Solving advanced trigonometric problems can be very difficult if you're having trouble remembering some exact values of sine, cosine and tan. I've found the following diagrams to be very helpful.

Angles 60° and 30°
```
C         Let ABC be an equilateral triangle (ie,
^         all angles are 60°).
/|\
1 / | \1      CD is the perpendicular bisector of AB
/  |  \      (ie, ang ADC = ang BDC = 90°; BD = AD).
/___|___\
B    D   A     As indicated, all sides of ABC have a
1         length of 1.

By Pythagorean Theorem, segment CD = sqrt (AC2 - AD2)
= sqrt (1 - 1/4)
= sqrt (3/4)

These facts allow us to find  sin 60° = opposite/hypotenuse
= sqrt (3/4) ÷ 1
= sqrt (3/4).

= 1/2 ÷ 1
= 1/2.

= sqrt (3/4) ÷ 1/2
= (sqrt (3/4)) / 2

Also use the fact that angle ACD = angle BCD = 30°.
sin 30° = opposite/hypotenuse
= 1/2 ÷ 1
= 1/2.

= sqrt (3/4) ÷ 1
= sqrt (3/4)

= 1/2 ÷ sqrt (3/4)
= 2 * sqrt (3/4)
```

Angle 45°
```
C
|\_               Let ABC be a right triangle.
|  \_ sqrt(2)     Angle BAC = Angle ACB = 45°.
1 |    \_
|______\
B      A
1

These facts allow us to find sin 45° = opposite/hypotenuse
= 1/sqrt(2)

= 1/sqrt(2)

= 1/1
= 1
```
Finding the values (or ratios) of sine and cosine for the unit circle is very easy given that you can memorize a simple table of values. The table for cosine goes like this:
cos0 = sqrt4/2 = 1
cos30 = sqrt3/2
cos45 = sqrt2/2
cos60 = sqrt1/2 = 1/2
cos90 = sqrt0/2 = 0
If you notice, is starts with sqrt4/2 and with each of these angles the value of the number having the square root decreases by one in each step until it reaches 0.

Likewise, here is the table for sine. Sine's table is the inverse of cosine's, so it should be easy to remember.
sin0 = sqrt0/2 = 0
sin30 = sqrt1/2 = 1/2
sin45 = sqrt2/2
sin60 = sqrt3/2
sin90 = sqrt4/2 = 1

Because we know that tangent is sine/cosine, we can find the value of tangent by putting the exact value of sine or that of cosine. The 2s of the denominator will always cancel out so you can just put the numerator of sine over cosine's. In the same way, we know that secant and cosecant are the inverse of cosine and sine so we can find these values also. Simply take the reciprocal of the sine, cosine, or even tangent (for cotangent) to find their inversed values. If the denominator need rationalizing, do so by multiplying by a form of 1.

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