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).
cos 60° = adjacent/hypotenuse
= 1/2 ÷ 1
= 1/2.
tan 60° = opposite/adjacent
= 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.
cos 30° = adjacent/hypotenuse
= sqrt (3/4) ÷ 1
= sqrt (3/4)
tan 30° = opposite/adjacent
= 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)
cos 45° = adjacent/hypotenuse
= 1/sqrt(2)
tan 45° = opposite/adjacent
= 1/1
= 1
| 
