TITLE: How to remember a particular class of trig identities.
QUESTION [5 upvotes]: 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...

REPLY [4 votes]: 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,
$$ \ \tan (\frac{\pi}{2} - x) \ = \ \frac{\sin (\frac{\pi}{2} - x )}{\cos (\frac{\pi}{2} - x)} \ = \ \frac{\cos x}{\sin x} \ = \ \frac{1}{\tan x} \ = \ \cot x \ \ . $$