PreCalc 2 · Trigonometric Functions
Swap a trig expression for a simpler one using the basic identities. Don’t jump to a calculator.
The identities you need
sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
1 + cot²θ = csc²θ
tan θ = sin θcos θ
csc θ = 1sin θ
sec θ = 1cos θ
cot θ = 1tan θ
Example
sec θ = 1cos θ
cos θ · 1cos θ = 1
Simplify cos θ · sec θ.
sec θ = 1cos θ
cos θ · 1cos θ = 1
Answer: 1
Simplify sin θcos θ.
That is the quotient identity for tangent.
Answer: tan θ
Simplify 1sin θ.
That is csc θ.
Answer: csc θ
Simplify sin²θ + cos²θ.
Pythagorean identity.
Answer: 1
Simplify tan θ · cot θ.
cot θ = 1tan θ
tan θ · cot θ = 1
Answer: 1
Simplify 1cos θ.
That is sec θ.
Answer: sec θ
Simplify 1 − sin²θ.
sin²θ + cos²θ = 1, so 1 − sin²θ = cos²θ.
Answer: cos²θ
Simplify sec²θ − 1.
1 + tan²θ = sec²θ, so sec²θ − 1 = tan²θ.
Answer: tan²θ
Simplify sin θ · csc θ.
csc θ = 1sin θ
sin θ · csc θ = 1
Answer: 1
Simplify cos θsin θ.
That is cot θ.
Answer: cot θ
Simplify 1 − cos²θ.
1 − cos²θ = sin²θ.
Answer: sin²θ
Simplify 1 + tan²θ.
That is sec²θ.
Answer: sec²θ
Simplify 1 − cos²θsin θ.
1 − cos²θ = sin²θ
sin²θsin θ = sin θ
Answer: sin θ
Simplify sec²θ − 1tan θ.
sec²θ − 1 = tan²θ
tan²θtan θ = tan θ
Answer: tan θ
Simplify (1 + tan²θ) cos²θ.
1 + tan²θ = sec²θ = 1cos²θ
1cos²θ · cos²θ = 1
Answer: 1
Simplify csc²θ − cot²θ.
1 + cot²θ = csc²θ, so csc²θ − cot²θ = 1.
Answer: 1
Simplify sin θ · cos θ · tan θ.
tan θ = sin θcos θ
sin θ · cos θ · sin θcos θ = sin²θ
Answer: sin²θ
Simplify 1 − sin²θcos θ.
1 − sin²θ = cos²θ
cos²θcos θ = cos θ
Answer: cos θ
Let sin t = a, cos t = b, and tan t = c. Rewrite sin(−t) − sin t in terms of a, b, c.
Sine is odd: sin(−t) = −sin t = −a.
−a − a = −2a.
Answer: −2a
Let sin t = a, cos t = b, and tan t = c. Rewrite sin(t + 2π) − cos(t − 4π) + tan(t + π).
Sine and cosine have period 2π: sin(t + 2π) = a and cos(t − 4π) = b.
Tangent has period π: tan(t + π) = c.
a − b + c.
Answer: a − b + c
Let sin t = a, cos t = b, and tan t = c. Rewrite cos t + cos(t + 6π) − tan(t + 3π) + 2 sin(t − 2π).
cos(t + 6π) = cos t = b.
tan(t + 3π) = tan(t + π) = tan t = c.
sin(t − 2π) = sin t = a.
b + b − c + 2a = 2a + 2b − c.
Answer: 2a + 2b − c
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