6.1.2 Trigonometric Identities
| 6.1.2 |
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| 1. Sine And Cosine Values Of Special Angles |
360o), and the negatives of these angles. We’ll find the sine and cosine values for the positive angles. Those for the
negative ones can be derived from those for the positive ones by the identities sin (–x) = –sin x and cos (–x) = cos x,
which we’ll discuss in Part 3. The values of the four remaining trigonometric functions for these angles can be readily
derived from these two functions.
Refer to Fig. 1.1. Clearly, recalling that the cosine and sine are the (u, v) coordinates of points on the unit circle (see
Section 6.1.1 Part 6), we have:
| Fig. 1.5 A Table Of Trigonometric Values. |
If you forget one or more of these values, you can build this table yourself as follows. The Radians and Degrees rows are obvious. For the Sine row, write down 0, 1, 2, 3, and 4, then take the square root of each number, and then divide each
by 2. For the Cosine row, reverse the order of the numbers in the Sine row.
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Let x be an arbitrary angle. The terminal arms of x and –x are symmetric with respect to the u-axis. See Fig. 3.1. Thus:
| sin (– x) = – sin x, |
It follows that sine is an odd function and cosine is an even function.
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| P and Q are symmetric with respect to the line v = u. |
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| P and Q are symmetric with respect to the v-axis. |
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Example 7.1
Solution
EOS
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| 8. The Pythagorean Identity |
The notation sin2 x means (sin x)2, ie, the square of sin x. This is true for any exponent and for all trigonometric functions.
Some examples are: cos3 x = (cos x)3, tan–2 x = (tan x)–2, secm x = (sec x)m. Note that sin x2 means sin (x2), ie, the sine
of x2, which is a different thing from sin2 x.
The point P = (u, v) = (cos x, sin x) lies on the unit circle u2 + v2 = 1. See Fig. 7.1. So cos2 x + sin2 x = 1. Thus:
This identity is called the Pythagorean identity because it’s the Pythagorean formula for the right triangle UPO as in
Fig. 7.1: UP2 + OU2 = OP2.
Example 8.1
Solution
EOS
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| 9. The Addition Identities For Sine And Cosine |
(cos (x – y) – 1)2 + (sin (x – y) – 0)2 = (cos x – cos y)2 + (sin x – sin y)2,
cos2 (x – y) – 2 cos (x – y) + 1 + sin2 (x – y) = cos2 x – 2 cos x cos y + cos2 y + sin2 x – 2 sin x sin y + sin2 y.
Now, cos2 (x – y) + sin2 (x – y) = 1, cos2 x + sin2 x = 1, and cos2 y + sin2 y = 1. It follows that:
2 – 2 cos (x – y) = 2 – 2 cos x cos y – 2 sin x sin y,
cos (x – y) = cos x cos y + sin x sin y.
That’s what we wish to get. It holds true for all values of x and all values of y (that’s why it’s called an identity). In
particular, it holds true for t = – y. That is:
cos (x + y) = cos (x – (– y)) = cos x cos (– y) + sin x sin (– y) = cos x cos y – sin x sin y.
Next, we want to express sin (x + y) in terms of the trigonometric functions of x and y. We have:
That’s what we want. Replacing y by – y and employing symmetry we get:
sin (x – y) = sin x cos y – cos x sin y.
We’ve obtained these four identities, called the addition identities:
| sin (x + y) = sin x cos y + cos x sin y, [9.1] |
Note that these identities express the sine and cosine of the sum and difference of 2 angles in terms of those of each
individual angle.
Remark 9.1
sin (x + y) is not identical to sin x + sin y,
cos (x – y) is not identical to cos x – cos y,
etc.
The relation sin (x + y) = sin x + sin y is an equation and not an identity, because it may be true for some values of x
and y, but it isn’t true for every value of x and every value of y.
Example 9.1
Solution
EOS
We expressed the given angles in terms of the special angles, at which the values of the trigonometric functions are
known, and we applied the trigonometric addition identities.
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| 10. Half-Angle Identities |
We have:
sin 2x = sin (x + x) = sin x cos x + cos x sin x = 2 sin x cos x,
cos 2x = cos (x + x) = cos x cos x – sin x sin x = cos2 x – sin2 x,
cos2 x – sin2 x = cos2 x – (1 – cos2 x) = 2 cos2 x – 1,
cos2 x – sin2 x = (1 – sin2 x) – sin2 x = 1 – 2 sin2 x,
from cos 2x = 2 cos2 x – 1 we get cos2 x = (1 + cos 2x)/2,
from cos 2x = 1 – 2 sin2 x we get sin2 x = (1 – cos 2x)/2.
| sin 2x = 2 sin x cos x, |
These identities are called half-angle identities. This is because the angle x is half of the angle 2x.
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| 11. Alternate Versions Of The Pythagorean Identity |
We’re now going to establish two alternate versions of the Pythagorean identity. They’re the Pythagorean identities for the
remaining four trigonometric functions: tangent, cotangent, secant, and cosecant.
Dividing the Pythagorean identity sin2 x + cos2 x = 1 by sin2 x we get 1 + ((cos x)/(sin x))2 = (1/(sin x))2, or 1 + cot2 x =
csc2 x. Similarly, division of sin2 x + cos2 x = 1 by cos2 x yields 1 + tan2 x = sec2 x.
| 1 + tan2 x = sec2 x, [11.1] |
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| 12. The Addition Identities For Tangent |
Note that tan (– s) = (sin (– s))/(cos (– s)) = – (sin s)/(cos s) = – tan s for any real number s. We have:
Replacing y by – y we obtain tan (x – y) = (tan x – tan y)/(1 + tan x tan y).
Note that these identities express the tangent of the sum and difference of 2 angles in terms of that of each individual
angle.
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| 13. The Sine And Cosine Laws |
| Sine Law: Cosine Law: a2 = b2 + c2 – 2bc cos A, b2 = c2 + a2 – 2ca cos B, c2 = a2 + b2 – 2ab cos C. |
| Fig. 13.1 |
| Fig. 13.2 |
Hence, in any case, utilizing the Pythagorean identity we get:
c2 = (b sin C )2 + (a – b cos C )2
= b2 sin2 C + a2 – 2ab cos C + b2 cos2 C
= a2 + b2(sin2 C + cos2 C ) – 2ab cos C
= a2 + b2 – 2ab cos C.
Example 13.1
Solution
EOS
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1. Find the values of the following quantities. Don’t use tables or calculators.
Solution
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2. Express the following quantities in terms of sin x or cos x or both.
Solution
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3. Prove the following identities.
Solution
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Solution
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5. Let ABC be an arbitrary triangle with sides a, b, and c opposite to angles A, B, and C respectively.
Solution
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