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, 03:16 AMdixAddition 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, 03:26 AMTKHunny
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, 03:28 AMmr fantastic
- Jun 4th 2011, 03:37 AMdix
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.

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, 03:45 AMmr fantastic
- Jun 4th 2011, 03:52 AMdix
- Jun 4th 2011, 04:29 AMHallsofIvy
Yes, that's exactly

**why**it is 3 and not 4. It is sin(a) you want, not $\displaystyle sin(\theta)$.

Compare $\displaystyle Rcos(\theta)sin(a)+ Rsin(\theta)cos(a)$

and $\displaystyle 3cos(\theta)+ 4sin(\theta)$. You should see immediately that you want to have Rsin(a)= 3 and Rcos(a)= 4. Set the coefficients of $\displaystyle sin(\theta)$ and $\displaystyle cos(\theta)$ in the two equations equal. - Jun 4th 2011, 05:24 AMdix
So in this one: $\displaystyle Rcos(\theta)sin(a)+ Rsin(\theta)cos(a)$ 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 $\displaystyle R(\sin (\theta) \cos (a) + \cos (\theta) \sin (a)$ and swap it around like you have and take it that way?

- Jun 4th 2011, 05:36 AMQuacky
Don't overthink this.

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

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

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

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

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

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

And then it's just an application of $\displaystyle \tan\alpha =\frac{\sin\alpha}{\cos\alpha}}$ - Jun 4th 2011, 06:19 AMdix
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, 06:21 AMQuacky
Not quite. Firstly, you forgot the 'R'

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

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

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

This means that $\displaystyle R\cos\alpha=12$ because you need to have the same amount of $\displaystyle \cos\theta$ on both sides. It doesn't have anything to do with opposites. - Jun 4th 2011, 07:29 AMdix
Thanks for the explanation Quacky, I think I finally get it :)