PreCalc 2 · Trigonometric Functions
Sine and cosine come back to the same value after a full turn around the circle (2π). Tangent comes back after a half turn (π). Peel off those extra turns, then simplify.
They come back around
Sine and cosine repeat every 2π — that’s one full trip around the unit circle.
sin(θ + 2π) = sin θ
cos(θ + 2π) = cos θ
Adding 4π or 6π is just more full trips, so you still get the same value.
Tangent repeats sooner: every π.
tan(θ + π) = tan θ
Adding 2π or 3π is more of those half turns, so tangent is still the same.
Pythagorean identities
Once the extra turns are gone, use these.
sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
1 + cot²θ = csc²θ
The last two are the first one in disguise: divide every term of sin²θ + cos²θ = 1 by cos²θ to get 1 + tan²θ = sec²θ. Divide by sin²θ to get 1 + cot²θ = csc²θ.
Even and odd
Cosine and secant are even: a minus inside does not change them.
cos(−θ) = cos θ
Sine, tangent, cotangent, and cosecant are odd: a minus inside flips the sign.
sin(−θ) = −sin θ
tan(−θ) = −tan θ
cot(−θ) = −cot θ
Example
Simplify tan(θ + π).
Tangent repeats every π.
tan(θ + π) = tan θ
Simplify sin(θ + 2π).
Sine repeats every 2π (one full circle).
sin(θ + 2π) = sin θ
Answer: sin θ
Simplify cos(θ + 2π).
Cosine repeats every 2π (one full circle).
cos(θ + 2π) = cos θ
Answer: cos θ
Simplify tan(θ + π).
Tangent repeats every π (a half turn).
tan(θ + π) = tan θ
Answer: tan θ
Simplify sin(θ + 4π).
4π is two full circles.
sin(θ + 4π) = sin θ
Answer: sin θ
Simplify cos(θ − 2π).
Subtracting 2π is still one full circle.
cos(θ − 2π) = cos θ
Answer: cos θ
Simplify tan(θ + 2π).
2π is two half-turns for tangent.
tan(θ + 2π) = tan θ
Answer: tan θ
Simplify sin²(θ + 2π) + cos²θ.
Sine repeats every 2π, so sin(θ + 2π) is just sin θ.
sin(θ + 2π) = sin θ
Now you have sin²θ + cos²θ = 1
Answer: 1
Simplify cos²(θ + 2π) + sin²θ.
Cosine repeats every 2π, so cos(θ + 2π) is just cos θ.
cos(θ + 2π) = cos θ
Now you have cos²θ + sin²θ = 1
Answer: 1
Simplify tan(θ + π) − tan θ.
Tangent repeats every π, so tan(θ + π) is just tan θ.
tan(θ + π) = tan θ
tan θ − tan θ = 0
Answer: 0
Simplify 1 + tan²(θ + π).
Tangent repeats every π, so tan(θ + π) is just tan θ.
tan(θ + π) = tan θ
1 + tan²θ = sec²θ
Answer: sec²θ
Simplify sin(θ + 2π) + sin θ.
Sine repeats every 2π, so sin(θ + 2π) is just sin θ.
sin(θ + 2π) = sin θ
sin θ + sin θ = 2 sin θ
Answer: 2 sin θ
Simplify cos(θ − 2π) − cos θ.
Cosine repeats every 2π, so subtracting a full circle does not change it.
cos(θ − 2π) = cos θ
cos θ − cos θ = 0
Answer: 0
Simplify tan²(θ + 3π) + cos²θ + sin²(θ + 4π).
Tangent repeats every π, and 3π is three of those half turns.
tan(θ + 3π) = tan θ
Sine repeats every 2π, and 4π is two full circles.
sin(θ + 4π) = sin θ
Now you have tan²θ + cos²θ + sin²θ.
sin²θ + cos²θ = 1
So 1 + tan²θ = sec²θ
Answer: sec²θ
Simplify tan²(x + π) + cos²x + sin²(x + 2π).
Tangent repeats every π.
tan(x + π) = tan x
Sine repeats every 2π (one full circle).
sin(x + 2π) = sin x
Now you have tan²x + cos²x + sin²x.
sin²x + cos²x = 1
So 1 + tan²x = sec²x
Answer: sec²x
Simplify sin²(θ − 2π) + tan²(θ + 2π) + cos²θ.
Sine repeats every 2π. Subtracting a full circle does not change it.
sin(θ − 2π) = sin θ
Tangent repeats every π, and 2π is two of those half turns.
tan(θ + 2π) = tan θ
Now you have sin²θ + tan²θ + cos²θ.
sin²θ + cos²θ = 1
So 1 + tan²θ = sec²θ
Answer: sec²θ
Simplify cos²(θ + 4π) + sin²θ + tan²(θ − π).
Cosine repeats every 2π, and 4π is two full circles.
cos(θ + 4π) = cos θ
Tangent repeats every π. Subtracting π does not change it.
tan(θ − π) = tan θ
Now you have cos²θ + sin²θ + tan²θ.
sin²θ + cos²θ = 1
So 1 + tan²θ = sec²θ
Answer: sec²θ
Simplify sin²(θ + 2π) + cos²(θ − 4π).
Sine repeats every 2π.
sin(θ + 2π) = sin θ
Cosine repeats every 2π, and 4π is two full circles.
cos(θ − 4π) = cos θ
sin²θ + cos²θ = 1
Answer: 1
Simplify 1 + tan²(θ + 5π).
Tangent repeats every π, and 5π is five of those half turns.
tan(θ + 5π) = tan θ
1 + tan²θ = sec²θ
Answer: sec²θ
Simplify 2 cot(−9θ) + 8 cot(9θ + π).
Cotangent is odd, so a negative inside flips the sign.
cot(−9θ) = −cot(9θ)
Cotangent repeats every π.
cot(9θ + π) = cot(9θ)
2(−cot(9θ)) + 8 cot(9θ)
−2 cot(9θ) + 8 cot(9θ)
6 cot(9θ)
Answer: 6 cot(9θ)
Simplify −5 cot(−θ) − 4 cot(−θ + π).
Cotangent is odd, so a minus inside flips the sign.
cot(−θ) = −cot θ
Cotangent repeats every π.
cot(−θ + π) = cot(−θ) = −cot θ
−5(−cot θ) − 4(−cot θ)
5 cot θ + 4 cot θ
9 cot θ
Answer: 9 cot θ
Simplify 3 cot(−θ) + 2 cot(−θ + π).
Cotangent is odd.
cot(−θ) = −cot θ
Cotangent repeats every π.
cot(−θ + π) = cot(−θ) = −cot θ
3(−cot θ) + 2(−cot θ)
−3 cot θ − 2 cot θ
−5 cot θ
Answer: −5 cot θ
Simplify −2 tan(−θ) + 6 tan(θ + π).
Tangent is odd.
tan(−θ) = −tan θ
Tangent repeats every π.
tan(θ + π) = tan θ
−2(−tan θ) + 6 tan θ
2 tan θ + 6 tan θ
8 tan θ
Answer: 8 tan θ
Simplify 4 sin(−θ) − 7 sin(θ + 2π).
Sine is odd.
sin(−θ) = −sin θ
Sine repeats every 2π.
sin(θ + 2π) = sin θ
4(−sin θ) − 7 sin θ
−4 sin θ − 7 sin θ
−11 sin θ
Answer: −11 sin θ
Simplify −6 cos(−θ) + 3 cos(θ + 2π).
Cosine is even, so a minus inside does not change it.
cos(−θ) = cos θ
Cosine repeats every 2π.
cos(θ + 2π) = cos θ
−6 cos θ + 3 cos θ
−3 cos θ
Answer: −3 cos θ
Simplify 5 cot(−2θ) − 3 cot(2θ + π).
Cotangent is odd.
cot(−2θ) = −cot(2θ)
Cotangent repeats every π.
cot(2θ + π) = cot(2θ)
5(−cot(2θ)) − 3 cot(2θ)
−5 cot(2θ) − 3 cot(2θ)
−8 cot(2θ)
Answer: −8 cot(2θ)
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