alpha \end{align*}$$

Terbukti : $ \cos 2\alpha = 1 - 2\sin ^2 \alpha $

$ \bigstar$ Tangent Double Angle Formula : $ \tan 2\alpha = \frac{2\tan \alpha }{1 - \tan ^2 \alpha } $

Remember the formula : $ \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} $

$$\begin{align*} \tan 2\alpha &= \tan ( \alpha + \alpha ) \\ &= \frac{\tan \alpha + \tan \alpha }{1 - \tan \alpha \tan \alpha } \\ &= \frac{2\tan \alpha }{1 - \tan ^2 \alpha } \end{align*}$$

So it is proved: $ \tan 2\alpha = \dfrac{2\tan \alpha }{1 - \tan ^2 \alpha } $

Examples of Compound Angle Questions

  1. Simplify the following shapes

    1. $2 \sin 22.5^\circ \cos 22.5^\circ$
    2. $2 \cos^2 67,5^\circ – 1$
    3. $\dfrac{2\tan 3\alpha}{1-\tan^2\alpha}$

    Solution

    1. $2 \sin 22.5^\circ \cos 22.5^\circ$
      gunakan rumus $\sin 2A=2\sin A \cos A$
      $$\begin{align*}2 \sin 22,5^\circ \cos 22,5^\circ&=\sin 2(22,5^\circ)\\&=\sin 45^\circ \\ &=\frac12 \sqrt{2}\end{align*}$$
    2. $2 \cos^2 67,5^\circ – 1$
      use the formula $\cos 2A=2\cos^2 A -1$
      $$\begin{align*}2 \cos^2 67,5^\circ – 1&=\cos 2(67,5^\circ)\\&=\cos 135^\circ = \cos (180^\circ - 45^\circ) \\ &=-\cos 45^\circ =-\frac12 \sqrt{2}\end{align*}$$
    3. $\dfrac{2\tan 3\alpha}{1-\tan^2\alpha}$
      gunakan rumus $\tan 2A=\dfrac{2\tan A}{1-\tan^2A} $
      $$\begin{align*}\frac{2\tan 3\alpha}{1- \tan^2 \alpha }&=\tan 2(3\alpha) \\ &=\tan 6\alpha \end{align*}$$

  2. Given $\sin A =\dfrac{5}{13}$, where A is sharp. Calculate the values ​​of $\sin 2A$, $\cos 2A$, and $\tan 2A$ !

    Solution

    Because $\sin A$ is already known, first find $\cos A$. We can use trigonometric identity formulas or draw trigonometric comparison triangles.

    We just use the identity formula, namely $\cos A = \sqrt{1-\sin^2A}$

    $\sin A =\dfrac{5}{13}$ for you

    $$\begin{align*} \cos A &= \sqrt{1-\sin^2A}\\ &= \sqrt{1-\left( \frac{5}{13} \right)^2} \\ &=\sqrt{1- \frac{25}{169} } \\ &=\sqrt{\frac{144}{169}} \\ \cos A &= \frac{12}{13}\end{align*}$$

    1. Nilai $\sin 2A$

      $$\begin{align*}\sin 2 A &=2\sin A \cos A \\ &= 2 \cdot \frac{5}{13} \cdot \frac{12}{13} \\ \sin 2A&= \frac{120}{169}\end{align*} $$

    2. Nilai $\cos 2A$

      $$\begin{align*}\cos 2 A &=\cos^2 A - \sin^2 A \\ &= \left(\frac{12}{13} \right)^2- \left(\frac{5}{13} \right)^2\\ &= \frac{144}{169}-\frac{25}{169}\\ \cos 2A &= \frac{119}{169} \end{align*} $$

    3. Nilai $\tan 2A$

      $$\begin{align*}\tan 2A &=\frac{\sin 2A}{\cos 2A} \\ &=\frac{\frac{120}{169}}{\frac{119}{169}} \\ \tan 2A &=\frac{120}{119} \end{align*} $$

2. Trigonometry Formulas for Half Angles

From the double angle trigonometry formula, the trigonometry formula for half an angle, namely by setting $\dfrac12 \alpha$ as a single angle and $\alpha$ as a double angle.

Proof of the Angle Formula $ \frac{1}{2} A $ :

Misalkan $ 2\alpha = A \rightarrow \alpha = \frac{1}{2} A $

Substitute the form of the example above into the double angle trigonometric equation that will be proven.

$\bigstar$ Rumus : $ \sin \frac{1}{2} A = \sqrt{\frac{1- \cos A}{2}} $

use the formula: $ \cos 2 \alpha = 1 - 2\sin ^2 \alpha $

$$\begin{align*}\cos 2 \alpha &= 1 - 2\sin ^2 \alpha \\ \cos A &= 1 - 2\sin ^2 \frac{1}{2} A\\2\sin ^2 \frac{1}{2} A &= 1 - \cos A \\\sin ^2 \frac{1}{2} A &= \frac{1 - \cos A}{2} \\\sin \frac{1}{2} A &= \sqrt{\frac{1 - \cos A}{2} } \end{align*}$$

Until proven : $ \sin \frac{1}{2} A = \sqrt{\frac{1- \cos A}{2}} $

$\bigstar$ Rumus : $ \cos \frac{1}{2} A = \sqrt{\frac{1 + \cos A}{2}} $

use the formula: $ \cos 2 \alpha = 2\cos ^2 \alpha - 1 $

$$\begin{align*} \cos 2 \alpha &= 2\cos ^2 \alpha - 1 \\\cos A &= 2\cos ^2 \frac{1}{2}A - 1 \\2\cos ^2 \frac{1}{2}A &= 1 + \cos A \\\cos ^2 \frac{1}{2}A &= \frac{1 + \cos A}{2} \\\cos \frac{1}{2}A &= \sqrt{\frac{1 + \cos A}{2} } \end{align*}$$

Until proven : $ \cos \frac{1}{2} A = \sqrt{\frac{1 + \cos A}{2}} $

$\bigstar$ Rumus : $ \tan \frac{1}{2} A = \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A }{1+ \cos A} = \frac{1- \cos A}{\sin A } $

gunakan rumus : $ \tan \frac{1}{2}A = \frac{\sin \frac{1}{2}A }{\cos \frac{1}{2}A } , \sin \frac{1}{2} A = \sqrt{\frac{1- \cos A}{2}} $, dan $ \cos \frac{1}{2} A = \sqrt{\frac{1 + \cos A}{2}} $

to). Rumus Pertama: $ \tan \frac{1}{2}A = \frac{\sin \frac{1}{2}A }{\cos \frac{1}{2}A }$

$$\begin{align*}\tan \frac{1}{2} A &= \frac{\sin \frac{1}{2}A }{\cos \frac{1}{2}A } \ &= \frac{ \sqrt{\frac{1- \cos A}{2}} }{ \sqrt{\frac{1 + \cos A}{2}} } \ \tan \frac{1}{2}A &= \sqrt{ \frac{1- \cos A}{1 + \cos A} } \end{align*}$$

b). Second formula:

$$\begin{align*} \tan \frac{1}{2}A &= \sqrt{ \frac{1- \cos A}{1 + \cos A} } \\ \tan \frac{1}{2}A &= \sqrt{ \frac{1- \cos A}{1 + \cos A} \times \frac{1 + \cos A}{1 + \cos A} } \\ &= \sqrt{ \frac{1- \cos ^2 A}{(1 + \cos A)^2} } \\ &= \sqrt{ \frac{\sin ^2 A }{(1 + \cos A)^2} } \\ &= \frac{\sin A}{1+ \cos A} \end{align*}$$

c). Third formula:

$$\begin{align*} \tan \frac{1}{2}A &= \sqrt{\frac{1- \cos A}{1 + \cos A} } \\ \tan \frac{1}{2}A &= \sqrt{\frac{1- \cos A}{1 + \cos A} \times \frac{1 - \cos A}{1 - \cos A} } \\ &= \sqrt{\frac{(1- \cos A)^2}{1 - \cos ^2 A} } \\ &= \sqrt{\frac{(1- \cos A)^2}{\sin ^2 A } } \\ &= \frac{1- \cos A}{\sin A } \end{align*}$$

Sehingga terbukti: $ \tan \frac{1}{2} A = \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1+ \cos A } = \frac{1- \cos A}{\sin A } $

Example of a Half-Angle Question:

Calculate the value of:

  1. $ \sin 15^\circ $
  2. $ \cos 67,5^\circ $
  3. $ \tan 22.5^\circ $

Alternative Solution:

  1. The value of $ \sin 15^\circ $

    Let $ \frac{1}{2}A = 15^\circ \rightarrow A = 30^\circ $

    $$\begin{align*}\sin \frac{1}{2} A &= \sqrt{\frac{1 - \cos A}{2} } \\ \sin 15^\circ &= \sqrt{\frac{1 - \cos 30^\circ}{2} } \\ &= \sqrt{\frac{1 - \frac{1}{2} \sqrt{3} }{2} } \\ &= \sqrt{\frac{2 - \sqrt{3} }{4} } \\ &= \frac{1}{2} \sqrt{2 - \sqrt{3} } \end{align*}$$

    So, the value $ \sin 15^\circ = \frac{1}{2} \sqrt{2 - \sqrt{3} } $

  2. The value of $ \cos 67.5^\circ $

    Let $ \frac{1}{2}A = 67.5^\circ \rightarrow A = 135^\circ $

    nilai $ \cos 135^\circ = \cos ( 180^\circ - 45^\circ ) = -\cos 45^\circ = -\frac{1}{2} \sqrt{2} $

    $$\begin{align*}\cos \frac{1}{2} A &= \sqrt{\frac{1 + \cos A}{2} } \\ \cos 67,5^\circ &= \sqrt{\frac{1 + \cos 135^\circ}{2} } \\ &= \sqrt{\frac{1 + (-\frac{1}{2} \sqrt{2} )}{2} } \\ &= \sqrt{\frac{2 - \sqrt{2} }{4} } \\ &= \frac{1}{2} \sqrt{2 - \sqrt{2} } \end{align*}$$

    So, the value of $ \cos 67.5^\circ = \frac{1}{2} \sqrt{2 - \sqrt{2} } $

  3. The value of $ \tan 22.5^\circ $

    Let $ \frac{1}{2}A = 22.5^\circ \rightarrow A = 45^\circ $

    $$\begin{align*} \tan \frac{1}{2} A &= \frac{\sin A}{1+ \cos A} \\ \tan 22,5^\circ &= \frac{\sin 45^\circ}{1+ \cos 45^\circ} \\ &= \frac{\frac{1}{2} \sqrt{2} }{1+ \frac{1}{2} \sqrt{2} } \\ &= \frac{ \sqrt{2} }{2+ \sqrt{2} } \\ &= \frac{ \sqrt{2} }{2+ \sqrt{2} } \times \frac{2 - \sqrt{2} }{2 - \sqrt{2} } \\ &= \frac{ 2\sqrt{2} - 2 }{4 - 2} \\ &= \frac{ 2\sqrt{2} - 2 }{2} \\ &= \sqrt{2} - 1 \end{align*}$$

    Thus, the value $ \tan 22.5^\circ = \sqrt{2} - 1 $