How to remember trigonometric ratios for allied angles? I just started studying trigonometry in unit circle and I want to know if there is some intuitive way to remember the value of $\sin(n\cdot\frac{\pi}{2}\pm\theta) \text{ and }\cos(n\cdot\frac{\pi}{2}\pm\theta)$, $n\in \mathbb N$. Sure I can use the double angle formula to derive it, but I don't think that it would be a good idea to do so during a timed test. 
 A: If $n$ is odd, the function changes into its cofunction. Sine becomes cosine and vice versa, tangent in cotangent and so on. If $n$ is even, it does not.
To get the sign, suppose $\theta$ is a small angle, check in which quadrant $n\cdot \frac{\pi}{2}+\theta$ is and what is the sign of the function in question there.
Suppose you want to calculate $ \sin(x+\pi/2) $, $n$ is odd, so sine becomes cosine. $x+\pi/2$ is in the second quadrant, where the sine is positive, therefore $\sin(x+\pi/2)=\cos(x) $.
$\tan(x+\pi)$. $n$ is even in this case, so tangent will remain tangent; $x+\pi$ is in the third quadrant, where the tangent is positive, therefore $\tan(x+\pi)=\tan(x) $
One more example: $\sin(x+3\pi/2)$: $n$ is odd, so sine becomes cosine; $x+3\pi/2$ is in the fourth quadrant, where the sine is negative. Therefore $\sin(x+3\pi/2)=-\cos(x)$
A: *A-S-T-C rule:
If the terminal side of an angle lies in


*

*Quadrant I, all trigonometric functions are positive for that angle.

*Quadrant II, only sin and cosec trigonometric functions are positive whereas other trigonometric functions are negative for that angle.

*Quadrant III, only tan and cot trigonometric functions are positive whereas other trigonometric functions are negative for that angle.

*Quadrant IV, only cos and sec trigonometric functions are positive whereas other trigonometric functions are negative for that angle.

*Simplification of a trigonometric function when the angle is negative


*

*sin (-A) = -sin A

*cos(-A) = cos A

*tan(-A) = -tan A

*cosec(-A) = -cosec A

*sec(-A) = sec A

*cot(-A) = -cot A

If any given angle A, is expressed in terms of the angle that a horizontal line makes with the X-axis (the expression consisting of two terms) then the trigonometric function does not change whereas the angle changes to the second term and the sign is determined in accordance with the A-S-T-C rule.  
If A is the given angle and A =90°. m + B, where m is an even whole number and B is an acute angle, then A is associated to the angle that a horizontal line makes with the X-axis. The quadrant in which the terminal side of angle A lies is given by m%4+1 if the second term is added. If the second term is subtracted then it is given by m%4.
sin A = sin (90°. m + B) = ±sin B, where ± indicates that the value of the trigonometric function can be positive or negative depending upon the quadrant in which the terminal side of angle A lies and is generally determined by the A-S-T-C rule.

If any given angle A, is expressed in terms of the angle that a vertical line makes with the X-axis (the expression consisting of two terms) then the trigonometric function changes to the complementary function and the angle changes to the second term and the sign is determined in accordance with the A-S-T-C rule. 
If A is the given angle and A =90°. n + B, where n is an odd whole number and B is an acute angle, then A is associated to the angle that a vertical line makes with the X-axis. The quadrant in which the terminal side of angle A lies is given by n%4+1 if the second term is added. If the second term is subtracted then it is given by n%4.
sin A = sin (90°. n + B) = ±cos B, where ± indicates that the value of the trigonometric function can be positive or negative depending upon the quadrant in which the terminal side of angle A lies and is generally determined by the A-S-T-C rule.
A: I base the  exact values on the following descending/ascending rule which I personally find quite easy to remember once you know it:
$$\cos(\pi/2)=\sin(0)=\sqrt0/2=0$$
$$\cos(\pi/3)=\sin(\pi/6)=\sqrt1/2=1/2$$
$$\cos(\pi/4)=\sin(\pi/4)=\sqrt2/2$$
$$\cos(\pi/6)=\sin(\pi/3)=\sqrt3/2$$
$$\cos(0)=\sin(\pi/2)=\sqrt4/2=1$$
After that use that the sign of sine goes with $y$, whereas cosine goes with x. Hence in the quadrant $++$, cosine and sine will be positive, in $-+$ cosine will be negative and sine positive and so on...
A: My method won’t work for everybody, but I remember what the graphs of the sine and cosine look like, and that there really are only three nonobvious values: $1/2$, $\sqrt2/2$, and $\sqrt3/2$. Couple that with the three interesting angles in the first quadrant, namely $30^\circ$, $45^\circ$, and $30^\circ$. I suppose that if you want to be mystical about it, you can fit $0$ and $90^\circ$ into the pattern by saying that the values of the sine are $\sqrt0/2$, $\sqrt1/2$, $\sqrt2/2$, $\sqrt3/2$ and $\sqrt4/2$.
