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Making Sense of Trigonometry and Trigonemtric Identities
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Appendix D has a trigonometric review. This material is meant to help you remember the values of the trig functions
at special values, and help you see how the trig identities are related. You will not be tested on this material directly;
you mainly need to have certain trig identities memorized, or know how to derive them if you need them. Remember–
memorized means memorized correctly, not just that you are familiar with something! If you use an identity in class or
on the homework that means it is important and might show up again.
Right Angle Triangles
The six basic trigonometric functions relate the angle θ to ratios of the length of the sides of the right triangle.
sin θ =
opposite
hypotenuse
cos θ =
adjacent
hypotenuse
opp
hyp
θ
adj
tan θ =
hypotenuse
opposite
hypotenuse
sec θ =
adjacent
adjacent
cot θ =
opposite
csc θ =
opposite
adjacent
From the definition of the trig functions, we immediately get the following identities:
csc θ =
1
sin θ
sec θ =
1
cos θ
cot θ =
1
tan θ
tan θ =
sin θ
cos θ
sin θ =
1
csc θ
cos θ =
1
sec θ
tan θ =
1
cot θ
cot θ =
cos θ
sin θ
For certain triangles, the trig functions of the angles can be found geometrically. These special triangles occur frequently
enough that it is expected that you learn the value of the trig functions for the special angles.
A 45-45-90 Triangle
Consider the square given below.
1
π/4
1
√
1
2
π/4
1
The angle here must be π/4 radians, since this triangle is half of a square of side length 1.
Now, we can write down all the trig functions for an angle of π/4 radians = 45 degrees:
opposite
1
=√
4
hypotenuse
2
π
adjacent
1
cos
=
=√
4
hypotenuse
2
sin
π
=
csc
π
4
sec
π
4
=
=
1
sin(
π
4
)
1
cos( π
4)
=
=
√
2
√
2
opposite
1
= =1
4
adjacent
1
π
1
=1
cot
=
4
tan π4
tan
π
=
Barry McQuarrie’s Calculus II
Making Sense of Trigonometry and Trigonemtric Identities
Page 2 of 4
A 30-60-90 Triangle
Consider the equilateral triangle given below. Geometry allows us to construct a 30-60-90 triangle:
We can now determine the six trigonometric functions at two more angles!
60o =
π
radians:
3
√
2
3
π/3
1
30o =
π
radians:
6
2
π
opposite
1
=
6
hypotenuse
2
√
π
3
adjacent
cos
=
=
6
hypotenuse
2
π opposite
1
=√
tan
=
6
adjacent
3
sin
1
π/6
√
√
opposite
3
sin
=
=
3
hypotenuse
2
π
adjacent
1
cos
=
=
3
hypotenuse
2
√
π opposite
√
3
=
= 3
tan
=
3
adjacent
1
π
3
=
1
2
√
=
π =
3
sin 3
3
π
1
=2
sec
=
3
cos π3
π
1
1
=√
cot
=
3
tan π3
3
csc
csc
π
π
6
=
1
sin
π
3
=2
2
1
=√
π
6
cos 3
3
√
π
3 √
1
=
cot
= 3
=
π
6
1
tan 3
sec
π
=
Obtuse Angles
y
P (x, y)
6
I
@
@
@ θ
x
-
If wep
label the point at the end of the terminal side as P (x, y), and if we let
r = x2 + y 2 , we can construct the following relationships between the six
trig functions and our diagram:
x
,
r
r
csc θ = , y =
6 0,
y
cos θ =
II
6
S A
T C
III
I
IV
y
,
r
r
sec θ = , x 6= 0,
x
sin θ =
y
, x 6= 0
x
x
cot θ = , y 6= 0
y
tan θ =
The CAST diagram tells us the sign of sine, cosine and tangent in the quadrants.
Quadrant IV: Cosine is positive, the other two are negative.
Quadrant I:
All are positive.
Quadrant II:
Sine is positive, the other two are negative.
Quadrant III: Tangent is positive, the other two are negative.
Barry McQuarrie’s Calculus II
Making Sense of Trigonometry and Trigonemtric Identities
Page 3 of 4
Identities
You will need to be able to know the basic trig identities, or derive them. I recommend memorizing a few, and deriving
others that you will need when necessary. I would memorize the following three:
cos2 x + sin2 x = 1
cos(u − v) = cos u cos v + sin u sin v
sin(u + v) = sin u cos v + cos u sin v
You will no doubt memorize others identities as you work with them, but if you do forget them you can derive them using
ideas like what is below.
Deriving Other Identities
• Derive some identities from cos2 θ + sin2 θ = 1:
Divide by cos2 θ:
sin2 θ
1
cos2 θ
+
=
−→ 1 + tan2 θ = sec2 θ .
2
cos θ cos2 θ
cos2 θ
Divide by sin2 θ:
cos2 θ sin2 θ
1
+
=
−→ cot2 θ + 1 = csc2 θ .
2
2
sin θ
sin θ
sin2 θ
• Derive some identities from cos(u − v) = cos u cos v + sin u sin v:
Evaluate at u = π/2: cos(π/2 − v) = sin v .
Evaluate at u = 0: cos(−v) = cos v .
Evaluate at v = −v: cos(u + v) = cos u cos v − sin u sin v ,
⇒ Now evaluate at v = u: cos(2u) = cos2 u − sin2 u ,
⇒ Now use sin2 u + cos2 u = 1: cos(2u) = cos2 u − 1 + cos2 u −→ cos2 u = 21 (1 + cos(2u)) .
⇒ Now use sin2 u + cos2 u = 1: cos(2u) = 1 − sin2 u − sin2 u −→ sin2 u = 21 (1 − cos(2u)) .
• Derive some identities from sin(u + v) = sin u cos v + cos u sin v:
Evaluate at v = u: sin(2u) = 2 sin u cos u .
Evaluate at u = π/2: sin(π/2 + v) = cos v .
Evaluate at v = −v: sin(u − v) = sin u cos v − cos u sin v . Now evaluate at u = 0: sin(−v) = − sin v
Barry McQuarrie’s Calculus II
Making Sense of Trigonometry and Trigonemtric Identities
Page 4 of 4
• Product to Sum Identities (very useful in certain types of trig integrals):
Add cos(u − v) = cos u cos v + sin u sin v and cos(u + v) = cos u cos v − sin u sin v:
cos u cos v =
1
cos(u − v) + cos(u + v)
2
Subtract cos(u − v) = cos u cos v + sin u sin v from cos(u + v) = cos u cos v − sin u sin v:
sin u sin v =
1
cos(u − v) − cos(u + v)
2
Add sin(u − v) = sin u cos v − cos u sin v and sin(u + v) = sin u cos v + cos u sin v:
sin u cos v =
1
sin(u − v) + sin(u + v)
2
Barry McQuarrie’s Calculus II