In the previous section, we introduced the "circle" function, C(t) = et i, and discussed how to compute approximate values for it, using the concept of radian angle measure. We then introduced trigonometric functions by showing how cos and sin can be defined algebraically as the real and complex parts of C(t) = cos(t) + sin(t) i. The remaining functions, tan, cot, sec, csc, were defined algebraically as:
csc(t) = 1/sin(t)
sec(t) = 1/cos(t)
cot(t) = cos(t)/sin(t)
tan(t) = sin(t)/cos(t)
We gave geometric descriptions of cos, sin, tan, and sec that explained the meaning behind their names. We also explained the naming of cot and csc, as co-functions. In this section, we will see how rules of exponents lead to a large number of algebraic formulas involving these trigonometric functions.
As applications, we will show how these formulas describe basic symmetries and relationships between the graphs of trigonometric functions. We will see how they can be used to compute exact values for certain angles. By remembering the decimal approximation to these values, it is a simple matter to deduce exact values for a large number of other angles. Finally, we will show how these identities lead to some convenient algebraic approximations to sin and cos.
The first pair of identities we want to consider follow directly from Euler's formula, et i = cos(t) + sin(t) i, via conjugation. If we conjugate both sides of this equation, , this becomes e-t i = cos(t) - sin(t) i. Applying Euler's formula again to the left-hand side gives cos(-t) + sin(-t) i = e-t i = cos(t) - sin(t) i, so that:
cos(-t) = cos(t) and sin(-t) = -sin(t)
That is, cos is an even function, while sin is odd. Note: This is not so surprising, given the formulas for these two functions: cos is a sum of even powers, while sin is a sum of odd powers. Remember what these means geometrically: the graph of cos is unchanged if it is flipped horizontally; flipping the graph of sin horizontally or vertically gives the same result.
Another fundamental pair of trigonometric identities follow directly from rules of exponents. We know that:
et ies i = et i + s i = e(t + s) i
so that:
(cos(t) + i sin(t))(cos(s) + i sin(s)) = cos(t + s) + i sin(t + s)
multiplying out and simplifying the left-hand side gives:
cos(t)cos(s) + i sin(t)cos(s) + i cos(t)sin(s)
+ i2sin(t)sin(s)
= (cos(t)cos(s) - sin(t)sin(s)) + i (sin(t)cos(s)
+ cos(t)sin(s))
cos(t + s) = cos(t)cos(s) - sin(t)sin(s) and sin(t + s) = sin(t)cos(s) + cos(t)sin(s)
These are the most basic identities, and are naturally referred to as the angle sum identities. Using the even/odd identities, we can derive corresponding difference identities. For example:
cos(t - s) = cos(t + -s) = cos(t)cos(-s) - sin(t)sin(-s) = cos(t)cos(s) - -sin(t) sin(s) = cos(t)cos(s) + sin(t)sin(s)
cos(t - s) = cos(t)cos(s) + sin(t)sin(s) and sin(t - s) = sin(t)cos(s) - cos(t)sin(s)
Notice that these identities allow us to algebraically verify the cofunction identities. For example:
sin(p/2 - t) = sin(p/2)cos(t) - cos(p/2)sin(t) = 1·cos(t) - 0·sin(t) = cos(t)
Three more fundamental identities follow immediately, by simply setting t = s in the angle sum and difference identities:
cos(t + t) = cos(t)cos(t)
- sin(t)sin(t),
sin(t + t) = sin(t)cos(t) + cos(t)sin(t),
and
cos(t - t) = cos(t)cos(t) + sin(t)sin(t)
In other words:
cos(2t) = (cos(t))2
- (sin(t))2,
sin(2t) = 2sin(t)cos(t), and
1 = cos(0) = (cos(t))2 + (sin(t))2.
Note: The fourth identity: sin(t - t) = sin(t)cos(t) - cos(t) sin(t), simplifies to 0 = 0, and so is not useful. The first two identities are known as the double angle identities, while the last is called the Pythagorean identity, since it also can be derived from the Pythagorean theorem (using our geometric description of sin and cos). We should mention that, when dealing with trigonometric functions, Euler established the practice of writing the exponent over the function, so that these identities are normally written:
cos(2t)
= cos2(t) - sin2(t),
sin(2t) = 2sin(t)cos(t), and
1 = cos2(t) + sin2(t).
By using the Pythagorean identity, we can either eliminate the cos2(t) or sin2(t) term in the first double angle formula to get two other useful variants:
cos(2t) =
cos2(t) - (1 - cos2(t)) = 2cos2(t)
- 1 and
cos(2t) = (1 - sin2(t)) - sin2(t) = 1
- 2sin2(t).
We will use these identities to derive even more subtle identities in the next section.
In the Exercises from the previous section, we recognized the values and showing up in the values of the trigonometric functions for particular angles. Using these identities, it is now possible to explain why this happens. Specifically, by using the double angle identities, and the symmetry of the circle, we will compute the exact values of cos(p/3), sin(p/3), cos(p/4), sin(p/4), cos(p/5), cos(2p/5), cos(p/6), and sin(p/6).
We begin by observing, from the symmetry of the circle, that:
sin(2p/3)
= sin(p/3),
cos(2p/3)
= -cos(p/3),
cos(p/4)
= sin(p/4),
cos(4p/5)
= -cos(p/5),
sin(p/3)
= cos(p/6), and
cos(p/3)
= sin(p/6).
To
make this easier to see, we have shown each trigonometric value as the length of
a correspondingly colored line segment, with equal segments having the same
color.
If we let a = sin(p/3) = sin(2p/3), b = cos(p/3), and apply the double angle formula for sin with t = p/3, we get:
sin(2p/3) =
2sin(p/3)cos(p/3),
or
a = 2ab
Since a ¹ 0, this means that 1/2 = b = cos(p/3).
Likewise, using the double angle formula for cos with t = p/3, since cos(2p/3) = -cos(p/3) = -b = -1/2, we get:
cos(2p/3)
= 1 - 2sin2(p/3)
or
-1/2 = 1 - 2a2
so that a2 = 3/4, and (since a > 0) sin(p/3) = sin(2p/3) = a = . Notice that, since » 0.866025, these two values are consistent with a simple visual estimate of C(p/3) » 0.5 + 0.9i. By symmetry of the circle, we automatically obtain the exact value of C(p/6) = + 0.5i, that is, cos(p/6) = and sin(p/6) = 1/2.
Similarly, if we let c = sin(p/4) = cos(p/4) and apply the double angle formula for sin with t = p/4, we get:
sin(p/2)
= sin(2p/4) = 2sin(p/4)cos(p/4),
or
1 = 2c2
so that c = . Again, since » 0.707107, these two values are consistent with a simple visual estimate of C(p/4) » 0.7 + 0.7i.
Finally, if we let d = cos(p/5) and e = cos(2p/5), and apply the double angle formula for cos with t = p/5 and 2p/5, we get:
cos(2p/5)
= 2cos2(p/5) - 1
or e = 2d2 - 1, and
cos(p/5)
= -cos(4p/5)
= -(2cos2(2p/5) - 1)
or d = -2e2 + 1.
If we plug the first equation into the second, we obtain a fourth degree polynomial: d = -2(2d2 - 1)2 + 1 = -2(4d4 - 4d2 + 1) + 1 = -8d4 + 8d2 - 1, or 0 = 8d4 - 8d2 + d + 1. We will eventually learn general techniques for solving such a polynomial equation. For now, we will simply say that we can show how this equation has solutions d = -1, 1/2, and , with corresponding values of e = 1, -1/2, and , respectively. Since C(p/5) is further to the right than C(2p/5), d > e > 0, so that the correct solutions must be cos(p/5) = and cos(2p/5) = . Since » 0.809017 and » 0.309017, these two values are consistent with the simple visual estimates of cos(p/5) » 0.8 and cos(2p/5) » 0.3.
To summarize, using the symmetry of the circle and some algebra, we have computed the following exact values for sin and cos:
| x | sin(t) | cos(t) |
| p/6 | 1/2 | |
| p/5 | ||
| p/4 | ||
| p/3 | 1/2 | |
| 2p/5 |
In addition, using the Pythagorean identity, we can then solve for sin(p/5) and sin(2p/5). For example:
» 0.6
We will leave the computation of sin(2p/5) as an Exercise.With a bit of imagination, we can use the sum and difference identities to derive even more values. For example, we may compute sin(p/12) as:
sin(p/12) = sin(p/3
- p/4) = sin(p/3)cos(p/4)
- cos(p/3)sin(p/4)
= »
0.26
More generally, for small inputs, the higher power terms in the formula for sin become negligible, so that sin(t) » t. For example, sin(p/12) » 0.258819, while p/12 » 0.261799. This approximation is also evident in the geometric definition of sin, since small arcs are relatively straight, so that the vertical coordinate is approximately equal to the distance along the circle. It is also clear from the graphs of y = sin(x) and y = x, as seen in the Example in the following applet:
Note: This approximation is commonly used in Physics when studying periodic motion, such as that of springs and pendulums.
We can then use this approximation and the double angle formula for cos to obtain another approximation formula cos(2t) = 1 - 2sin2(t) » 1 - 2t2. Replacing t by t/2 and simplifying, gives the approximation cos(t) » 1 - 2(t/2)2 = 1 - t2/2. This is a good approximation for small values of t, as seen in the Example in the following applet:
Practice using and deriving trigonometric identities and approximations by
completing the following Exercises.
We can use the Pythagorean and double angle identities to derive even more interesting and useful identities. For example, by dividing the Pythagorean identity by sin2 or cos2, we can derive corresponding identities for the more advanced trigonometric functions:
csc2(t) = 1 + cot2(t) and sec2(t) = 1 + tan2(t).
These are useful for determining the values of tan or cot directly (and hence more accurately) from the values of cos and sin alone. For example, since cos(p/6) = , we know that sec2(p/6) = 4/3, so that tan2(p/6) = 1/3 and tan(p/6) = .
By solving the double angle formulas:
cos(2t) = 2cos2(t) - 1 and cos(2t) = 1 - 2sin2(t)
for sin2 and cos2:
(cos(2t) + 1)/2 = cos2(t) and (cos(2t) - 1)/(-2) = sin2(t)
taking square roots, and replacing t by t/2, we obtain the half-angle formulas for sin and cos:
and .
Notice that in order to use these formulas, we must first estimate the value from the unit circle in order to choose the correct sign outside the radical. For example, since we can visually estimate that sin(p/12) is between 0.2 and 0.3, the half-angle formula for sin gives:
While this looks different that the value of sin(p/12) = that we obtained before, if you plug both expressions into your calculator, you will find that they both simplify to exactly the same value (about 0.258819). Note: You can prove the two are equal by squaring both expressions and comparing the results.
Using the double angle identities in a different way, we can derive a corresponding half-angle formula for tan:
Notice how we first multiplied the numerator and denominator by 2sin(t), added and subtracted 1 from the numerator, and used the double angle identities for sin and cos. In a similar manner, we can also show that:
As you can see, the trigonometric functions satisfy a rich collection of algebraic identities. While these identities prove to be useful in a number of advanced mathematics courses, they are surprising and beautiful in their own right.
Practice working with and deriving more trigonometric identities by completing
the following Exercises.
Go to Graphing and Trigonometric Functions
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