simplify to one trigonometric expression: 2 sin (pi/4) 2 cos (pi/4)

[math_knight]

Got stuck on this one even after looking at a solution online:

2 sin (pi/4) 2 cos (pi/4)

Looks familiar, like the double angle sin

sin (2X) = 2 sin x cos x

if we let x = pi/4 we get

sin ((2(pi/4) = 2 sin (pi/4) cos (pi/4)

2(pi/4) = pi/2

so sin (pi/2) = 2 sin (pi/4) cos (pi/4)

the last step shown says that therefore

2 sin (pi/2) = 2 sin (pi/4) 2 cos (pi/4)

So the dark green part is part of the double angle formula

sin(2θ) = 2 sin θ cos θ

which we can re-arrange...

2 sin θ = cos θ - sin(2θ) putting that back into...? bleh, my brain is fried today. I think factoring out 1/2 comes next somehow? Any help would be appreciated.

 

[Dr.Peterson]

Got stuck on this one even after looking at a solution online:

2 sin (pi/4) 2 cos (pi/4)

Looks familiar, like the double angle sin

sin (2X) = 2 sin x cos x

if we let x = pi/4 we get

sin ((2(pi/4) = 2 sin (pi/4) cos (pi/4)

2(pi/4) = pi/2

so sin (pi/2) = 2 sin (pi/4) cos (pi/4)

the last step shown says that therefore

2 sin (pi/2) = 2 sin (pi/4) 2 cos (pi/4)

So the dark green part is part of the double angle formula

sin(2θ) = 2 sin θ cos θ

which we can re-arrange...

2 sin θ = cos θ - sin(2θ) putting that back into...? bleh, my brain is fried today. I think factoring out 1/2 comes next somehow? Any help would be appreciated.

The green part is the answer. Essentially, you have doubled the double angle formula to get 2 sin(pi/2) = 2 sin(pi/4) 2 cos(pi/4).

But why did you then subtract cosθ? That would be valid if it were being added, but you don't undo multiplication by subtracting.

Furthermore, there is no need to undo anything! What you started with was the right side of the equation above; you have shown that it is equal to the left side, so the left side is the simplified answer: 2 sin(pi/2). What answer did this site give? Wasn't it that?

And if you haven't actually stated the original problem you are working on, can you quote it exactly for us? I have to say, "one trigonometric expression" sounds like there should be no coefficient on the outside of the answer, which can't happen here as far as I can see; and the 2 in the middle seems odd for an expression as given.

 

[math_knight]

Hey Dr. Peterson, thank you for your reply. Sorry about any confusion. To sort of start over the problem gives us the expression "2 sin (2x) = 2 sin (x) 2 cos (x)" and then asks us to "simplify to one trigonometric expression"

So to your point, I guess they meant go from have two trig expressions (2sinxCosx) to one, and that it's ok to have a coefficient.

And the problem is an odd numbered one so the solution in the back of the book is indeed

2 sin (pi/2)

The solution online shows most of the work, but not all of it because of course the person solving it takes it for granted that you know what he's thinking.

Anyway my question is about what you said " Essentially, you have doubled the double angle formula". How did they get the 2 in front of the sin (pi/2) AND the 2 in front of the cos

See, I only follow the solution as to this point:

sin (pi/2) = 2 sin (pi/4) cos (pi/4)

So far there is no 2 coefficient on the left side, NOR before the cos. I see that this is a a double angle formula with x= pi/4, so the double angle became "hidden" on the left side. Pi/2 is (2(pi/4)

The next step just makes the 2 coefficient appear on the left side and on the right side before the cos. Please note the orange "2"s. Poof.

2 sin (pi/2) = 2 sin (pi/4) 2 cos (pi/4)

In general form

sin (2x) = 2 sin x cos x

So multiplying the left side by 2....multiplies the cos on the right side (and not the sin) by 2?

2 sin (2x) = 2 sin (x) 2 cos (x)

Why does multiplying the left side by 2 make a 2 appear in front of the cos on the right side? Best I can guess is that it's a transform? 2 sin x would double the amplitude? And....I have no idea..I'm just guessing. I have no idea how the cos x on the right side is related to an amplitude transform.

edit: it's probably somewhere in the proof of the double angle formula? which I will go over

edit #2: what would tripling the double angle formula do? this? 3 sin (2x) = 2 sin (x) 3 cos (x) ?

 

[ksdhart2]

So multiplying the left side by 2....multiplies the cos on the right side (and not the sin) by 2?

Okay, so it seems to me like the fact that there's trig functions involved is obfuscating what's really going on here, which is nothing more than the normal rules of multiplication you learned oh so long ago. To attempt to alleviate this, let's work with another example without trig functions and see if that helps. Suppose we had the following equation:

$\displaystyle x = 2sc$

Now what would happen if we multiplied both sides by 2? Well, clearly we'd get:

$\displaystyle 2x = 4sc$

Or we can also write it in an unsimplified and expanded form:

$\displaystyle 2 \cdot x = 2 \cdot 2 \cdot s \cdot c$

From there, can you see that we can use the commutative property of multiplication (Refresher here if needed) to freely move terms around, such that all three of the following equations are exactly the same?

$\displaystyle 2 \cdot x = 2 \cdot 2 \cdot s \cdot c$

$\displaystyle 2 \cdot x = 2 \cdot s \cdot 2 \cdot c$

$\displaystyle 2 \cdot x = 2 \cdot s \cdot c \cdot 2$

Now, make a few substitutions. Let $\displaystyle x = \sin \left( \frac{\pi}{2} \right)$, $\displaystyle s = \sin \left( \frac{\pi}{4} \right)$, and $\displaystyle c = \cos \left( \frac{\pi}{4} \right)$. Can you see why exactly the same principle that we just used with "ordinary" numbers and variables, also applies when dealing with trig functions?

If I had to hazard a guess, I'd say the problem statement used the form $\displaystyle 2 \cdot \sin \left( \frac{\pi}{4} \right) \cdot 2 \cdot \cos \left( \frac{\pi}{4} \right)$ just to see if you were paying attention and test if you'd get tripped up by the unusual (but still mathematically valid) notation... and it seems to have worked

 

[Dr.Peterson]

Hey Dr. Peterson, thank you for your reply. Sorry about any confusion. To sort of start over the problem gives us the expression "2 sin (2x) = 2 sin (x) 2 cos (x)" and then asks us to "simplify to one trigonometric expression"

So to your point, I guess they meant go from have two trig expressions (2sinxCosx) to one, and that it's ok to have a coefficient.

And the problem is an odd numbered one so the solution in the back of the book is indeed

2 sin (pi/2)

The solution online shows most of the work, but not all of it because of course the person solving it takes it for granted that you know what he's thinking.

Anyway my question is about what you said " Essentially, you have doubled the double angle formula". How did they get the 2 in front of the sin (pi/2) AND the 2 in front of the cos

See, I only follow the solution as to this point:

sin (pi/2) = 2 sin (pi/4) cos (pi/4)

This is why you need to first state the problem as given, before anything else! What you said was quite confusing, and you can ignore everything I said, which was not about the problem you were given, as far as I can tell.

But also, it is confusing for the problem to give an equation and say to simplify it, and expect an expression as the answer!

If it gave you the equation 2 sin(2x) = 2 sin(x) 2 cos(x), they must really want you to solve it, finding the value of x.

But the solution is not x = 2 sin(pi/2). It's an identity, which is always true.

Did they say somewhere to replace x with pi/4? I don't see a reason to do so.

So, what is the problem really? Copy it exactly as given, and then, if necessary, copy (or give a link to) the solution you found. Giving us little snippets only makes things worse.

 

[math_knight]

Thanks ksdhart2, yeah they succeeded in throwing me off, lol. Your explanation fixed that, thanks so much.

Hey Dr. Peterson, I think ksdhart2 hit the nail on the head, but in case you are curious the problem was stated as folllows:

"For the following exercises, simplify to one trigonometric expression. "

11.) 2 sin (pi/4) 2 cos (pi/4)

Sorry I didn't make that clear from the beginning.

Thank you guys both