1
Reciprocal and quotient identities are definitions.
csc θ = 1sin θ
sec θ = 1cos θ
cot θ = 1tan θ
tan θ = sin θcos θ
cot θ = cos θsin θ
the quotient pair
the reliable move
rewrite everything in sin and cos
when nothing else is obvious
Sine and cosine are the only two functions that really exist here. The other four are
abbreviations, and turning them back is the move that unsticks most problems.
2
All three Pythagorean identities come from one.
sin²θ + cos²θ = 1
the unit circle, x² + y² = 1
divide every term by cos²θ
tan²θ + 1 = sec²θ
divide every term by sin²θ
1 + cot²θ = csc²θ
Three identities to remember, or one to remember and two lines of division. The second
is more reliable under pressure, because a misremembered identity is worse than none.
3
Cosine is even; sine and tangent are odd.
cos(−θ)
= cos θ
even
sin(−θ)
= −sin θ
odd
tan(−θ)
= −tan θ
odd
Reflecting an angle across the x-axis keeps the x-coordinate and flips the y. Since
cosine IS the x-coordinate it survives unchanged, and sine does not.
4
Work on ONE side until it becomes the other.
start with the left: cot θ · sin θ
the messier side
= (cos θsin θ) · sin θ
rewrite in sin and cos
= cos θ ✓
matches the right side
Never move terms across the equals sign
Proving an identity is not solving an equation. Cross-multiplying or adding to both
sides assumes the statement is true, which is the thing being proved. Transform one
side only, and let it arrive at the other.
5
Variations: a value with a sign to decide, two fractions, and a negative angle.
The same identities, used to find a value, to combine fractions and to handle −θ.
A value given: sin θ = 35, θ in quadrant II. Find cos θ and tan θ.
cos²θ = 1 − (35)² = 1 − 925 = 1625
the Pythagorean identity
cos θ = ±45; quadrant II, so cos θ = −45
cosine is negative on the left
tan θ = (35) ÷ (−45) = −34
sin over cos
The identity gives the size; the quadrant gives the sign. Answering +45 forgets the second half.
Two fractions: 11 − sin θ + 11 + sin θ
= (1 + sin θ) + (1 − sin θ)(1 − sin θ)(1 + sin θ)
common denominator
= 21 − sin²θ
the sines cancel on top; difference of squares below
= 2cos²θ = 2 sec²θ
1 − sin²θ = cos²θ
Multiplying (1 − sin θ)(1 + sin θ) out to 1 − sin²θ is what lets the identity in.
A negative angle: tan(−θ) · cot θ
tan(−θ) = −tan θ
tangent is odd
−tan θ · (1tan θ)
cot θ = 1tan θ
Deal with the −θ first, using the even and odd rules; the minus sign survives to the answer.
6
Look for a Pythagorean pair hiding in the expression.
sin θ (csc θ − sin θ)
expand the bracket
= sin θ · (1sin θ) − sin²θ
csc θ = 1sin θ
= 1 − sin²θ
the sines cancel in the first term
= cos²θ
from sin² + cos² = 1
sin²θ + cos²θ + tan²θ = sec²θ
Rearranged, the first identity gives sin²θ = 1 − cos²θ and cos²θ = 1 − sin²θ. Those
two forms appear far more often in practice than the original, so learn to spot
"1 minus a square" as a signal.
Remember the flow
Rewrite in sin and cos · Look for 1 − sin² or 1 − cos² · Work one side only · Stop when it matches
An identity is transformed, never solved. Nothing crosses the equals sign.
On the exam
Tests ask to simplify an expression to a single function, to pick the equivalent expression, or to find cos θ or tan θ from a given sin θ and quadrant. The fastest route: rewrite everything in sin and cos, then look for 1 − sin²θ or 1 − cos²θ. A 10-second check: put θ = 30° (or 45°) into the original and your answer; the two values must agree.
