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

1. ## 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.

2. 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.

3. 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.

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.

4. 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 but it sounds intersting if if could help with these types of questions, I can never get them right.
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

5. 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?

6. 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 $\theta$"

7. 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.

8. 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?

9. 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 $\tan\alpha =\frac{\sin\alpha}{\cos\alpha}}$

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

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

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.

12. Thanks for the explanation Quacky, I think I finally get it