Worked examples
Every line says which identity it uses, so the jump from "they substituted something" to "they used the sine sum formula" is not left as an exercise. Answers are on the page — the point is the route, not the quiz.
Verify or refute: sin²θ · sec²θ − 1 = tan²θ
sin²θ · sec²θ − 1 = tan²θ, an identity (true for every θ where defined).
verification · 5 steps ·sec θ = 1/cos θ, tan θ = sin θ / cos θ
Simplify sin²θ + cos²θ + 3cos²θ
1 + 3cos²θ — and with the power-reducing formula, 5/2 + (3/2)cos 2θ.
rewriting · 4 steps ·sin²θ + cos²θ = 1, cos²x = (1 + cos 2x)/2
Verify (1 + tan²θ)cos θ = sec θ
Left side reduces to 1/cos θ = sec θ, so the equation is an identity.
verification · 5 steps ·1 + tan²θ = sec²θ, sec θ = 1/cos θ
Verify cot θ + csc θ = (1 + cos θ)/sin θ
Both sides are the same single fraction once cot and csc are written over sin θ.
verification · 4 steps ·cot θ = cos θ / sin θ, csc θ = 1/sin θ
Verify tan θ + cot θ = sec θ csc θ
Convert to sin and cos, take the common denominator, and the Pythagorean identity closes it.
verification · 5 steps ·tan θ = sin θ / cos θ, cot θ = cos θ / sin θ, sec θ = 1/cos θ, csc θ = 1/sin θ, sin²θ + cos²θ = 1
Simplify sin(−x)cos(−x) + tan(−x)
−sin x cos x − tan x
rewriting · 5 steps ·sin(−θ) = −sin θ, cos(−θ) = cos θ, tan(−θ) = −tan θ
If sin 32° = 0.53, what is cos 58°?
cos 58° = 0.53 — the two angles are complementary.
exact value · 4 steps ·sin θ = cos(90° − θ), cos θ = sin(90° − θ)
Find the exact value of sin 75°
(√6 + √2)/4
exact value · 4 steps ·sin(A + B) = sin A cos B + cos A sin B
Find the exact value of sin 15°
(√6 − √2)/4
exact value · 4 steps ·sin(A − B) = sin A cos B − cos A sin B, sin(A + B) = sin A cos B + cos A sin B
Find the exact value of cos 15° without a half-angle formula
(√6 + √2)/4
exact value · 4 steps ·cos(A − B) = cos A cos B + sin A sin B
Find the exact value of tan 75°
2 + √3
exact value · 4 steps ·tan(A + B) = (tan A + tan B) / (1 − tan A tan B)
Two lines have slopes 3 and 1/2. What is the tangent of the angle between them?
1 — so the angle is 45°.
exact value · 4 steps ·tan(A − B) = (tan A − tan B) / (1 + tan A tan B)
Verify tan A + tan B = sin(A + B)/(cos A cos B)
Common denominator, then recognise the sine sum.
verification · 4 steps ·tan A + tan B = sin(A + B) / (cos A cos B), sin(A + B) = sin A cos B + cos A sin B, tan θ = sin θ / cos θ
Derive the double-angle formulas from the sum formulas
Set B = A in each sum formula; cosine then yields two further forms via sin² + cos² = 1.
rewriting · 6 steps ·sin(A + B) = sin A cos B + cos A sin B, cos(A + B) = cos A cos B − sin A sin B, sin 2x = 2 sin x cos x, cos 2x = cos²x − sin²x, cos 2x = 2 cos²x − 1, cos 2x = 1 − 2 sin²x
∫ sin 3x cos x dx
−cos 4x/8 + cos 2x/4 + C
calculus preparation · 5 steps ·sin A cos B = ½[sin(A + B) + sin(A − B)]
Rewrite as a sum: (a) cos 2θ cos 4θ (b) sin θ sin 3θ
(a) ½[cos 2θ + cos 6θ] (b) ½[cos 2θ − cos 4θ]
rewriting · 4 steps ·cos A cos B = ½[cos(A − B) + cos(A + B)], sin A sin B = ½[cos(A − B) − cos(A + B)]
∫ sin²x dx
x/2 − sin 2x/4 + C
calculus preparation · 4 steps ·sin²x = (1 − cos 2x)/2
∫₀^π cos²x dx
π/2
calculus preparation · 4 steps ·cos²x = (1 + cos 2x)/2
∫ sin x cos x dx — two ways
sin²x/2 + C = −cos 2x/4 + C (the forms differ by a constant)
calculus preparation · 4 steps ·sin 2x = 2 sin x cos x
Solve sin x + sin 3x = 0 on [0, 2π)
x = 0, π/2, π, 3π/2 (and 2π excluded from the interval)
rewriting · 5 steps ·sin A + sin B = 2 sin((A + B)/2) cos((A − B)/2)
A triangle has a = 7, b = 5 and included angle C = 60°. Find the area and then angle A via the sine rule
Area = 35√3/4; and the same area written with another base gives a sin C = c sin A
exact value · 5 steps ·½ a b sin C = ½ b c sin A
Why does cos 2x have three formulas and how do you choose?
They are one formula plus the Pythagorean identity; choose by which square is already in your expression.
rewriting · 4 steps ·cos 2x = cos²x − sin²x, cos 2x = 2 cos²x − 1, cos 2x = 1 − 2 sin²x, sin²θ + cos²θ = 1
Solve tan 2x = 1 on [0, π)
x = π/8 and x = 5π/8
rewriting · 4 steps ·tan 2x = 2 tan x / (1 − tan²x)
Given cos x = −3/5 with 180° < x < 270°, find sin(x/2) and cos(x/2)
sin(x/2) = √(4/5) = 2/√5 and cos(x/2) = −1/√5
exact value · 6 steps ·sin(x/2) = ±√((1 − cos x)/2), cos(x/2) = ±√((1 + cos x)/2)
Simplify √(1 − cos x) so it can be integrated
√2·|sin(x/2)| — on 0 < x < 2π the absolute value is sin(x/2)
calculus preparation · 4 steps ·sin(x/2) = ±√((1 − cos x)/2)
Show that the three half-angle tangent forms agree at x = 60°
All three give 1/√3 = tan 30°
verification · 4 steps ·tan(x/2) = (1 − cos x)/sin x, tan(x/2) = sin x/(1 + cos x), tan(x/2) = csc x − cot x
∫ sin³x dx
−cos x + cos³x/3 + C
calculus preparation · 5 steps ·sin³x = (3 sin x − sin 3x)/4, sin²θ + cos²θ = 1
∫ cos³x dx
sin x − sin³x/3 + C
calculus preparation · 4 steps ·cos³x = (3 cos x + cos 3x)/4, sin²θ + cos²θ = 1
Ready to try without the answer next to you?
Ten practice problems with answers and hints — built from the same verified identities.