# Addition Formula question expressing function in terms of Rsin(a+b)

Printable View

• Jun 4th 2011, 04:16 AM
dix
Addition Formula question expressing function in terms of Rsin(a+b)
I don't know where I'm going wrong. The answer at the end is supposed to be "tan a= 3/4" but going through it step by step I've gone wrong but I don't know how.

http://i.imgur.com/jRhWC.png
• Jun 4th 2011, 04:26 AM
TKHunny
Draw a right triangle.
Designate an acute angle, Angle a.
Label the legs consistent with the given value of the tangent of 'a'.
Contemplate the hypotenuse.
• Jun 4th 2011, 04:28 AM
mr fantastic
Quote:

Originally Posted by dix
I don't know where I'm going wrong. The answer at the end is supposed to be "tan a= 3/4" but going through it step by step I've gone wrong but I don't know how.

http://i.imgur.com/jRhWC.png

You are blindly using a formula.

Surely if you compare $R(\sin (\theta) \cos (a) + \cos (\theta) \sin (a))$ with the expression in the question, it is obvious that R sin(a) = 3 and R cos(a) = 4 and so tan(a) = 3/4.
• Jun 4th 2011, 04:37 AM
dix
Quote:

Originally Posted by TKHunny
Draw a right triangle.
Designate an acute angle, Angle a.
Label the legs consistent with the given value of the tangent of 'a'.
Contemplate the hypotenuse.

Thanks TKHunny, I only got this far with your explanation http://i.imgur.com/OkSmo.png but it sounds intersting if if could help with these types of questions, I can never get them right.
Quote:

Originally Posted by mr fantastic
You are blindly using a formula.

Surely if you compare $R(\sin (\theta) \cos (a) + \cos (\theta) \sin (a))$ with the expression in the question, it is obvious that R sin(a) = 3 and R cos(a) = 4 and so tan(a) = 3/4.

I know that I should surely be able to understand but I don't, I wouldn't be asking if I did. I don't get it :)
• Jun 4th 2011, 04:45 AM
mr fantastic
Quote:

Originally Posted by dix
[snip]
I know that I should surely be able to understand but I don't, I wouldn't be asking if I did. I don't get it :)

What part of what I said do you not understand? Do you understand how to compare the two expressions?
• Jun 4th 2011, 04:52 AM
dix
Quote:

Originally Posted by mr fantastic
What part of what I said do you not understand? Do you understand how to compare the two expressions?

I don't understand where I've went wrong in my step by step thing. I don't understand how R sin(a) = 3 and not 4. It says in the function '...4sin $\theta$"
• Jun 4th 2011, 05:29 AM
HallsofIvy
Yes, that's exactly why it is 3 and not 4. It is sin(a) you want, not $sin(\theta)$.

Compare $Rcos(\theta)sin(a)+ Rsin(\theta)cos(a)$
and $3cos(\theta)+ 4sin(\theta)$. You should see immediately that you want to have Rsin(a)= 3 and Rcos(a)= 4. Set the coefficients of $sin(\theta)$ and $cos(\theta)$ in the two equations equal.
• Jun 4th 2011, 06:24 AM
dix
Quote:

Originally Posted by HallsofIvy
Yes, that's exactly why it is 3 and not 4. It is sin(a) you want, not $sin(\theta)$.

Compare $Rcos(\theta)sin(a)+ Rsin(\theta)cos(a)$
and $3cos(\theta)+ 4sin(\theta)$. You should see immediately that you want to have Rsin(a)= 3 and Rcos(a)= 4. Set the coefficients of $sin(\theta)$ and $cos(\theta)$ in the two equations equal.

So in this one: $Rcos(\theta)sin(a)+ Rsin(\theta)cos(a)$ you put a/b (3/4) not b/a (4/3) because sin (a) comes before cos (a)? So the method for doing these questions is to take the general formula of $R(\sin (\theta) \cos (a) + \cos (\theta) \sin (a)$ and swap it around like you have and take it that way?
• Jun 4th 2011, 06:36 AM
Quacky
Quote:

Originally Posted by dix
So in this one: $Rcos(\theta)sin(a)+ Rsin(\theta)cos(a)$ you put a/b (3/4) not b/a (4/3) because sin (a) comes before cos (a)? So the method for doing these questions is to take the general formula of $R(\sin (\theta) \cos (a) + \cos (\theta) \sin (a)$ and swap it around like you have and take it that way?

Don't overthink this.

You have essentially said at the beginning of the question (correctly):

Let $3\cos\theta + 4\sin\theta = R\sin\alpha\cos\theta + R\cos\alpha\sin\theta$

In order for both sides of the equation to be equal, both sides have to have $3\cos\theta$

This means that, looking at the right hand side, $R\sin\alpha =3$.

Both sides have to have $4\sin\theta$.

So, looking at the right hand side, $R\cos\alpha=4$

And then it's just an application of $\tan\alpha =\frac{\sin\alpha}{\cos\alpha}}$
• Jun 4th 2011, 07:19 AM
dix
Thanks Quacky. So you just replace whatever the opposites are? I've done another using you method, is it right?

http://i.imgur.com/Hg9Q9.png
• Jun 4th 2011, 07:21 AM
Quacky
Quote:

Originally Posted by dix
Thanks Quacky. So you just replace whatever the opposites are? I've done another using you method, is it right?

http://i.imgur.com/Hg9Q9.png

Not quite. Firstly, you forgot the 'R'

$12\cos\theta+5\sin\theta=R\cos\alpha\cos\theta+R \sin\alpha\sin\theta$

Now look at the left hand side. You have $12\cos\theta$

Now look at the right hand side. You have $R\cos\alpha\cos\theta$

This means that $R\cos\alpha=12$ because you need to have the same amount of $\cos\theta$ on both sides. It doesn't have anything to do with opposites.
• Jun 4th 2011, 08:29 AM
dix
Thanks for the explanation Quacky, I think I finally get it :)