Using Trig Identities to Actually Simplify an Expression
Common Trigonometric Identities You Should Know lists the formulas; this article is about the separate skill of actually reaching for the right one at the right moment. Simplification is less about memorising more identities and more about pattern recognition — spotting a Pythagorean shape, a reciprocal in disguise, or a quotient hiding inside a fraction — and that recognition only really develops by working through enough concrete examples to see the same handful of shapes recur again and again. Every step below is checked against real output from the trig function & identity calculator, so each simplification is confirmed numerically, not just asserted algebraically.
A caution before the examples: simplifying can change the domain
An identity is true everywhere both sides are defined, but "both sides are defined" is doing real work in that sentence. Simplifying (1 − cos2θ) / sinθ down to sinθ, worked through below, is only valid where the original expression is defined in the first place — which means sinθ ≠ 0, since the original has sinθ in a denominator. The simplified form, plain sinθ, is perfectly well defined at sinθ = 0 (it just equals 0 there), so writing "= sinθ" without the caveat quietly extends the expression's domain by patching over a hole that was really there in the original. For most practical purposes this goes unnoticed, but it matters the moment a problem specifically asks about behaviour at the excluded points, and it's exactly the kind of detail that separates a careful simplification from an approximate one.
Example 1: (1 − cos²θ) / sinθ
The numerator, 1 − cos2θ, should immediately register as a rearranged Pythagorean identity: since sin2θ + cos2θ = 1, subtracting cos2θ from both sides gives sin2θ = 1 − cos2θ. Substituting that in:
(1 − cos2θ) / sinθ = sin2θ / sinθ = sinθ.
Checking at θ = 50°: the calculator gives cos 50° ≈ 0.6428 and sin 50° ≈ 0.766. Computing (1 − 0.64282) / 0.766 gives ≈ 0.7661, a hair above the calculator's own sin 50° ≈ 0.766. That tiny gap is not an error in the identity — it's the rounding already baked into the calculator's four-decimal cos and sin values compounding slightly once they're squared and divided. The identity itself is exact for every θ where sinθ ≠ 0; the rounding only shows up because real numbers, not infinite-precision ones, were fed into the check.
Example 2: sinθ × cotθ
Cotangent is a quotient in disguise: cotθ = cosθ / sinθ. Substituting turns the product into a cancellation:
sinθ × cotθ = sinθ × (cosθ / sinθ) = cosθ.
Checking at θ = 73°: sin 73° ≈ 0.9563 and cot 73° ≈ 0.3057, and their product is 0.9563 × 0.3057 ≈ 0.2923 — matching cos 73° ≈ 0.2924 to within the same one-digit rounding noise as the previous example. Recognising cot, sec, and csc as reciprocals or quotients of sin, cos, and tan, rather than as unrelated functions to memorise separately, is very often the single fastest route to a simplification.
Example 3: tanθ + cotθ
This one is worth doing fully in terms of sine and cosine rather than guessing at a shortcut. Substitute both quotient identities and find a common denominator:
tanθ + cotθ = sinθ/cosθ + cosθ/sinθ = (sin2θ + cos2θ) / (sinθ cosθ) = 1 / (sinθ cosθ),
using the Pythagorean identity to collapse the numerator to exactly 1. Since 1/sinθ = cscθ and 1/cosθ = secθ, that final expression can also be written as secθ × cscθ, giving the tidy identity tanθ + cotθ = secθ cscθ. Checking at θ = 35°: tan 35° ≈ 0.7002 and cot 35° ≈ 1.4281, summing to 2.1283. Separately, sec 35° ≈ 1.2208 and csc 35° ≈ 1.7434, multiplying to 2.1283 as well — an exact match this time, since no squaring was involved to amplify the rounding.
Example 4: (sinθ + cosθ)²
Expanding a squared binomial is ordinary algebra, but it's easy to forget that trig terms simplify further once expanded. Distributing the square:
(sinθ + cosθ)2 = sin2θ + 2 sinθ cosθ + cos2θ = (sin2θ + cos2θ) + 2 sinθ cosθ = 1 + 2 sinθ cosθ,
using the Pythagorean identity again to collapse the first two terms. Checking at θ = 20°: sin 20° ≈ 0.342 and cos 20° ≈ 0.9397, so (0.342 + 0.9397)2 ≈ 1.6428, and separately 1 + 2 × 0.342 × 0.9397 ≈ 1.6428 — matching exactly. As a bonus, the double-angle identity sin 2θ = 2 sinθ cosθ means this result can be written even more compactly as 1 + sin 2θ; at θ = 20°, sin 40° ≈ 0.6428, and 1 + 0.6428 = 1.6428, the same value reached by two completely different routes.
Example 5: sec²θ − tan²θ
This one uses the Pythagorean identity in its secant form rather than its original sine-cosine form. Starting from tan2θ + 1 = sec2θ and rearranging: sec2θ − tan2θ = 1, with no further work needed — the entire expression collapses to the constant 1 for every θ where both functions are defined. Checking at θ = 60°: sec 60° = 2 exactly, and tan 60° ≈ 1.7321, so sec260° − tan260° ≈ 4 − 3.0002 ≈ 0.9998 — extremely close to 1, with the small gap coming entirely from tan 60°'s irrational true value (√3) being rounded to four decimals before squaring. This is a good reminder that "checks out numerically" always means "checks out to the precision you calculated with," not "is exactly equal down to infinite digits" — the identity itself has no error in it at all.
A common trap: assuming trig functions distribute over addition
It is worth seeing an invalid "simplification" alongside the valid ones, since the mistake is common and the counterexample is quick. sin(A + B) is not equal to sin A + sin B — there is no shortcut that lets a trig function distribute over a sum the way multiplication distributes over addition. Checking at A = 30° and B = 45°: sin 30° + sin 45° ≈ 0.5 + 0.7071 = 1.2071, while sin(30° + 45°) = sin 75° ≈ 0.9659 — two clearly different numbers, not a rounding-sized gap like the ones above. The correct identity, sin(A + B) = sin A cos B + cos A sin B, is covered in Common Trigonometric Identities You Should Know; the point here is narrower: whenever an expression involves a trig function applied to a sum or difference, resist the urge to split it apart term by term the way you would with an ordinary linear function, and reach for the actual angle-sum identity instead.
Example 6: rationalising 1 / (1 − sinθ)
Some expressions resist the moves above until you borrow a plain algebra trick: multiplying by a conjugate. The denominator 1 − sinθ has a natural conjugate, 1 + sinθ, and multiplying the fraction by (1 + sinθ) / (1 + sinθ) — which is just 1, so nothing about the value changes — sets up a difference of squares in the denominator:
1 / (1 − sinθ) × (1 + sinθ) / (1 + sinθ) = (1 + sinθ) / (1 − sin2θ).
The denominator 1 − sin2θ is exactly the Pythagorean identity again, rearranged to equal cos2θ, giving a final simplified form of (1 + sinθ) / cos2θ. Checking at θ = 40°: 1 / (1 − sin 40°) ≈ 1 / (1 − 0.6428) ≈ 1 / 0.3572 ≈ 2.7996, and separately (1 + 0.6428) / 0.7662 ≈ 1.6428 / 0.5868 ≈ 2.7998 — matching to three decimal places, with the tiny remaining gap again coming from the calculator's four-decimal rounding of sin 40° and cos 40° before they were squared. Multiplying by a cleverly disguised form of 1 is not a trig-specific trick — it's the same conjugate technique used to rationalise a denominator in ordinary algebra — but recognising 1 − sinθ as an invitation for that particular conjugate is a pattern that only becomes obvious with practice.
A general strategy, not just a list of examples
The six worked simplifications above all lean on the same short list of moves, and running through them roughly in this order clears most simplification problems: first, rewrite every reciprocal function (csc, sec, cot) in terms of sin, cos, and tan, since a mixed bag of six functions is much harder to pattern-match than an expression using only two or three. Second, look explicitly for a Pythagorean shape — anything resembling 1 − sin2, 1 − cos2, sec2 − tan2, or sin2 + cos2 sitting somewhere in the expression, since those almost always simplify to something shorter. Third, if the expression is a sum of fractions, find a common denominator before doing anything else, the same way you would with plain algebraic fractions. Fourth, if a denominator looks like 1 plus or minus a single trig function, consider multiplying through by its conjugate, exactly as in Example 6, to expose a Pythagorean shape that wasn't visible before. Finally, if none of that moves things forward, try expanding any squared or multiplied binomials, since a hidden Pythagorean pattern is often only exposed after expansion, exactly as in Example 4 — and throughout, resist the temptation to split a trig function apart across a sum the way the trap above warns against.
Numerically checking each step, the way every example above did, is not just a classroom habit — it is the fastest way to catch a sign error, a misapplied identity, or an accidental domain restriction before it propagates through a longer derivation. None of the five simplifications above needed anything beyond the core Pythagorean, reciprocal, and quotient identities — the same short list that generates almost everything else in the subject, which is exactly why it is worth being fluent in that short list rather than memorising simplified results by rote. Fluency here means being able to recognise the pattern the moment it appears, not being able to recite the identity from memory in isolation — the two skills feel similar but only one of them actually helps when a genuinely new expression shows up on a problem set the specific examples above never covered and never could have anticipated in advance. The trig function & identity calculator is built for exactly that kind of spot-check, evaluating all six functions and the Pythagorean identity at once for any angle you choose.