Trigonometry
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Real trigonometry questions asked by students, solved step by step Every question below was submitted by a real student and answered step by step.
The solutions are x = 5 pi/4 + 2 pi n and x = 7 pi/4 + 2 pi n. A negative sine puts the angle in quadrants three and four, both with reference angle pi/4.
The reference angle is pi/3 and sine is positive in quadrants I and II, so the solutions are x = pi/3 + 2 pi n and x = 2 pi/3 + 2 pi n for every integer n.
The angles are about 143.13° and 216.87° (2.498 and 3.785 rad). Take arccos(0.8) as the reference angle, then place it in quadrants II and III where cosine is negative.
The angles are about 9.49 degrees, 85.26 degrees and 85.26 degrees. With three sides and no angles, the law of cosines is the only way in.
The arc is 5pi/2 = 7.854 cm and the area is 45pi/4 = 35.343 cm^2. Change 50 degrees into 5pi/18 radians, then apply s = r theta and A = (1/2) r^2 theta.
The arc is 14pi/3 = 14.661 yd and the area is 28pi = 87.965 yd^2. Convert 70 degrees to 7pi/18 radians before using s = r theta and A = (1/2) r^2 theta.
Both come out to 2pi/3 = 2.094. With r = 2 the factor (1/2)r^2 equals r, so the arc length in ft and the sector area in ft^2 share the same number - a coincidence of r = 2.
The arc is 2pi/3 = 2.094 m and the area is 4pi/3 = 4.189 m^2. Halving the angle while doubling the radius keeps the arc length the same but doubles the sector area.
Convert 120 degrees to 2pi/3 radians, then s = r theta = 3 x 2pi/3 = 2 pi = 6.283 m - exactly one third of the circle's circumference.
The sum is pi/2 when x > 0 and -pi/2 when x < 0, and undefined at x = 0. Both angles carry the sign of x, so the value is (pi/2) times sgn(x) and never 0.
Even: -7x^2 + 1, cos(-x) and tan x / x. Odd: sin x / cos x and csc x. Neither: 4x^3 - 3x^2 + 5, whose mixed-parity terms cancel no symmetry.
The height is 1.75 root 3 = 3.03 km, below the 4.5 km comparison. Same method as the 600 km/h case but the shorter time flips the conclusion.
Aircraft Q climbs to 3.71231 km, below the 5 km of aircraft P, so the answer is "lower". Vertical speed is 450 sin 45 deg = 318.198 km/h and 0.7 minutes is 0.0116667 hours.
Aircraft A climbs to 5 km, below the 7 km of aircraft B, so the answer is "lower". Vertical speed is 500 sin 30 deg = 250 km/h and 1.2 minutes is 0.02 hours.
The height is 9 root 3 = 15.59 km, higher than the 10 km comparison. Only the vertical component v sin 60 counts, and 1.8 minutes must become 0.03 hours.
Aircraft E climbs to 7.7 km, below the 8 km of aircraft F, so the answer is "lower". Vertical speed is 440 sin 30 deg = 220 km/h and 2.1 minutes is 0.035 hours.
Answer: [-5pi/6, 5pi/6]. The inequality reduces to cos x >= -sqrt(3)/2, and on one full period centred at 0 that is the single middle band of the interval.
The two cities are about 695.8 miles apart. Subtract the latitudes to get a central angle of 10 degrees 4 minutes, convert to radians, then use s = r theta.
arccos(0.33) = 1.234493 rad = 70.7312 degrees. Because 0.33 is not a special cosine value, the exact answer stays arccos(0.33) and only a decimal follows.
The answer is about 41.41 degrees, or 0.7227 radians. The input is not a special angle, so a calculator is needed - and it must be in the right angle mode.
arccos(1/sqrt(3)) is about 0.9553 radians, or 54.7356 degrees. The value 0.57735 is not a special-angle cosine, so the angle has no closed form in terms of pi.
The answer is pi/4, a positive angle. sqrt(2)/2 is the cosine of 45 degrees, and the inverse cosine range 0 to pi forces the first-quadrant answer.
arcsin(0.375) = arcsin(3/8) ≈ 0.384397 rad ≈ 22.0243°. Only sines of 0, ±1/2, ±√2/2, ±√3/2 and ±1 give closed-form angles; 3/8 is not among them.
arcsin(0.395) ≈ 0.406068 rad ≈ 23.2660°. The value 0.395 = 79/200 is not a special-angle sine, so the exact answer stays in inverse-sine form.
The answer is pi/3, or 60 degrees. Although sin(2pi/3) is also sqrt3/2, arcsin returns only the value in [-pi/2, pi/2], so 2pi/3 is not the answer.
The answer is about 0.002999991 radians, or 0.1719 degrees. For tiny inputs arctan(x) is very nearly x, differing here only in the eighth decimal place.
The answer is about 12.68 degrees, or 0.2213 radians. Reduce the ratio to 0.2250 first, then take the inverse tangent - a small ratio means a small angle.
The value is about 0.75666 rad or 43.35 degrees. The fraction reduces to 929/984, which is not a special tangent value, so no exact angle exists.
The value is about 0.46365 radians or 26.565 degrees. It is the angle whose tangent is 1/2, and it is not a special angle, so no exact closed form exists.
The exact value is -sqrt(2)/2, about -0.7071. The angle lies in the third quadrant where cosine is negative, and its reference angle is 45 degrees.
The answer is -1/2. Since 240 = 180 + 60, the angle lies in quadrant three where cosine is negative, so cos 240 equals minus cos 60, which is -1/2.
The answer is -1/8. The angles double each time, so the identity cos(x)cos(2x)cos(4x) = sin(8x)/(8 sin x) collapses the whole product in a single step.
The value is about 0.9925. Converting 7 degrees to 0.12217 radians and using the series 1 - x^2/2 + x^4/24 reproduces the same four decimal places.
The answer is about 0.99006. sin(3 radians) equals 0.14112 because 3 is just under pi, and the cosine of that small number sits very close to 1.
The answer is about 0.3420. 20 degrees is not a constructible special angle, so no expression in ordinary radicals exists and a calculator is genuinely required.
The answer is -root 2 over 2, about -0.7071. 225 degrees sits in quadrant three where sine is negative, and its reference angle is a clean 45 degrees.
The answer is -1. Since 315 = 360 - 45, the angle sits in quadrant four where tangent is negative, so tan 315 equals minus tan 45, which is -1.
The exact value is sqrt(5 - 2 sqrt 5), about 0.72654. Tangent has period pi, so 4.2 pi reduces to 0.2 pi = pi/5, which is the 36 degree angle.
The value is about 0.1228. In radians 7 degrees is 0.12217, and for small angles tan x is close to x, so the answer barely exceeds the radian measure itself.
The value is about 0.8794. Since cos(20 degrees) is 0.9396926, doubling gives 1.8793852, and subtracting 1 leaves 0.8793852, which rounds to 0.8794.
The answer is 5pi/6, not -pi/6. The arccos range is [0, pi], so a negative input lands in the second quadrant at pi minus the reference angle pi/6.
The answer is plus or minus sqrt((7 sqrt145 - 79)/18), about 0.5422. Substituting u = sin^2 x gives a quadratic, and only the positive root can be a square.
Answer: sin 2x = +/- 2*sqrt(2)/3. Squaring gives 6u^2 + u - 1 = 0 with u = sin^2 x = 1/3, and cos x = sqrt(6)/3 is forced positive while sin x keeps both signs.
sin A = 3/4. The key move is cos B = -cos(A + C), which collapses the numerator to sin A sin C and turns the condition into tan C = sqrt 3, so C = 60 degrees.
sin A = 3/4. Replacing B by pi - A - C turns the numerator into sin A sin C, the ratio collapses to tan C = root 3, and the sine rule finishes the job.
The answer is about 41.30 degrees, or 0.7208 radians. A one-hundredth drop in the sine from 0.67 moves the angle by roughly three quarters of a degree.
The answer is about 42.07 degrees, or 0.7342 radians. Since 0.67 is just under sin(45) = 0.7071, the angle has to land just under 45 degrees too.
The answer is about 53.13 degrees, or 0.9273 radians. This is the 3-4-5 triangle angle, since sin(theta) = 4/5 = 0.8 exactly - a useful accuracy check.
From s = r theta, theta = s/r = 3/5 = 0.600 radians, about 34.38 degrees. The units cancel, which is why radian measure is dimensionless.
Rearranging A = (1/2) r^2 theta gives theta = 2A/r^2 = 6/25 = 0.240 radians, about 13.75 degrees.
From s = r theta, r = s/theta = 2 / (1/3) = 6 feet. Dividing by a fraction means multiplying by its reciprocal.
Rearrange A = (1/2) r^2 theta into r = sqrt(2A/theta). With A = 2 and theta = 1/3 this gives r^2 = 12, so r = 2 sqrt(3) = 3.464 ft.
It reaches 5 km, so it is still below an aircraft cruising at 7 km. Multiply the 500 km/h speed by sin 30 to get the vertical rate, then by the time in hours.
Answer: True. Cosine is the x-coordinate on the unit circle, and 180 degrees lands on (-1, 0), so cos(180 degrees) = -1 exactly, not an approximation.
True. Since 7π/6 = π + π/6 lies in quadrant III, sin(7π/6) = −1/2, and cosecant is the reciprocal of sine, so csc(7π/6) = 1/(−1/2) = −2 exactly.
sin x / x is even and cos x / x is odd. Dividing an odd function by the odd function x gives an even result, while dividing an even function by x gives an odd one.
Answer: true. Since tan = sin/cos and sin(pi) = 0 while cos(pi) = -1, the quotient is 0/(-1) = 0. A zero numerator with a nonzero denominator is simply zero.
A ladder reaching 24 ft up a wall with its base 7 ft out is 25 ft long and meets the ground at about 73.7°. Pythagoras gives the length, arctan the angle.
Two triangles fit: a = 75sqrt3 + 15sqrt39 = 223.58 or a = 75sqrt3 - 15sqrt39 = 36.23. The unknown side appears squared and linearly, giving a quadratic.
A is about 69.6 degrees and x is about 24.03. sin A = 23 sin 32 / 13 = 0.9375; the note that A appears acute rules out the obtuse companion angle 110.4 degrees.
The period is 2 pi / 3. Only the coefficient of x matters, so divide 2 pi by 3 - the amplitude 2 and the phase shift -2 pi leave the period untouched.
The period is 2 pi. Cosecant is the reciprocal of sine, so it inherits sine period of 2 pi, and the amplitude factor 3 has no effect on period at all.
Both sides reduce to tan²A + cos²A. The left gives tan²A + cos²A directly; the right becomes sec²A − sin²A, and sec²A = 1 + tan²A with 1 − sin²A = cos²A closes the gap.
The reference angles are 30°, 45°, π/3 and π/4. Identify the quadrant first, then measure back to the nearest half of the x-axis — one short rule per quadrant.
For x > 0 it equals pi/2 - arctan x; for x < 0 it equals -pi/2 - arctan x. The sign of x decides the constant, because the principal range only spans pi radians.
Convert 120 degrees to 2pi/3 radians, then A = (1/2)(9)(2pi/3) = 3 pi = 9.425 m^2 - exactly one third of the whole circle's area.
The answer is cos(2x) + 5. The co-function shift sin(theta + pi/2) = cos(theta) applies with theta = 2x, and the added 5 just raises the midline.
Answer: sin(9x)tan(12x). Dividing by cot(12x) = cos(12x)/sin(12x) means multiplying by its reciprocal sin(12x)/cos(12x), and that ratio is just tan(12x).
Answer: sin^2(A)/cos(A) - 1. Rewriting cot A as cos A/sin A turns the first quotient into a tangent-like ratio, and csc A times sin A collapses to 1.
Group the Pythagorean pair first: sin^2 + cos^2 = 1, leaving 1 + tan^2, which is exactly the identity for sec^2 theta.
The expression equals sqrt(1 - sin^2 a cos^2 b)/|cos a|. Writing tan a as sin a / cos a puts everything over cos^2 a before the root is taken.
Set the slope length to 1 and eliminate the horizontal distance between the two sightings. Using tan 15 = 2 - sqrt 3 gives tan theta = (6 + 9 sqrt 3)/23, about 0.9386 (theta ≈ 43.2 degrees).
The solutions are 7pi/24, 23pi/24, 31pi/24 and 47pi/24. Substituting t for the whole argument stretches the interval to [pi/4, 17pi/4), which holds four solutions.
The answer is x = pi/6 + 2pi*k or x = 5pi/6 + 2pi*k for any integer k. Sine equals 1/2 twice per revolution, so pi/6 alone is only half the solution set.
Two families solve it: y = π for every x, and sin(x + y/2) = 2 sin(y/2), which needs |sin(y/2)| ≤ 1/2. The y = π branch is easy to lose when cos(y/2) is cancelled.
The answer is x = 6 + 2arcsin(2/3) + 4pi*k and x = 6 + 2pi - 2arcsin(2/3) + 4pi*k. Isolate the sine, then use both branches of the general sine solution.
The general solution is x = -pi/6 + pi k. Isolating gives tan x = -sqrt(3)/3, a tangent value whose reference angle is pi/6, and tangent repeats every pi.
Answer: x = +/- pi/6 + k*pi. Taking square roots gives cos x = +/- sqrt(3)/2, whose four base angles in one turn merge into two families spaced pi apart.
x = n*pi plus or minus arcsin(sqrt((sqrt 145 - 1)/12)), about n*pi +/- 1.2843 radians. Only one root of 6u^2 + u - 6 = 0 lies in the allowed range 0 <= u <= 1.
The solution is x ≈ 440.67. Squaring both sides and applying sec²θ − 1 = tan²θ reduces the equation to x·tan 20° = √25725.16, so x = √25725.16 / tan 20°.
The solution is x ≈ 284.44. The constant under the root is 26.8² + 100² = 10718.24, and dividing its square root by tan 20° = 0.36397 gives x directly.
The solution is x ≈ 345.26, not 345.1. After squaring, sec²20° − 1 = tan²20°, so x = √15791.21 / tan 20°, which is 125.6631 divided by 0.3639702.
Answer: a = 6.32857 degrees (and 172.52555 degrees). Divide by the common factor 89.474, then use the auxiliary-angle form R sin(a + phi) with R = root(1.0001).
Answer: x = sqrt(5)/5, about 0.4472. Taking sine of both sides turns the equation into 2x = sqrt(1-x^2), whose sign condition rules out the negative root.
Answer: x = 1.227111 and x = 5.818835 radians. Sum-to-product turns the left side into 2cos(3x/2 - pi/6)cos(x/2 - pi/6), which has no closed-form roots.
Answer: x = 1.227111 and x = 5.818835 radians. Expand the compound angle, reduce to the quartic 4c^4 + 2c^3 - 4c^2 - 2c + 1 = 0 in c = cos x, and discard extraneous roots.
x = plus or minus arctan(sqrt(-cos t)) + k*pi, and only when cos t <= 0. Since tan^2 x is never negative, a positive cos t leaves the equation with no real root.
The solutions are x = 2π/3 + 2πk and x = 4π/3 + 2πk. Multiplying by sin x gives 2cos²x − cos x − 1 = 0, and the root cos x = 1 must be rejected since sin x ≠ 0.
The answer is x = 3pi/8 + (pi/2)k. Multiplying both sides by 2 turns sin 2x cos 2x into sin 4x, collapsing a product of two functions into one sine equation.
There are four solutions: x = π/6, π/3, 7π/6 and 4π/3. Solve for 2x first over [0, 4π], then halve every angle, which doubles the count compared with sin x.
Answer: sin 2x = +/- 2*sqrt(2)/3. Put u = sin^2 x to get 6u^2 + u - 1 = 0, discard the negative root, then use the identity sin^2 2x = 4 sin^2 x cos^2 x.
Answer: x = pi/4 + k*pi/2 for any integer k. Taking square roots gives sin x = +/- sqrt(2)/2, and the four resulting families merge into one arithmetic sequence.
The solutions on [0, 2 pi) are 0, pi and 3 pi / 2. Factor the equation to sin x (sin x + 1) = 0, then solve sin x = 0 and sin x = -1 separately.
The answer is x = 7pi/6 + 2pi*k or 11pi/6 + 2pi*k. Sine is negative in quadrants three and four, and pi/6 is the reference angle in both of them.
The solutions are sin x = 3/4 and sin x = -1. Use cos 2x = 1 - 2 sin^2 x to turn the equation into 4t^2 + t - 3 = 0, then solve each sine equation.
The answer is theta = pi/4 + k*pi, or 45 + 180k degrees. Tangent repeats every half turn, so both pi/4 and 5pi/4 solve it - one family covers them all.
Answer: y = sin(2x + pi/3). Replacing x by x + pi/12 adds 2 x pi/12 = pi/6 to the phase, so a shift of pi/12 moves the phase by exactly twice that amount.
With alpha = 36 deg, gamma = 50 deg and b = 78.2, the area is about 1380.1. Use K = b^2 sin(alpha) sin(gamma) / (2 sin(beta)) after finding the third angle.
Answer: a = k*pi - pi/4 for any integer k, the smallest positive value being 3pi/4. A shifted sine is odd exactly when its total phase is a multiple of pi.
The equation is false, so it has no solution. Since cos(2 pi) = 1 the right side equals 5, and 10 is not 5, which also follows from the bound 5 cos x is at most 5.
Answer: tan(-pi/6). Both equal -1/root3, because tangent has period pi and is odd; tan(7pi/6) and tan(-5pi/6) are both +1/root3, and cot(5pi/6) is -root3.
Only y = cos(5 pi x / 4) is even. Testing f(-x) shows the sine, tangent and cosecant options all flip sign, so each of those three is odd instead.
Only tan(m) = 2 and sin(m) = -0.9 are possible. The rest fail because sine and cosine live in [-1, 1] or because cot(0), csc(pi) and sec of pi/2 are undefined.
tan(30 degrees) = 1/root 3 = root 3/3, about 0.5774. It comes from sin 30 over cos 30 with the denominator rationalised; root 3 itself is tan(60 degrees).