Trigonometry Identities
The trigonometric functions of angles are the ratios of the
various sides of a triangle. Consider a right-angled triangle
ABC as shown in the figure below.
Hypotenuse: The side opposite to the right angle in a triangle
is called the hypotenuse. Here the side AC is the hypotenuse.
Opposite Side: The side opposite to the angle in consideration
is called the opposite side. In this case because we are using
angle A, then the opposite side is CB.
Adjacent: If A is the angle we are using, then the adjacent
side is side AB
sine = opposite / hypotenuse
cosecant = hypotenuse / opposite
cosine = adjacent / hypotenuse
secant = hypotenuse / adjacent
tangent = opposite / adjacent
cotangent = adjacent / opposite
OR in terms of sine and cosine
sine = opposite / hypotenuse
cosecant = 1 / sine
cosine = adjacent/hypotenuse
secant = 1 / cosine
tangent = sine / cosine
cotangent = 1 / tangent
a2 + b2 = c2
cos2x + sin2x= 1
1 + tan2x= sec2x
angle A + angle B = 90o
1 + cot2x= csc2x
secx
cscx
cotx
tanx
cotx
=
1
----cosx
=
1
-----sinx
=
1
-----tanx
=
sinx
-----cosx
=
cosx
-----sinx
Law of Sines
a / sin(A) = b / sin(B) = c / sin(C)
The ratio of any side of a general triangle when divided by the
sine value of the angle opposite the side is equal to the ratio
of any other side of a general triangle when divided by the
sine value of the angle opposite it.
Law of Cosines
c2 = a2 + b2 - 2ab cos(C)
b2 = a2 + c2 - 2ac cos(B)
a2 = b2 + c2 - 2bc cos(A)
The square of any side of a general triangle is equal to the
sum of the squares of the other two sides decreased by 2
times the multiplication product of the (two not included)
sides multiplied times the cosine of the angle between the two
(not included) sides.
Example 1 - Solve for a and b in the given triangle.
Solution: Since the sum of the angles of the triangle is 180o,
we get that B = 60o. Now we can find a by using the
trigonometric ratio sine = opposite / hypotenuse.
sin = opp/hyp
sin30o = a/12
a = 12 x sin30o
12(.454)
5.45
To find b, we can use cos30o or sin60o, we get the same
answer. Let's use
cos30o = b/12
b = 12 x cos30o
12(/2)
6
Example 2 - Use identities to help simplify each expression.
A) 1/secx
B) 1/tanx
C) 5cos2x+ 5sin2x
D) (1 – cos2x)/sinx
Solution:
A) cosx (Since secx = 1/cosx, then the reciprocal of sec is cos.
Which means that 1/secx = cosx)
B) cotx (Since tanx = 1/cotx, then the reciprocal of tanx is
cotx. Which means that 1/tanx = cotx)
C) We know that cos2x+ sin2x= 1. So you factor the 5 out and
you get:
5cos2x + 5sin2x
5(cos2 x + sin2x)
5(1)
5
D) We know (1 - cos2x= sin2x) so we can write:
(1-cos2x)/sinx
sin2x/sinx
One of the most fascinating constructions in mathematics is
the unit circle. It is a relatively simple object, but it can be
analyzed from a great number of very different viewpoints.
Most people first learn about the unit circle in trigonometry in
either high school or college. It is the circle with center at
the origin and with radius one. It can be thought of as the set
of points in the plane that satisfy the equation x2 + y2 = 1.
The unit circle helps a trig student remember the values of
the sine and cosine functions for a variety of standard angles.
Since the radius of the circle is 1, the cosine of an angle in
standard position corresponds exactly to the x coordinate of
the point where the terminal ray of the angle intersects the
unit circle. Likewise, the sine of the angle corresponds to the
y coordinate of that point.
Quiz:
Use laws of sines:
1. In triangle ABC, a = 4.56, A = 43o, and C = 57o. Solve for
side c.
2. In triangle ABC, a = 15, b = 25, and angle A = 47o. Solve the
triangle.
Use laws of cosines:
3. In triangle ABC, a = 24, c = 32, and angle B = 115o. Solve for
side B.
4. Problem: Solve for x: 2sin x = 1.
5. Simplify (sin2x)(cos2x) + cos4x .
Answers to Quiz:
1.
Angle B = 180o - (43o + 57o) = 80o
Now, we use the law of sines to find the other sides lengths.
a / sin(A) = b / sin(B) = c / sin(C)
c
a
----- = ----sin C sin A
=
c= 4.56(sin 57) / sin 43
c=
5.61
2. A/ sin a = B /sin b
Sin b = 25 (sin 47) / 15
Sin B = 1.219
Since an angle cannot have a sine greater than 1, there is no
solution for this triangle.
3. We know two sides. Find the third using the law of cosines.
b2 = a2 + c2 - 2ac(cos B)
Plug in any known information. b2 = 242 + 322 - 2 * 24 *
32(-.4226) b2 = 2249
Solve for b by taking the square root of each side.
b = SQRT(2249) = 47.4
4. Solve for sin x by dividing by 2.
sinx = (1/2)
5. Factor. (cos2x)(sin2x + cos2x)
Using a Pythagorean Identity replace
sin2x + cos2x with 1. (cos2x)(1)
=cos2x
Bibliography
http://www.zaimoni.com/Trig.htm
http://www.acts.tinet.ie/trigonometry_645.html
http://mathforum.com/dr.math/tocs/trig.high.html
http://www.quickmath.com/www02/pages/modules/graphs/equ
ations/basic/index.shtml
http://www.asu.edu/lib/noble/math/trigonom.htm
http://www.sparknotes.com/math/trigonometry/trigonometric
equations/