Double Angle Formulas

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In this activity, you will discover formulas for $\sin (2\theta )$ , $\cos (2\theta )$ , $\sin \left (\frac {\theta }{2}\right )$ and $\cos \left (\frac {\theta }{2}\right )$ , using your knowledge of the Angle Sum Formulas.

Recall the Pythagorean Theorem:

$\sin ^2 \theta + \cos ^2 \theta = 1$

We can use the Pythagorean theorem to derive expressions for $\sin ^2\theta$ and for $\cos ^2\theta$ .

Subtract $\cos ^2 \theta$ from both sides to get a formula for $\sin ^2 \theta$ :

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

Subtract $\sin ^2 \theta$ from both sides to get a formula for $\cos ^2 \theta$ :

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

We can derive double-angle formulas using the angle-sum formulas from a previous activity.

Rewrite the following values as a multiple of 2:

$4=2\times \answer {2}$
$8=2\times \answer {4}$
$70=2\times \answer {35}$
$\theta = 2 \times \answer {\frac {\theta }{2}}$
$2\theta = 2\times \answer {\theta }$

Rewrite the following multiples of $2$ as the sum of a value with itself:

$4=2+\answer {2}$
$8=\answer {4}+\answer {4}$
$70=\answer {35}+\answer {35}$
$\theta = \answer {\frac {\theta }{2}}+\answer {\frac {\theta }{2}}$
$2\theta =\answer {\theta }+\answer {\theta }$

Describe the relationship between doubling a number and adding a number to itself.

We can now combine all of these ideas to derive the double-angle formulas for sine and cosine.

Recall the angle-sum formula for sine: $\sin \left (A+B\right ) = \sin A \cos B + \cos A \sin B$ We seek an expression for $\sin \left (2\theta \right )$ .

Use a substitution you made above to rewrite $\sin \left (2\theta \right )$ as the sine of the sum of two angles:

$\sin \left (2\theta \right )=\sin \left (\answer {\theta }+\answer {\theta }\right )$

What substitutions can you make for $A$ and $B$ in order to apply the angle-sum formula for sine to $\sin \left (2\theta \right )$ ?

$A=\answer {\theta }$
$B=\answer {\theta }$

Rewrite the angle-sum formula using your substitutions from Problem 7: $\sin \left (\theta +\theta \right )=\sin \answer {\theta }\cos \answer {\theta }+\cos \answer {\theta }\sin \answer {\theta }$

Hence, $\sin \left (2\theta \right )=\answer {2\sin \theta \cos \theta }$

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

You will now discover the double-angle formula for cosine.

Recall the angle-sum formula for cosine: $\cos \left (A+B\right ) = \cos A \cos B - \sin A \sin B$ We seek an expression for $\cos \left (2\theta \right )$ .

Rewrite $\cos \left (2\theta \right )$ as the cosine of the sum of two angles:

$\cos \left (2\theta \right )=\cos \left (\answer {\theta }+\answer {\theta }\right )$

What substitutions can you make for $A$ and $B$ in order to apply the angle-sum formula for cosine to $\cos \left (2\theta \right )$ ?

$A=\answer {\theta }$
$B=\answer {\theta }$

Plug your values from above into the angle-sum formula for cosine: $\cos \left (\theta +\theta \right ) = \cos \answer {\theta }\cos \answer {\theta }-\sin \answer {\theta }\sin \answer {\theta }$

Hence, $\cos \left (2\theta \right )=\answer {\cos ^2\theta -\sin ^2\theta }$

$\cos \theta \cos \theta =\cos ^2\theta$

You can find two more formulas for $\cos \left (2\theta \right )$ by using the Alternate Versions of the Pythagorean Theorem from the start of this activity.

Write an expression for $\cos \left (2\theta \right )$ that does not involve $\cos \theta$ :

$\cos \left (2\theta \right ) = \answer {1-2\sin ^2\theta }$

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

Write an expression for $\cos \left (2\theta \right )$ that does not involve $\sin \theta$ :

$\cos \left (2\theta \right ) = \answer {2\cos ^2\theta -1}$

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

Using these same techniques, you can discover a double-angle formula for tangent. Recall from a previous activity that $\tan \left (A+B\right ) = \displaystyle {\frac {\tan A + \tan B}{1-\tan A \tan B}}$

Substitute $A=\theta$ and $B=\theta$ to derive an expression for $\tan \left (2\theta \right )=\tan \left (\theta +\theta \right )$ . $\tan \left (\theta +\theta \right )=\frac {\tan \answer {\theta }+\tan \answer {\theta }}{1-\tan \answer {\theta }\tan \answer {\theta }}$

Hence, $\tan \left (2\theta \right )=\answer {\frac {2\tan \theta }{1-\tan ^2\theta }}$

$\tan \theta +\tan \theta =2\tan \theta$

$\tan \theta \tan \theta =\tan ^2\theta$

Be mindful of parentheses when writing fractions! $a/1-b$ and $\frac {a}{1-b}$ are not the same expression.

If you can take the sine, cosine, and tangent of two-times-an angle, it might be natural to ask if you take the sine, cosine, and tangent of half of an angle.

Rewrite the following values as half of another value.

$2=\frac {\answer {4}}{2}$
$4=\frac {\answer {8}}{2}$
$35=\frac {\answer {70}}{2}$
$\frac {\theta }{2} = \frac {\answer {\theta }}{2}$
$\theta =\frac {\answer {2\theta }}{2}$

First you will derive the half-angle formula for sine.

Begin with $\cos \left (2A\right ) = 1 - 2 \sin ^2\left (A\right )$ . Let $2A = \theta$ . Rewrite the previous equation in terms of $\theta$ :

$\cos \left (\theta \right ) = \answer {1-2\sin ^2\left (\frac {\theta }{2}\right )}$

Your answer should include $\sin \left (\frac {\theta }{2}\right )$ .

Solve for $\sin \left (\frac {\theta }{2}\right )$ in the previous expression: $\sin \left (\frac {\theta }{2}\right )=\answer {\pm \sqrt {\frac {1-\cos \theta }{2}}}$

Subtract $1$ from both sides.

Divide by $-2$ and distribute the $-$ sign.

Take the square root of both sides. Don’t forget the $\pm$ sign!

Next you will derive a half-angle formula for cosine.

Begin with $\cos \left (2A\right ) = 2 \cos ^2\left (A\right ) - 1$ . Let $2A = \theta$ . Rewrite $\cos \left (2A\right )$ in terms of $\theta$ :

$\cos \left (\theta \right )=\answer {2\cos ^2\left (\frac {\theta }{2}\right )-1}$

Your answer should include $\cos \left (\frac {\theta }{2}\right )$ .

Solve for $\cos \left (\frac {\theta }{2}\right )$ in the previous expression: $\cos \left (\frac {\theta }{2}\right )=\answer {\pm \sqrt {\frac {\cos \theta +1}{2}}}$

Subtract $1$ from both sides.

Divide by $2$ .

Take the square root of both sides. Don’t forget the $\pm$ sign!

Lastly, you will discover the half-angle formula for tangent.

Recalling that $\tan \theta =\frac {\sin \theta }{\cos \theta }$ , derive the half-angle formula for tangent:

$\tan \left (\frac {\theta }{2}\right ) = \frac {\sin \left (\frac {\theta }{2}\right )}{\cos \left (\frac {\theta }{2}\right )} = \answer {\pm \sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}$

Use your expressions for $\sin \left (\frac {\theta }{2}\right )$ and $\cos \left (\frac {\theta }{2}\right )$ as the numerator and denominator of a fraction.

Square roots distribute over division, e.g., $\frac {\sqrt {a}}{\sqrt {b}}=\sqrt {\frac {a}{b}}$

Dividing by a fraction is the same as multiplying by the reciprocal of that fraction.