Easy - Prove sine subtraction formula with cosine addition formula

[Nazariy]

Hello,

I need a bit of help with the following problem. I am asked to prove the sine subtraction formula using cosine addition formula and the following identities:

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I suppose that A = P/2 - theta, and then substitute the former for the latter in the cosine addition formula to obtain the following:

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My question is, how do I proceed from here? How do I rewrite the first part of the identity? Thanks

 

[stapel]

Maybe you'd have better luck if you let A = pi/2 and B = theta?

 

[srmichael]

Hello,

I need a bit of help with the following problem. I am asked to prove the sine subtraction formula using cosine addition formula and the following identities:

View attachment 3969

​

I suppose that A = P/2 - theta, and then substitute the former for the latter in the cosine addition formula to obtain the following:

View attachment 3970

​

My question is, how do I proceed from here? How do I rewrite the first part of the identity? Thanks

You are overthinking this. Let A be π/2 and B = Θ. Then apply the difference identities for each using these two angles.

 

[Nazariy]

emmm...

Got this...

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[stapel]

Now evaluate the sine and cosine of the specific angle value. Simplify to get the desired result.

 

[Nazariy]

Now evaluate the sine and cosine of the specific angle value. Simplify to get the desired result.

Ahhh, dear Mr Stapel, you will be displeased with my progress I am afraid. No one in the whole universe probably spent this much time trying to prove such a simple thing like this one. I hope you will spoon feed my just a slight bit more. This is what I have done:

Assume P/2 = A , and –theta = B
Also sin(-theta) = -sin(theta), thus as before

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Now substitute cos (theta) and sin (theta) for the identities give:

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I can further simplify this:

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And now I have no idea what to do next. I have done a bit of reverse thinking (i.e. cheating) to see what I am supposed to obtain, and it seems like I should get:

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And it doesn’t seem like I am close, and I have no clue how to make my result to appear as above…

 

[stapel]

What are the numerical values of the sine and the cosine of $\displaystyle \frac{\pi}{2}?$ Plug those in.

 

[Nazariy]

What are the numerical values of the sine and the cosine of $\displaystyle \frac{\pi}{2}?$ Plug those in.

If I substitute the numeric values of angles into here, I get a very neat identity

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Oh dear.. haha

 

[Nazariy]

What are the numerical values of the sine and the cosine of $\displaystyle \frac{\pi}{2}?$ Plug those in.

I had it right in the very first post... All I had to do is to rewrite the LHS of the equiation!

Check this out...

If

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Then

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If you rewrite

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You get

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Since

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We have

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That simple... I am still interested in Mr Stapel's way of proving it.

 

[Deleted member 4993]

emmm...

Got this...

View attachment 3971
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cos(π/2 - Θ) = cos(π/2)*cos(Θ) + sin(π/2)*sin(Θ) = 0 * cos(Θ) + 1 * sin(Θ) = sin(Θ) ...........done

 

[Nazariy]

cos(π/2 - Θ) = cos(π/2)*cos(Θ) + sin(π/2)*sin(Θ) = 0 * cos(Θ) + 1 * sin(Θ) = sin(Θ) ...........done

And what is the point of doing that? I am supposed to prove sine subtraction rule > sin (A-B)= sinAcosB-cosAsinB