# Deriving the rest of trigonometric identities from the formulas for $\sin(A+B)$, $\sin(A-B)$, $\cos(A+B)$, and $\cos (A-B)$

I am trying to study for a test and the teacher suggest we memorize $\sin(A+B)$, $\sin(A-B)$, $\cos(A+B)$, $\cos (A-B)$, and then be able to derive the rest out of those. I have no idea how to get any of the other ones out of these, it seems almost impossible. I know the $\sin^2\theta + \cos^2\theta = 1$ stuff pretty well though. For example just knowing the above how do I express $\cot(2a)$ in terms of $\cot a$? That is one of my problems and I seem to get stuck half way through.

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What textbook do you use? I'm sure that in your textbook all the formulas are proved using only the four formulas mentioned above. –  Beni Bogosel Jul 1 '11 at 20:25
I don't have a text book, it got lost in the mail. I am more concerned with learning how to actually do it though, it seems impossibly hard to me. –  Adam Jul 1 '11 at 20:49
–  Beni Bogosel Jul 2 '11 at 9:30

Since $\displaystyle\cot(2a) = \frac{\cos(2a)}{\sin(2a)}$, you would have (assuming you know the addition formulas for sines and cosines): \begin{align*} \cos(2a) &= \cos(a+a) = \cos(a)\cos(a) - \sin(a)\sin(a)\\ &= \cos^2(a) - \sin^2(a);\\ \sin(2a) &= \sin(a+a) = \sin(a)\cos(a) + \cos(a)\sin(a)\\ &= 2\sin(a)\cos(a), \end{align*} and therefore \begin{align*} \cot(2a) &= \frac{\cos(2a)}{\sin(2a)} = \frac{\cos^2(a) - \sin^2(a)}{2\sin(a)\cos(a)}\\ &= \frac{1}{2}\left(\frac{\cos^2(a)}{\sin(a)\cos(a)}\right) - \frac{1}{2}\left(\frac{\sin^2(a)}{\sin(a)\cos(a)}\right)\\ &= \frac{1}{2}\left(\frac{\cos(a)}{\sin(a)} - \frac{\sin(a)}{\cos(a)}\right)\\ &= \frac{1}{2}\left(\cot(a) - \tan(a)\right)\\ &= \frac{1}{2}\left(\cot(a) - \frac{1}{\cot(a)}\right)\\ &= \frac{1}{2}\left(\frac{\cot^2(a)}{\cot(a)} - \frac{1}{\cot(a)}\right)\\ &= \frac{1}{2}\left(\frac{\cot^2(a) - 1}{\cot (a)}\right). \end{align*}

P.S. Now, as it happens, I don't know the formulas for double angles, nor most identities involving tangents, cotangents, etc. I never bothered to memorize them. What I know are:

1. The definitions of tangent, cotangent, secant, and cosecant in terms of sine and cosine;
2. That sine is odd ($\sin(-x) = -\sin(x)$) and cosine is even ($\cos(-x)=\cos(x)$);
3. The addition formulas for sine and cosine;
4. The values of sine and cosine at $0^{\circ}$, $30^{\circ}$, $45^{\circ}$, $60^{\circ}$, and $90^{\circ}$.

(I can derive $\sin^2\theta + \cos^2\theta = 1$ from the above, but in all honesty that one comes up so often that I do know it as well). I do not know the addition or double angle formulas for tangents nor cotangents, so the above derivation was done precisely "on the fly", as I was typing. I briefly thought that I might need to $\cos(2a)$ with one of the following equivalent formulas: $$\cos^2(a)-\sin^2(a) = \cos^2(a) + \sin^2(a) - 2\sin^2(a) = 1 - 2\sin^2(a)$$ or $$\cos^2(a) - \sin^2(a) = 2\cos^2(a) - \cos^2(a) - \sin^2(a) = 2\cos^2(a) - 1,$$ if the first attempt had not immediately led to a formula for $\cot(2a)$ that involved only $\cot(a)$ and $\tan(a) = \frac{1}{\cot(a)}$.

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I don't quit understand the result of the cot2a part and how that is equal to cot2a, is it somehow an inverse of the tan2a formula? Is deriving forumals mostly about factoring? I am very bad at that. –  Adam Jul 1 '11 at 21:55
@Adam: cotangent is, by definition, cosine over sine. So $\cot(2a) = \frac{\cos(2a)}{\sin(2a)}$. Then I substituted the values of $\cos(2a)$ and $\sin(2a)$ using the addition formulas; the rest is just algebra, except for using the fact that $\cot(x) = \frac{1}{\tan(x)}$ in the antepenultimate step. –  Arturo Magidin Jul 1 '11 at 22:53
@Adam: You are memorizing too many things, in the hope that a memorized formula will be a perfect fit. In an earlier post, you mentioned having to memorize many cosine laws. There really is only one cosine law, which can be memorized and then manipulated as needed. Also, I remember a test you posted. There were some hard questions on that test, but I am sure that most of the questions were variants of problems that had been done in class and/or assigned. You have difficulty recognizing clones of things you have seen before. –  André Nicolas Jul 1 '11 at 23:37
Yes, I have difficulty telling the difference between problems that are meant to be manipulated in a certain way against other. When I do the homework one section is all double angle, the next half. On the test it is all mixed and I have trouble telling what I need to use, for example something with the number 3 doesn't obviously fit half or double. –  Adam Jul 2 '11 at 0:20
@Adam: You are referring to the $\sin(3x)$ problem. I am confident that in class or homework, the very similar $\cos(3x)$ was done, or even $\sin(3x)$ itself. You also tend to have algebra accidents, so even when you know how to do a problem, things go wrong. What major are you interested in that has this course as prerequisite? –  André Nicolas Jul 2 '11 at 1:20

Three examples of algebraic derivations of trigonometric identities as an application of the addition and subtraction formulas.

1. Example on how to deduce the logarithmic transformation formulas (sum to product formulas) from

$$\sin (a+b)=\sin a\cos b+\sin b\cos a,\qquad (1)$$

$$\sin (a-b)=\sin a\cos b-\sin b\cos a.\qquad (2)$$

If you write

$$\left\{ \begin{array}{c} a=\frac{p+q}{2}, \\ b=\frac{p-q}{2},\end{array}\right. \Leftrightarrow \left\{ \begin{array}{c} a+b=p, \\ a-b=q,\end{array}% \right.$$

you get

$$\sin (a+b)+\sin (a-b)=2\sin a\cos b,$$

$$\sin (a+b)-\sin (a-b)=2\sin b\cos a,$$

and thus

$$\sin p+\sin q=2\sin \frac{p+q}{2}\cos \frac{p-q}{2},\qquad (3)$$

$$\sin p-\sin q=2\sin \frac{p-q}{2}\cos \frac{p+q}{2}.\qquad (4)$$

You can use $(3)$ to solve the equation

$$\sin (5x)+\sin x=\sin (3x)$$

that appeared in my exam in 1968. (See a comment of mine to this post ).

2. As for the example in your question we present the following derivation. From

$$\sin (a+b)=\sin a\cos b+\sin b\cos a,\qquad (5)$$

$$\cos (a+b)=\cos a\cos b-\sin a\sin b,\qquad (6)$$

we get

$$\cot (a+b)=\frac{\cos (a+b)}{\sin (a+b)}=\frac{\cos a\cos b-\sin a\sin b}{\sin a\cos b+\sin b\cos a},$$

or dividing the numerator and denominator by $\sin a\cos b$

$$\begin{eqnarray*} \cot (a+b) &=&\dfrac{\dfrac{\cos a\cos b-\sin a\sin b}{\sin a\cos b}}{\dfrac{% \sin a\cos b+\sin b\cos a}{\sin a\cos b}}=\dfrac{\dfrac{\cos a\cos b}{\sin a\cos b}-\dfrac{\sin a\sin b}{\sin a\cos b}}{\dfrac{\sin a\cos b}{\sin a\cos b}% +\dfrac{\sin b\cos a}{\sin a\cos b}} \\ &=&\dfrac{\dfrac{\cos a}{\sin a}-\dfrac{\sin b}{\cos b}}{1+\dfrac{\sin b\cos a}{% \sin a\cos b}}=\dfrac{\cot a-\tan b}{1+\tan b\cot a}=\dfrac{\cot a-\dfrac{1}{% \cot b}}{1+\dfrac{\cot a}{\cot b}} \\ &=&\dfrac{\dfrac{\cot a\cot b-1}{\cot b}}{\dfrac{\cot b+\cot a}{\cot b}}=\dfrac{% \cot a\cot b-1}{\cot b+\cot a},\qquad (7) \end{eqnarray*}$$

a particular case of which (for $a=b$) is given by

$$\cot (2a)=\dfrac{\cot ^{2}a-1}{2\cot a}.\qquad (8)$$

3. The summation and subtraction formula for the tangent. From

$$\sin (a\pm b)=\sin a\sin b\pm \sin b\cos a$$

we get

$$\tan a\pm \tan b=\frac{\sin a}{\cos a}\pm \dfrac{\sin b}{\cos b}=\frac{\sin a\cos b\pm \sin b\cos a}{\cos a\cos b}=\frac{\sin (a\pm b)}{\cos a\cos b}.\qquad (9)$$

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I am having difficulty understand where step 7 came from for the example problem. I am trying to understand how to derive these formulas better but I can't see what you did. Also in #3 I don't understand why sin(a+b) is not equal to sincos + cossin? I am having a lot of trouble with the third one, I am assuming sin turns into tan by dividing everything by cos but then why does the second set of sinbcosa dissapear? –  Adam Jul 2 '11 at 15:57
@Adam: I derived the addition formula $\cot (a+b)$, which for $a=b$ reduces to the duplication formula $\cot (2a)$. Formula (7) comes from (5) and (6) and the definition of $\cot (a+b)$. –  Américo Tavares Jul 2 '11 at 16:12
In more detail $$\cot (a+b)=\frac{\cos (a+b)}{\sin (a+b)}=\frac{\cos a\cos b-\sin a\sin b}{% \sin a\cos b+\sin b\cos a}=\frac{\frac{\cos a\cos b-\sin a\sin b}{\sin a\cos b}}{\frac{\sin a\cos b+\sin b\cos a}{\sin a\cos b}}$$ $$=\frac{\frac{\cos a\cos b}{\sin a\cos b}-\frac{\sin a\sin b}{\sin a\cos b}}{% \frac{\sin a\cos b}{\sin a\cos b}+\frac{\sin b\cos a}{\sin a\cos b}}=\frac{% \frac{\cos a}{\sin a}-\frac{\sin b}{\cos b}}{1+\frac{\sin b\cos a}{\sin a\cos b}}=\frac{\cot a-\tan b}{1+\tan b\cot a}=\frac{\cot a-\frac{1}{\cot b}% }{1+\frac{\cot a}{\cot b}}$$ –  Américo Tavares Jul 2 '11 at 16:12
I understand now, not sure how that went over my head. Thanks. I am going to be working on these for most of today. Is there anything else really important that I should learn before I take calculus? I have a little over a month to study. –  Adam Jul 2 '11 at 16:20
@Adam: I think you need to know functions and polynomials. –  Américo Tavares Jul 2 '11 at 17:31

If you understand complex numbers there is a very nice mnemonic. We know that $$e^{it} = \cos(t) + i\sin(t).$$

Take real and imaginary parts on the identity $$e^{i(A + B)} = e^{iA}e^{iB} = (\cos(A) + i\sin(A))(\cos(B) + i\sin(B)).$$

Disassemble and the the sum formulae for $\sin$ and $\cos$. To get the differences, use the assignment $B\leftarrow -B$ and the fact that $\cos(-B) = \cos(B)$ and $\sin(-B) = -\sin(B)$.

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This is exactly how I console myself that I could work out the trig identities if trapped on a desert island. If I had known Euler in high school trig I wonder how many times I would have run out of time running everything through $e^{it}$ –  Ben Jackson Jul 2 '11 at 6:15

Maybe this will help? cot(x) = cosx / sinx -> cot(2a) = cos(a + a) / sin(a + a) and then I assume you know these two.

Edit: Had it saved as a tab and didnt see the posted answer, but I still think it would have been best to let you compute the rest by yourself so that you could learn it by doing instead of reading.

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I'm not certain, but I think perhaps you need to revisit the basic trigonometric definitions. $$\begin{array}{lll} \sin\theta = \frac{opposite}{hypotenuse}&\csc\theta=\frac{hypotenuse}{opposite}\\ \cos\theta = \frac{adjacent}{hypotenuse}&\sec\theta=\frac{hypotenuse}{adjacent}\\ \tan\theta = \frac{opposite}{adjacent}&\cot\theta=\frac{adjacent}{opposite} \end{array}$$

Having done that, we can now manipulate the definitions. For example

$$\cot\theta = \frac{adjacent}{opposite}\cdot\frac{\frac{1}{hypotemnuse}}{\frac{1}{hypotemnuse}}=\frac{\frac{adjacent}{hypotenuse}}{\frac{opposite}{hypotenuse}}=\frac{\cos\theta}{\sin\theta}$$

or alternatively

$$\cot\theta = \frac{adjacent}{opposite}=\frac{\frac{1}{opposite}}{\frac{1}{adjacent}}=\frac{\frac{1}{opposite}}{\frac{1}{adjacent}}\cdot\frac{hypotenuse}{hypotenuse}=\frac{\frac{hypotenuse}{opposite}}{\frac{hypotenuse}{adjacent}}=\frac{\csc\theta}{\sec\theta}$$

After that, move on to making 3 triangles that are similar to the adjacent-opposite-hypotenuse triangle. These triangles will rotate which side is equal to 1. This is where we get the familiar Pythagorean identities.

The next step is to derive $\cos (\alpha+\beta)$ and $\sin (\alpha+\beta)$ by setting the lengths (I would use the squares of the lengths though) of the green lines in the diagram equal to each other. It's easy to do, since the endpoints are well known.

Finally, lets derive $\cot(A+B)$ $$\begin{array}{lll} \cot(A+B)&=&\frac{\sin(A+B)}{\cos(A+B)}\\ &=&\frac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}\\ &=&\frac{\frac{\sin A\cos B + \cos A\sin B}{\sin A\sin B}}{\frac{\cos A\cos B - \sin A\sin B}{\sin A\sin B}}\\ &=&\frac{\frac{\cos B}{\sin B}+\frac{\cos A}{\sin A}}{\frac{\cos A\cos B}{\sin A\sin B}-1}\\ &=&\frac{\cot B+\cot A}{\cot A\cot B-1} \end{array}$$ Note that $\cot 2A$ is just $\cot(A+B)$ where $A=B$, so $$\cot 2A = \cot(A+A)=\frac{\cot A+\cot A}{\cot A\cot A-1}=\frac{2\cot A}{\cot^2 A-1}$$ Of course this list isn't comprehensive, for example we didn't derive the sine and cosines laws, nor did we consider the double angle and half angle formulas. But I hope that what little I did present here was helpful.

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