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MATH 135: PRECALCULUS
TRIGONOMETRIC IDENTITES
Establish the following identities.
1. cot ϑ sec ϑ = csc ϑ
A good strategy here is to write every function in terms of sine and cosine. This
illuminates cancellations we may not have seen otherwise.
cot ϑ sec ϑ =
2.
cos ϑ 1
1
=
= csc ϑ.
sin ϑ cos ϑ
sin ϑ
1
1
+
= 2 csc2 ϑ
1 − cos ϑ 1 + cos ϑ
This time, get common denominators.
1
1
1 + cos ϑ
1 − cos ϑ
2
+
=
−
=
.
1 − cos ϑ 1 + cos ϑ
1 − cos2 ϑ 1 − cos2 ϑ
1 − cos2 ϑ
Recall from the Pythagorean identity that cos2 ϑ + sin2 ϑ = 1. Then, we have
2
2
=
= 2 csc2 ϑ.
1 − cos2 ϑ
sin2 ϑ
3.
cos ϑ
= tan ϑ(1 + sin ϑ)
csc ϑ − 1
Multiply and and divide the left side by the conjugate csc ϑ + 1.
cos ϑ
cos ϑ csc ϑ + 1
cos ϑ(csc ϑ + 1)
=
=
.
csc ϑ − 1
csc ϑ − 1 csc ϑ + 1
csc2 ϑ − 1
Addition or subtraction in the denominator is bad. Use the Pythagorean theorem: csc2 ϑ − 1 = cot2 ϑ. Then, write everything with sines and cosines.
cos ϑ sin2 ϑ sin1 ϑ + 1
cos ϑ sin1 ϑ + 1
cos ϑ(csc ϑ + 1)
=
=
.
cos2 ϑ
csc2 ϑ − 1
cos2 ϑ
sin2 ϑ
Keeping in mind you want a tan ϑ in your result, cancel and distribute accordingly.
cos ϑ sin2 ϑ sin1 ϑ + 1
sin ϑ
=
(1 + sin ϑ) = tan ϑ(1 + sin ϑ).
2
cos ϑ
cos ϑ
4.
sec ϑ − 1
tan ϑ
−
= −2 cot ϑ
tan ϑ
sec ϑ − 1
Get common denominators.
sec ϑ − 1
tan ϑ
(sec ϑ − 1)2
tan2 ϑ
−
=
−
tan ϑ
sec ϑ − 1
tan ϑ(sec ϑ − 1) tan ϑ(sec ϑ − 1)
Simplify the numerator.
tan2 ϑ
sec2 ϑ − 2 sec ϑ + 1 − tan2 ϑ
(sec ϑ − 1)2
−
=
tan ϑ(sec ϑ − 1) tan ϑ(sec ϑ − 1)
tan ϑ(sec ϑ − 1)
Use the Pythagorean identity to get sec2 ϑ − tan2 ϑ = 1. Tip: if you see a sum
or difference of squares but are unsure which Pythagorean identity to use, start
with sin2 ϑ + cos2 ϑ = 1 and massage it until you get the one you want (in this
case, divide by cos2 ϑ and subtract tan2 ϑ).
sec2 ϑ − 2 sec ϑ + 1 − tan2 ϑ
2 − 2 sec ϑ
2
=
=−
= −2 cot ϑ.
tan ϑ(sec ϑ − 1)
tan ϑ(sec ϑ − 1)
tan ϑ