I got a maths problem and just checking whether my answer and method is correct:
Tan(A+B/2) / Tan (A-B/2) = SinA + SinB / SinA - SinB
I started to solve it:
Tan(A+B/2) = Sin(A+B)/2Cos
Tan(A-B/2) = Sin(A-B)/2Cos
2Cos'es cancel out each other and leaves
Sin(A+B)/Sin(A-B)
SinA + SinB / SinA - SinB
Is my method correct? Just checking whether Tan(A+B/2) = Sin(A+B)/2Cos
~Thanks.
Your strategy is erroneous in a number of ways: $\;\cos\;$ without an angle means nothing, just like $\,\sqrt{\;\;\;}$ means nothing without an argument, and certainly, neither is a number by itself, that can be canceled! Likewise, $\sin(A + B) \neq \sin A + \sin B$, etc.
Instead, rely on the trigonometric identities you are learning, and use the correct definition of the tangent of $\theta$: $$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$$ which is not the same thing as $\tan\theta = \dfrac{\sin}{\cos}(\theta)$
In the numerator, the angle $\theta_1$ of question is $\theta_1 = \left(\frac{A+B}{2}\right)$, and in the denominator, the angle $\theta_2 = \left(\frac{A-B}{2}\right)$, giving us:
$$\tan\theta_1 = \tan\left(\frac{A + B}{2}\right) = \dfrac{\sin \left( \frac{A+B}{2} \right)}{\cos \left( \frac{A+B}{2} \right)} \tag{1} $$
$$\tan \theta_2 = \tan\left(\frac{A-B}{2}\right) = \dfrac{\sin \left( \frac{A-B}{2} \right)}{\cos \left( \frac{A-B}{2} \right)} \tag{2}$$
Now divide $(1)$ by $(2)$ and use trigonometric identities to simplify. (Not necessarily in that order!)
Suggestion: review the identities for the sum and difference of angles, and half-angle identities.
One approach, for example, is to recall that $\,\tan\left(\frac{\theta}{2}\right) = \csc\theta - \cot\theta$. Then you can express each of $\csc\theta$ and $\cot\theta$ in terms of $\sin\theta$ and $\cos\theta$, and apply identities on the sum and difference of angles. Try to simplify $(1)$ and $(2)$ sufficiently before dividing what you get for $(1)$ by what you get for $(2)$.
There seem to be quite a few misunderstandings in your treatment of trig functions.
It looks almost like you're making up rules at random; it's quite reminiscent of this example of a student doing the same thing. When you're manipulating mathematical expressions you have to be very careful with what you can and can't do. Bulldozing through writing $=$ signs is only okay if the expressions that appear on either side of the $=$ sign are actually equal! As illustrated above, expressions are often not equal even though they look similar.
Instead of telling you exactly how to solve the problem, I'll try and clear up some of your more fundamental errors.
When you write $\cos$ here, it needs an argument, i.e. something you apply $\cos$ to. Remember that when you write something like $\tan \theta$, it refers to a number; specifically, it refers to the number obtained by applying the function $\tan$ to its argument $\theta$. Without the argument, the function can't be taken to have a numerical value, and so it makes no sense to include it in an expression involving numbers.
The identity $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$ applies for all values of $\theta$. Notice that writing something like $\dfrac{\sin \theta}{\cos}$ makes no sense: what does the $\cos$ at the bottom of the fraction mean? The cosine of what?
Here you need to set $\theta = \dfrac{A+B}{2}$ for one of your tangent functions and $\theta = \dfrac{A-B}{2}$ for the other. Doing so gives, for instance,
$$\tan \left( \frac{A+B}{2} \right) = \frac{\sin \left( \frac{A+B}{2} \right)}{\cos \left( \frac{A+B}{2} \right)}$$
Notice also that, for instance, $\sin \left( \dfrac{A+B}{2} \right) \ne \dfrac{\sin (A+B)}{2}$ and so on; and $\sin (A+B) \ne \sin A + \sin B$. (For example, set $A=B=90^{\circ}$, then $\sin(A+B)=0$ and $\sin A + \sin B = 2$.)
Here's an explanation about why your method is wrong (assuming you've taken a basic algebra course):
Let's say you wanted to graph a line. You might have seen: $$ f(x) = 3x + 2 $$
That means we plug in $x$ into the equation in order to get $f(x)$. Simple enough. So $f(1) = 5$, because $f(1) = 3(1) + 2 = 5$.
If you had: $$ \frac{f(2)}{f(1)} $$
The answer is not $\frac{2}{1}$. You need to plug both of these in to your equation $f(x) = 3x + 2$: $$ \frac{3 \cdot 2 + 2}{3 \cdot 1 + 2} $$ $$ \frac{6 + 2}{5} $$ $$ \frac{8}{5} $$
Often people write $3(2)$ to detonate 3 times 2. 3 is not a function, so you can just multiply there.
What does this have to do with sine and cosine? Well actually, sine and cosine are functions too! For some reason, the parenthesis which showed you that they were a function got lost somewhere along the way, so now most people just write $ \sin 3 $ as opposed to $\sin(3)$. (Sometimes they are used, sometimes they aren't, but they're the same either way.)
Going back to our function, $f(x) = 3x+2$, what is the result of: $$ \frac{f}{f} $$
Alright, so we'll just plug in a ... wait? What do we plug in to our equation? There's nothing there! We can't plug in zero, because it doesn't say $f(0)$. And we can't plug in $x$, because it doesn't say $f(x)$! That equation makes no sense. However, this does: $$ \frac{f(x)}{f(x)} $$
Because we have: $$ \frac{3x + 2}{3x + 2} $$
And anything divided by itself is 1. Now let's go back to sine and cosine, what is1: $$ \frac{\sin}{\sin} $$
Well that makes no sense! However2: $$ \frac{\sin \theta}{\sin \theta} = 1$$
Alright, now we realize that $ \cos $ makes no sense, because it's not attached to any function.
Let's try and prove a similar identity. Cover up the steps, and see if you can figure out what to do next: $$ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} $$
We can start by ONLY focusing on the right side, and trying to make it the same as the left side: $$\frac{1 - \cos 2\theta}{2}$$
There is a special identity to use here. It is called the Cosine Sum Formula:
$$ \cos (A + B) = \cos A \cos B - \sin A \sin B $$
How does that apply here? There's no sum inside that cosine! Actually, there is. $$ \theta + \theta = 2\theta $$
So we can rewrite that above equation as: $$\frac{1 - \cos(\theta + \theta)}{2}$$
Now apply the Cosine Sum Formula (note the use of the parenthesis!): $$\frac{1 - (\cos \theta \cos \theta - \sin \theta \sin \theta)}{2} $$
Remember that $\cos \theta \cdot \cos \theta = \cos^2 \theta$.
So we have (remember to "distribute" the negative sign to make $\sin^2 \theta$ positive): $$\frac{1 - \cos^2 \theta + \sin^2 \theta}{2}$$
Well, that's great, but it looks like we just made the problem more complex! There's another formula, though:
The Pythagorean Identity: $$ \sin^2 x + \cos^2 x = 1 $$ and if we subtract $ \cos^2 x $ from both sides: $$ 1 - \cos^2 x = \sin^2 x $$
So substitute $1 - \cos^2 \theta$ for $\sin^2 \theta$ to get: $$\frac{\sin^2 \theta + \sin^2 \theta}{2}$$
Add!
$$\frac{2\sin^2 \theta}{2}$$
And divide: $$\sin^2 \theta$$
That's the same as the left side, and that's what we wanted to show!
1 Sometimes people do this anyway, because they're too lazy to write out $ \theta $. Generally it just means that they meant to write $ \sin \theta $. This is an abuse of notation.
2 Where $ \sin \theta \neq 0 $, because division by zero is undefined.
