MHB How to find the Double Angle formula for Sin given only a triangle

[bsmithysmith]

View attachment 2423

I know the Double Angle for Sine is:
$$\sin(2x) = 2\sin(x) \cos(x)$$

but from the triangle given, how do I figure it out? We did this in class, but the teacher just told a small amount of things, and then let us talk amongst each other to solve it. Nearly all the students were glossy eyed and did not know how to find it.

Another question Is how do I find what $\cos(3x)=?$ Same thing but we did it today, and as always, none of the students know how to do it and the teacher didn't seem to help a lot. We had an origami folding paper to help us and made 3 triangles within 3x, so a trisect, if that's what it's called. Still not understanding the triple angle formula, but he did say that it's a combination of $\sin(2x), \sin, \cos(2x),$ or $\cos$; something like that. And he's looking for single thetas (whatever that means).

 

[MarkFL]

I would first look at $$\angle OCD$$. Can you express this angle as a function of $\theta$? We know $\angle CAD$ and $\angle ACD$ must be complementary...

 

[bsmithysmith]

I would first look at $$\angle OCD$$. Can you express this angle as a function of $\theta$? We know $\angle CAD$ and $\angle ACD$ must be complementary...

Since I don't know the line OD, then I was assume that the line OD = Sin $\theta$

Then what I also know is that $$\angle BAO$$ and $$\angle BCO$$ are definitely the same, so the $\theta$ of both the angles seem to be Sin $\theta$ as well.

Since $\theta$ is equal to Sin $\theta$, $$Sin(θ)=x/1$$ which is the same as $$Sin(θ)=x$$, Line BO should be Sin $\theta$. And looking at the diagram, AO is 1, and the hypotenuse, and proves the identity $$Sin^2(θ)+Cos^2(θ)=1$$.

So right now it seems that I need to know the Triangle OCD, where OC = 1. But if the hypotenuse is 1, wouldn't CD = $$Sin^2(θ)$$ and OD = $$Cos^2(θ)$$?

 

[MarkFL]

What I am suggesting is to observe that we must have (where $\alpha=\angle OCD$):

$$2\theta+\alpha=90^{\circ}$$

Hence:

$$\alpha=90^{\circ}-2\theta$$

Now, with respect to triangle $\Delta OCD$, we see that (where $\beta=\angle COD$):

$$\sin(\alpha)=\overline{OD}$$

$$\cos(\beta)=\overline{OD}$$

And so we may conclude that:

$$\sin(\alpha)=\cos(\beta)$$

Now, substitute for $\alpha$ and apply a co-function identity, and what is your conclusion?

 

[soroban]

Hello, bsmithysmith!

View attachment 2423

\angle COD is an exterior angle of \Delta AOC.

It is equal to the sum of the two nonadjacent angles:
. . .\angle COD \:=\:2\theta

 

[bsmithysmith]

What I am suggesting is to observe that we must have (where $\alpha=\angle OCD$):

$$2\theta+\alpha=90^{\circ}$$

Hence:

$$\alpha=90^{\circ}-2\theta$$

Now, with respect to triangle $\Delta OCD$, we see that (where $\beta=\angle COD$):

$$\sin(\alpha)=\overline{OD}$$

$$\cos(\beta)=\overline{OD}$$

And so we may conclude that:

$$\sin(\alpha)=\cos(\beta)$$

Now, substitute for $\alpha$ and apply a co-function identity, and what is your conclusion?

I don't think I'm following this well enough;

$$\sin(\alpha)=\cos(\beta)$$ where I substituted $$\sin(\alpha)$$ to $$1 - \sin(2θ)=\cos(\beta)$$, and choosing a cofunction identity of sine, which is $$sin (90° – x) = cos x$$...

So $$1 - \sin(2θ)= sin (90° – x)$$?

I'm really sorry, this is just a little confusing to me at the moment, I've been working on it since noon and it's frustrating.

Hello, bsmithysmith!


\angle COD is an exterior angle of \Delta AOC.

It is equal to the sum of the two nonadjacent angles:
. . .\angle COD \:=\:2\theta

I understand that AOD is 180 degrees, so the angle AOC - 180 = COD,

but still, how can I get from there to $$sin(2θ)=2sinθcosθ$$ (Got from a "Cheat Sheet" the professor gave to everybody, but of course, I'd rather understand it than just use it without knowing about it.

 

[MarkFL]

I don't think I'm following this well enough;

$$\sin(\alpha)=\cos(\beta)$$ where I substituted $$\sin(\alpha)$$ to $$1 - \sin(2θ)=\cos(\beta)$$, and choosing a cofunction identity of sine, which is $$sin (90° – x) = cos x$$...

So $$1 - \sin(2θ)= sin (90° – x)$$?

I'm really sorry, this is just a little confusing to me at the moment, I've been working on it since noon and it's frustrating.

Okay, look at angle $\alpha$ in triangle $\Delta OCD$. We may state:

$$\sin(\alpha)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\overline{OD}}{1}=\overline{OD}$$

Now, recall we found:

$$\alpha=90^{\circ}-2\theta$$

And so we may state:

$$\sin\left(90^{\circ}-2\theta\right)=\overline{OD}$$

But, recall a co-function identity says:

$$\sin\left(90^{\circ}-x\right)=\cos(x)$$ and so we may now write:

$$\cos(2\theta)=\overline{OD}$$

Going back to triangle $\Delta OCD$, we may also state:

$$\cos(\beta)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\overline{OD}}{1}=\overline{OD}$$

And so we may conclude:

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

or:

$$\beta=2\theta$$

A much simpler approach would be two equate the two pairs of complementary angles:

$$2\theta+\alpha=\beta+\alpha$$

$$2\theta=\beta$$

 

[Opalg]

View attachment 2423

Step 1. The angle $COD$, labelled ? in the diagram, is equal to $2\theta$ (exterior angle of the triangle $AOC$).

Step 2. Therefore, in the right-angled triangle $OCD$, $\sin2\theta = \dfrac {CD}{OC} = CD$ (because $OC = 1$).

Step 3. From the triangle $ACD$, $\sin\theta = \dfrac{CD}{AC}$. But $B$ is the midpoint of $AC$, and so $AC = 2AB$. Therefore $\sin\theta = \dfrac{CD}{2AB}$. Plugging in $CD = \sin2\theta$ (from Step 1), you get $\sin\theta = \dfrac{\sin2\theta}{AB}$.

Step 4. From the triangle $ABO$, $\cos\theta = \dfrac{AB}{AO} = AB$ (because $AO = 1$). Now plug that into the last equation from Step 3 to get $\sin\theta = \dfrac{\sin2\theta}{\cos\theta}$.

Finally, multiply out the fraction at the end of Step 4 to see that $\sin2\theta = 2\sin\theta\cos\theta$.

 

[bsmithysmith]

Step 1. The angle $COD$, labelled ? in the diagram, is equal to $2\theta$ (exterior angle of the triangle $AOC$).

Step 2. Therefore, in the right-angled triangle $OCD$, $\sin2\theta = \dfrac {CD}{OC} = CD$ (because $OC = 1$).

Step 3. From the triangle $ACD$, $\sin\theta = \dfrac{CD}{AC}$. But $B$ is the midpoint of $AC$, and so $AC = 2AB$. Therefore $\sin\theta = \dfrac{CD}{2AB}$. Plugging in $CD = \sin2\theta$ (from Step 1), you get $\sin\theta = \dfrac{\sin2\theta}{AB}$.

Step 4. From the triangle $ABO$, $\cos\theta = \dfrac{AB}{AO} = AB$ (because $AO = 1$). Now plug that into the last equation from Step 3 to get $\sin\theta = \dfrac{\sin2\theta}{\cos\theta}$.

Finally, multiply out the fraction at the end of Step 4 to see that $\sin2\theta = 2\sin\theta\cos\theta$.

And from that how would I find the double angle formula for $$Cosθ$$? This helped a good amount, too. Below the diagram are 4 questions regarding the term of four lines in the triangle:

$$AB = Cosθ$$
$$BC = Cosθ$$
$$CD = Sin(2θ)$$
$$OD = Cos(2θ)$$

Now the thing is that I'm not confident on the line OD, which is messing me up. So I may have been working on a worthless problem the entire time to find Cosθ of ACD.