Identitas Sudut Ganda

The double angle formulas for the trig functions can often come in handy. I will present two proofs of them: a geometric one, and a purely algebraic one.

First I will present a geometric proof.

Consider the following diagram:

Let $\angle HAC = \theta, AC = 1, BC = b$. Note then that $\angle HCA = 90^\circ -\theta$ and $HC = \frac{b}{2}$. So, $\sin \theta = \frac{b}{2} \implies b = 2\sin \theta$. Looking at the entire triangle, the Law of Sines tells us that $$\dfrac{\sin \angle BAC}{b} = \dfrac{\sin \angle HCA}{1}$$ $$\implies \dfrac{\sin 2\theta}{b} = \sin (90^\circ-\theta) = \cos \theta.$$ Plugging in $b = 2\sin \theta$ gives $\dfrac{\sin 2\theta}{2\sin\theta} = \cos \theta \implies \sin 2\theta = 2\sin\theta\cos\theta$.

Using the same diagram, we will prove formula for $\cos 2\theta$. Now, however, we will let $\angle ACH = \alpha$.

This time, try to come up with an idea of the proof on your own.

We find that $\cos \alpha = \frac{b}{2}$.

What's next?

Now, we use the Law of Cosines: $$b^2 = 2 - 2\cos(180-2\alpha)$$ We know that $\cos(180-x) = -\cos x$, so we get$$b^2 = 2 +2\cos 2\alpha$$ Substituting in $b = 2\cos\alpha$ gives $$4\cos^2\alpha = 2 + 2\cos 2\alpha$$ $$\cos 2\alpha = 2\cos^2 \alpha - 1 = \cos^2\alpha - \sin^2\alpha = 1 - 2\sin^2\alpha$$

It's harder to find $\tan 2\alpha$ geometrically, but we can get it by dividing $\sin 2\alpha$ by $\cos 2\alpha$ anyways.

Now, the algebraic proof:

We know that $\sin(x+y) = \sin x \cos y + \sin y \cos x$. We want $\sin 2\theta$. Do you see how we can find the double angle formulas now?

If you figured it out, just continue. If you haven't gotten it yet, think harder--it's really simple!

Here's a hint: $2\theta = \theta + \theta$.

All we have to do is use the angle addition formula, where $x=y = \theta$. So, we have $$\sin(2\theta) = \sin(\theta + \theta) = \sin\theta\cos\theta + \sin\theta\cos\theta = 2\sin\theta\cos\theta.$$

Similarly, we know that $\cos(x+y) = \cos x\cos y - \sin x \sin y$. Substituting in $x=y=\theta,$ we get $$\cos(2\theta) = \cos(\theta + \theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta-1 = 1-2\sin^2\theta.$$ Finally, $\tan(x+y) = \dfrac{\tan x + \tan y}{1 -\tan x\tan y}$, so we get $$\tan(2\theta) = \dfrac{2\tan\theta}{1-\tan^2\theta}.$$

So we're done. Whichever proof you prefer, the double angle identities are great to have in mind in a lot of trig problems. Therefore, it's good to memorize them, especially since it isn't particularly hard to.

The double angle identities are really easy to derive. In this post, we'll show how to find formulas for $\sin{2\theta}$, $\cos{2\theta}$, and $\tan{2\theta}$. That means that we will find the double angle identities for sine, cosine, and tangent.

For $\sin{2\theta}$,...actually you think about it first! Click for a hint.

$2\theta = \theta + \theta$...

$\sin{2\theta} = \sin{\theta + \theta} = \sin{\theta}\cos{\theta} + \sin{\theta}\cos{\theta} = 2\sin{\theta}\cos{\theta}$ $$\boxed{\sin{2\theta} = 2\sin{\theta}\cos{\theta}}$$

Apply the same reasoning for $\cos{2\theta}$!

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

What's cool about this is that by the Pythagorean identity, $\sin^2{\theta} + \cos^2{\theta} = 1$, we have $\cos^2{\theta} = 1 - \sin^2{\theta}$ and $\sin^2{\theta} = 1 - \cos^2{\theta}$. Substitute these in:

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

Therefore we have:

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

Try doing something for $\tan{2\theta}$!

$\tan{2\theta} = \tan{\theta + \theta} = \frac{\tan{\theta} + \tan{\theta}}{1 - (\tan{\theta})(\tan{\theta})} = \frac{2\tan{\theta}}{1 - \tan^2{\theta}}$ $$\boxed{\tan{2\theta} = \frac{2\tan{\theta}}{1 - \tan^2{\theta}}}$$

The double-angle identities are identities that can be used to find the values of the trigonometric functions of double of an angle, given the values for the original angle.

The double-angle identities follow immediately from the general angle-addition identities: $$\sin(A+B) = \sin A\cos B + \cos A\sin B$$ $$\cos(A+B) = \cos A\cos B - \sin A\sin B$$ $$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$$ Setting $B = A$, we get the double-angle identities: $$\sin(2A) = 2\sin A\cos A$$ $$\cos(2A) = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$$ $$\tan(2A) = \frac{2\tan A}{1 - \tan^2 A}$$

Trigonometry: Double-Angle Identities

There are six negative angle identities you need to memorize!

• Negative angle identities
• Co-function identities
• Angle and sum differences identities
• Double angle identities (this lesson)
• Half angle identities

Double Angle Identities are trig formulas that help you to find the sine, cosine, and tangent of an angle that is twice as large as another angle. Instead of working with $2x$ directly, you would rewrite the expression using only $x$. Double angle identities come from the angle sum identities.

Double angle identities are shortcuts! They turn trig functions with larger angles into expressions using smaller, simpler angles. That's incredibly helpful as we navigate complex trig expressions.

Once you learn the main patterns, recognizing when to use them becomes much easier.

There are three very important formulas.

For sine, $$\sin(2x)=2\sin(x)\cos(x)$$

For cosine, $$\cos(2x)=\cos^2(x) - \sin^2(x)$$

For tangent, $$\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$$

Special Forms

The cosine double angle identity can also be rewritten into other useful ways.

Using $\sin^2(x) + \cos^2(x) = 1$, we can say,

$$\cos(2x) = 1 - 2\sin^2(x)$$

, and,

$$\cos(2x) = 2\cos^2(x)-1$$

You can use any of these, whichever is most helpful for you to tackle an expression. Think about which one matches the information you already are working with.

Common Mistakes

A common mistake is thinking that $\sin(2x) = 2\sin(x)$.

That's not true! The cosine term is important here. Here is the correct answer:

$$\sin(2x) = 2\sin(x)\cos(x)$$

Always use the full identity.

(Video) Double Angle Identities

by Randy Anderson

As we continue on our journey of trigonometry, we have learned about negative angle identities, co-function identities, and angle, sum, and difference identities, but now time to learn about double angle identities!

Check out this video by Randy Andersen to learn how to derive the double angle identities.

Watch the video to see the step-by-step derivation. In summary, the double angle identities are:

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

Let's work through the example problem in the video!

If given that $\cos{\theta}=-\frac{4}{5}$ and $\sin{\theta}>0$, we can find $\sin{2\theta}$.

Following along with the video...

$$\sin{2\theta}=2\sin{\theta}\cos{\theta}$$ $$=2(\frac{3}{5})(-\frac{4}{5})$$ $$=-\frac{24}{25}$$

Keep watching the video to try more practice problems!