# How to remember a particular class of trig identities.

Please how can I easily remember the following trig identities: $$\sin(\;\pi-x)=\phantom{-}\sin x\quad \color{red}{\text{ and }}\quad \cos(\;\pi-x)=-\cos x\\ \sin(\;\pi+x)=-\sin x\quad \color{red}{\text{ and }}\quad \cos(\;\pi+x)=-\cos x\\ \sin(\frac\pi2-x)=\phantom{-}\cos x\quad \color{red}{\text{ and }}\quad \cos(\frac\pi2-x)=\phantom{-}\sin x\\[12pt] \text{and similar things where we add a radian angle inside cos or sin as you can see}$$

So how can I remember all of them? They're pretty confusing! I can easily interchange some and do mistakes thus...

• Swapping trig identities in one's memory is a common problem. Keeping them straight is made more difficult because so many of them look similar to one another. That is why it's helpful to understand where they come from or how to derive them, and to practice using them. (Not many people have these completely in hand in just one semester...) – colormegone Apr 2 '14 at 23:05

I created the images that "inspired" the illustration of the Angle-Sum and Difference identities on Wikipedia, so I'm all in favor of learning those relations. However, this is how I remember your particular class of identities.

First, in the Unit Circle, I imagine the "sine-cosine-$1$" triangle for a generic first-quadrant angle $\theta$, taking $\theta$ small enough that the triangle is considerably wider than it is tall, creating an obvious "long leg" and "short leg". Because trig values are positive in the first quadrant, I can say that the "long leg" of the triangle has length $\cos\theta$, and that the "short leg" has length $\sin\theta$.

Then, I imagine what happens when I rotate that triangle through multiples of $90^\circ$, getting a nice windmill:

The "(co-)reference triangle" for each of the compound angles is a rotation of the original triangle, and we can read off sine and cosine values by paying attention to the positions of the short legs and long legs (and assigning signs, as appropriate). For instance, the point labeled "$\theta+90^\circ$" is at distance "$\cos\theta$" above the $x$-axis, and at distance "$\sin\theta$" to the left of the $y$-axis; therefore, $$\sin\left(\theta+\phantom{1}90^\circ\right) = \phantom{-}\color{red}{\cos\theta} \qquad\qquad \cos\left(\theta+\phantom{1}90^\circ\right) = -\color{blue}{\sin\theta}$$

Likewise, $$\sin\left(\theta+180^\circ\right) = -\color{blue}{\sin\theta} \qquad\qquad \cos\left(\theta+180^\circ\right) = -\color{red}{\cos\theta}$$ $$\sin\left(\theta-\phantom{1}90^\circ\right) = -\color{red}{\cos\theta} \qquad\qquad \cos\left(\theta-\phantom{1}90^\circ\right) = \phantom{-}\color{blue}{\sin\theta}$$

Similarly, there's a windmill for compound angles involving $-\theta$:

And we have $$\sin\left(\phantom{180^\circ}-\theta\right) = -\color{blue}{\sin\theta} \qquad\qquad \cos\left(\phantom{180^\circ}-\theta\right) = \phantom{-}\color{red}{\cos\theta}$$ $$\sin\left(\phantom{1}90^\circ-\theta\right) = \phantom{-}\color{red}{\cos\theta} \qquad\qquad \cos\left(\phantom{1}90^\circ-\theta\right) = \phantom{-}\color{blue}{\sin\theta}$$ $$\sin\left(180^\circ-\theta\right) = \phantom{-}\color{blue}{\sin\theta} \qquad\qquad \cos\left(180^\circ-\theta\right) = -\color{red}{\cos\theta}$$ $$\sin\left(-90^\circ-\theta\right) = -\color{red}{\cos\theta} \qquad\qquad \cos\left(-90^\circ-\theta\right) = -\color{blue}{\sin\theta}$$

It's worthwhile to point out that the identity $$\cos\theta = \sin\left(90^\circ - \theta\right)$$ is, for some (such as myself), definitional:

The co-sine of the angle is the sine of the co-angle.

where "co-angle" means "complementary angle", just as "co-sine" literally means "complementary sine".

Also, the identities

$$\cos(-\theta) = \cos\theta \qquad \sin(-\theta) = -\sin\theta$$

have significance in establishing that cosine is an "even function" (it acts on negative arguments ---killing the sign--- the way an even exponent would) and sine is an "odd function" (it acts on negative arguments ---preserving the sign--- the way an odd exponent would). These are handy properties, which say interesting things about how the graphs are drawn. So, you might want to reserve a special area of your brain-space for them.

• FANTASTIC! Thank you so much for all the effort you've put out here! That's incredible! – user139919 Apr 3 '14 at 11:56
• Those wikipedia images are fantastic. – littleO May 4 '14 at 9:16
• Neat explanation!. Certainly drawings do clear up any misconceptions when you don't get the idea from looking plain equations. – Chris Steinbeck Bell Jul 29 '19 at 17:51

I wouldn't try to remember them, but try to understand them by visualizing the graphs of the sine and cosine function. If you draw the functions (mentally), you can easily "derive" all of these identities.

• I have poor mental power, so I can't visualize that :// – user139919 Apr 2 '14 at 20:59
• If you sit down and draw the functions and try to see what it means to shift them by $\pi$ or $\pi/2$ you're done, and you will be able to visualize it later on. This is still easier than trying to learn them by heart without understanding them. – Matt L. Apr 2 '14 at 21:01

Attempting to memorize all the relations you'll see in trigonometry is a daunting task, at least until you've worked with them for some time. For the sort of identities you're talking about, it can be helpful to go back to the unit circle and picture the angles there.

Starting with the angle $\ x \ ,$ the unit radius pointing in that direction has its tip at a "vertical distance" from the $\ x-$ axis which we call $\ \sin x \$ , and a "horizontal distance" from the $\ y-$ axis which we call $\ \cos x \ .$ The supplementary angle $\ \pi - x \$ completes a semi-circle from the positive $\ x-$ direction to the negative $\ x-$ direction.

We can now flip this around to look at the corresponding distances for the angle $\ \pi - x \$ . We see that we have the same triangle as before, but now it is reversed, on the other side of the $\ y-$ axis. The "vertical distance" for the tip of the unit radius is the same as it is for angle $\ x \ ,$ so we say $\ \sin(\pi - x) \ = \ \sin x \ .$ The "horizontal distance", however, is now in the opposite direction, so it is the negative of what it is for angle $\ x \ ,$ or $\ \cos(\pi - x) \ = \ -\cos x \ .$ Since $\ \tan x \ = \ \frac{\sin x}{\cos x} \ ,$ we can also say that

$$\ \tan (\pi - x) \ = \ \frac{\sin (\pi - x)}{\cos (\pi - x)} \ = \ \frac{\sin x}{-\cos x} \ = \ -\tan x \ \ .$$

$$\\$$

The next identities you list apply to the angle $\ \pi + x \ ,$ for which the unit radius points in exactly the opposite direction to the angle $\ x \ .$ The triangle for $\ \pi + x \$ is "upside down and backwards" from the triangle for angle $\ x \$ so both the "vertical" and "horizontal distances" have their signs changed. Thus, we have $\ \sin(\pi + x) \ = \ -\sin x \$ and $\ \cos(\pi + x) \ = \ -\cos x \ ;$ the relation for tangent will then be

$$\ \tan (\pi + x) \ = \ \frac{\sin (\pi + x)}{\cos (\pi + x)} \ = \ \frac{-\sin x}{-\cos x} \ = \ \tan x \ \ .$$

$$\\$$

The third pair you have applies to the complementary angle to $\ x \ ,$ which is $\ \frac{\pi}{2} - x \ .$ (If two of the angles in a right triangle are $\ \frac{\pi}{2} \$ (or 90º) and $\ x \$ , the third angle is $\ \frac{\pi}{2} - x \ .$ ) In the unit circle, that angle fills in the corner of an imaginary box we can draw as shown in the graph above. Now we'll flip this box around in the unit circle like this:

This has the effect of making the "vertical distance" for angle $\ x \$ the "horizontal distance" for angle $\ \frac{\pi}{2} - x \ ,$ and vice versa. So this tells us that $\ \sin(\frac{\pi}{2} - x) \ = \ \cos x \$ and $\ \cos(\frac{\pi}{2} - x) \ = \ \sin x \ .$ (Because these formulas concern complementary angles, they are sometimes referred to as the " co- relations". This is also the origin of the "co-" prefix in the names of trig functions: the co-sine of an angle is the sine of the complementary angle, etc.) For tangent, we then get a complementary function,

$$\ \tan (\frac{\pi}{2} - x) \ = \ \frac{\sin (\frac{\pi}{2} - x )}{\cos (\frac{\pi}{2} - x)} \ = \ \frac{\cos x}{\sin x} \ = \ \frac{1}{\tan x} \ = \ \cot x \ \ .$$

$$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\sin\beta\cos\alpha\\\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$$ Logically just replace $\alpha$ and $\beta$ with the value and you'll get the equation you need.