Trigonometry without a calculator?
as a count of actuality you do no longer choose a $one hundred TI-80 4 calculator to do the trignometry that they coach you in school. The scientific calculator that they have got coded into Microsoft abode windows is sufficient. The graphing calculators do help you already know what the graphs appear as if, yet are not necessary for the widespread public of faculty paintings. or you or your mom can get an elementary $10 TI-30 and that must be sufficient on your math homework. I understand doing an arccos(40 two.7364°) is unreasonable and not employing a calculator, yet lower back, the calculator on your pc or an elementary $10 scientific calculator must be waiting to try this. do no longer place self assurance in calculators, too many stupid human beings do. in case you may furnish me with an occasion out of your text cloth e book the place the calculator it relatively is provided in abode windows is inadequate, i ought to alter my recommendations, yet I have been given with the aid of all my ode (elementary differential equations) and pde (partial differential equations) training with the aid of college with what's such as the TI-30s of at present.
[math]1^{\frac 13}[/math] has three answers. That's the fundamental reason why exponentiation with a non-integral power leads to multiple answers.You are solving [math]x^3=1[/math] when you consider the cube root. Since it is a cubic equation, you should expect there are three roots. If you cube the correct trigonometric answer, you will see that it equals 1, and in fact, the answer is one of the cube roots of unity, and it is denoted as [math]\omega[/math] in mathematics.If you have learnt DeMovire’s theorem, it is easier to understand this is the case.[math]e^\theta i=\cos \theta +i\sin \theta[/math]If you consider the polar coordinates on a complex plane, it is actually [math](1,\theta)[/math].If you multiply two complex numbers together[math](r_1,\theta _1)\times (r_2,\theta_2)=(r_1r_2,\theta_1+\theta_2)[/math]You can prove it using trigonometric identities, and proof is omitted here.Basically, when you solve[math](r^3,3\theta)=(1,0)[/math]which is solving the cubic equation in question.In real numbers r can only be 1, so r is 1 (distance cannot be complex), but[math]\theta=0[/math] or [math]\frac {2\pi}3[/math] or [math]\frac {4\pi}3[/math], since [math]2k\pi[/math] radians is also considered as 0.So the answers to [math]1^{\frac 13}[/math] are [math](1,0),(1,\frac {2\pi}3),(1,\frac {4\pi}3)[/math]and the second answer corresponds to [math]e^{\frac {2\pi}3}[/math], according to our previous discussion about how to turn exponential to polar coordinates.
Solve trig without calculator sin12sin48sin54?
sin(12) * sin(48) * sin(54) => sin(12) * sin(48) * sin(90 - 36) => sin(12) * sin(48) * (sin(90)cos(36) - sin(36)cos(90)) => sin(12) * sin(48) * cos(36) => sin(30 - 18) * sin(30 + 18) * cos(36) => (sin(30)cos(18) - sin(18)cos(30)) * (sin(30)cos(18) + sin(18)cos(30)) * cos(36) => (sin(30)^2 * cos(18)^2 - sin(18)^2 * cos(30)^2) * cos(36) => ((1/4) * cos(18)^2 - sin(18)^2 * (3/4)) * cos(36) => (1/4) * (cos(18)^2 - 3 * sin(18)^2) * (2cos(18)^2 - 1) => (1/4) * (cos(18)^2 - 3 * (1 - cos(18)^2) * (2cos(18)^2 - 1) => (1/4) * (cos(18)^2 - 3 + 3 * cos(18)^2) * (2cos(18)^2 - 1) => (1/4) * (4 * cos(18)^2 - 3) * (2 * cos(18)^2 - 1) => (1/4) * (8 * cos(18)^4 - 6 * cos(18)^2 - 4 * cos(18)^2 + 3) => (1/4) * (8 * cos(18)^4 - 10 * cos(18)^2 + 3) cos(5t) => cos(4t)cos(t) - sin(4t)sin(t) => (2cos(2t)^2 - 1) * cos(t) - 4 * sin(t) * cos(t) * cos(2t) * sin(t) => (2 * (2 * cos(t)^2 - 1)^2 - 1) * cos(t) - 4 * sin(t)^2 * cos(2t) * cos(t) => cos(t) * (2 * (4cos(t)^4 - 4cos(t)^2 + 1) - 1 - 4 * sin(t)^2 * (2cos(t)^2 - 1)) => cos(t) * (8 * cos(t)^4 - 8 * cos(t)^2 + 2 - 1 - 4 * (1 - cos(t)^2) * (2 * cos(t)^2 - 1)) => cos(t) * (8 * cos(t)^4 - 8 * cos(t)^2 + 1 + 4 * (cos(t)^2 - 1) * (2 * cos(t)^2 - 1)) => cos(t) * (8 * cos(t)^4 - 8 * cos(t)^2 + 1 + 4 * (2 * cos(t)^4 - 3 * cos(t)^2 + 1)) => cos(t) * (8 * cos(t)^4 - 8 * cos(t)^2 + 1 + 8 * cos(t)^4 - 12 * cos(t)^2 + 4) => cos(t) * (16 * cos(t)^4 - 20 * cos(t)^2 + 5) t = 18 degrees cos(5 * t) = cos(5 * 18) = cos(90) = 0 cos(18) * (16 * cos(18)^4 - 20 * cos(18)^2 + 5) = 0 We know that cos(18) is not equal to 0 16 * cos(18)^4 - 20 * cos(18)^2 + 5 = 0 cos(18)^2 = (20 +/- sqrt(400 - 4 * 16 * 5)) / (2 * 16) cos(18)^2 = (20 +/- sqrt(400 - 320)) / 32 cos(18)^2 = (20 +/- sqrt(80)) / 32 cos(18)^2 = (20 +/- 4 * sqrt(5)) / 32 cos(18)^2 = (5 +/- sqrt(5)) / 8 Since cos(18) is much closer to 0 than to 1, then it stands to reason that cos(18)^2 = (5 + sqrt(5)) / 8 (1/4) * (8 * cos(18)^4 - 10 * cos(18)^2 + 3) (1/4) * (8 * (1/8)^2 * (5 + sqrt(5))^2 - 10 * (1/8) * (5 + sqrt(5)) + 3) => (1/4) * ((1/8) * (25 + 10 * sqrt(5) + 5) - (1/8) * 10 * (5 + sqrt(5)) + (1/8) * 24) => (1/4) * (1/8) * (30 + 10 * sqrt(5) - 50 - 10 * sqrt(5) + 24) => (1/32) * (54 - 50) => (1/32) * 4 => 1/8
Doing trigonometry without a calculator: cos(-2pi/3)?
All of these "do in your head" trig things come from either on an axis or from a special triangle. ON AXIS quadrantal angles on the unit circle where "r" is always = 1 : recall the trig ratios are sin= y/r, cos=x/r and tan=y/x (know the reciprocals csc,sec and cot) 0°=360°=0π=2π where (1, 0), 90°=π/2 where (0, 1), 180°=π where(-1, 0), and 270° = 3π/2 where (0, -1) SPECIAL TRIANGLES: the signs will change in the different quadrants "All (all trig + in QI) Students (sin and it's reciprocal + in QII) Take (tan and its recip + in QIII) Calculus" (cos and its recip + in QIV) 30-60-90 Triangle sides are 1, √3, and 2 ( in order opposite the angles in order) 30°= π/6 (and all multiples of these i.e. 150°= 5/6π )sin=1/2, cos=√3/2, tan=√3/3 60°= π/3 (all mults. again i.e. 120° = 2/3π) sin=√3/2, cos= 1/2 and tan= √3 45-45-90 Triangle sides are 1, 1, √2 ( again in order) 45°= π/4 (all mults. again i.e. 135°= 3/2π) sin= √2/2, cos= √2/2 and tan= 1 cos (‒ 2π/3) well the" ‒ " means the rotation is backwards (counter clockwise) the n/3 tells me this is a multiple of 60° and 60*2= 120° so rotating from 0 backwards 120° puts me in QIII so cos is negative and the cos of all 60° and multiples of 60° angles = 1/2 so cos(-2π/3) = -1/2 tan(x) = 1/√3= √3/3 tells me the 30-60-90 triangle is involved and the tan 60°= tan (π/6) and its multiples gives this result specific answer. It is "+" so the tan is only positive in QI and QIII. QI= 60° or π/3 QIII= (180 + 60)° or (π + π/3)...............(1π*3 + 1π)/3= 4π/3 .....= 240° or 4π/3 S..|..A __|___....the 4 quadrants and who is positive where ....| T..|..C I hope this helps you.
How do you solve a trig function without using a calculator?
To every angle, θ, there corresponds a ray, starting at the origin and pointing out, like the minute hand on a clock.Pick ANY point, P, on that ray, except O = (0,0). It will have a coordinate, P = (x, y), and a length, r. Notice that x and y can be positive, negative, or 0 but r must be a positive number.It is true that cos(θ) = x/r, sin(θ) = y/r, and tan(θ) = y/x. But wait, there's more! From the point P = (x, y), draw a line to the point X = (x, 0) on the x-axis. The right triangle OXP is called the reference right triangle or usually just the reference triangle and the acute angle POX is called the reference angle. Certain reference angles can be associated with known right triangles. Like the 30-60-90 right triangles can be associated with the triangle with sides 1, √3, and 2 and the 45-45-90 right triangle can be associated with the triangle with sides 1, 1, and √2. Knowing this, you can sometimes find exact values for x, y, and r. Which means you can find exact values for sine, cosine, etc. For instance: tan 330° 330° is in the fourth quadrant. the reference angle is 30°. Draw a right triangle in the 4th quadrant with a reference angle of 30°. In the fourth quadrant x is positive and y is negative. The 30-60-90 right triangle has sides of 1, √3, and 2. In the 4th quadrant this means x = √3, y = -1, and r = 2. So tan 330° = y/x = -1/√3.
I’ve answered hundreds of trig question on Quora without a calculator. I’ll tell you why.A calculator gives you an approximation. An approximation is not the exact answer. It is therefore often not the correct answer. It is an approximation to the correct answer.As I prefer to give correct solutions, I generally solve trigonometric equations without a calculator. I don’t bother to calculate the approximation unless I’m curious.I’ve told you why I don’t use a calculator, but not how.How does I do it? I find Euler’s formula [math]e^{i\theta}=\cos \theta + i \sin \theta[/math] to be the ultimate shortcut to most trig identities. But that really just means I’ve been practicing so long I know most of the useful identities by heart and how to quickly derive the rest. So perhaps the real answer is practice, practice, practice.In math class the questions are usually constructed so that besides straight and right angles, we only get 30,60,90 and 45,45,90 right triangles. These correspond to Pythagorean equations of [math]1+3=4[/math] and [math]1+1=2.[/math] It’s rather pitiful that this is all we ask of our students, but because of the unnecessarily complicated way we’ve constructed Trigonometry it’s about all we can hope for.But as a student it’s useful to know if you’re stuck to try the multiples of [math]30^\circ[/math] and [math]45^\circ,[/math] and of course the exact values of the trig functions of these should be at your fingertips. The calculator often doesn’t help much unless you can recognize [math].707…[/math] is [math]1/\sqrt{2}[/math] and things like that.
How to solve Sin, Cos, Tan without using a calculator?
One way is to use a table. They're in the back of some math books. That's probably not what you want though. The other way only works for a few angles, like 30 degrees, 45 degrees, or 90 degrees. It's based on special triangles, like the 30-60-90 or 45-45-90. Sin is y/r, as you probably know. If you don't, it's the ratio of the y-coordinate to the hypotenuse if you draw a right triangle with the given angle at the origin and one side on the x-axis. Cos is x/r, and tan is y/x. If your angle is 90 degrees, for example, the point will end up right on the y-axis, having an x-coordinate of zero. Therefore, the cosine would be zero as well. If it's 30 degrees or 60 degrees, draw a 30-60-90 triangle. Since the sides of a 30-60-90 triangle are some multiple of 1, the square root of 3, and 2, and you only need ratios, you can just assume they are exactly those. If you want to find sin (30), just use the triangle. If you draw it, the hypotenuse will be 2 and the y-coordinate will be 1. Therefore, the answer is 1/2. 45 degree angles work the same way. If you have a larger angle, such as 150 degrees, you need to use a reference angle: the smallest angle from that to the x-axis. Draw the triangle in the new spot, remembering to include any negative x or y values.
In that case, C would not be undefined, it would be complex. Take the relatively elementary case[math] x^3 - x = 0 [/math]or[math] a=1,b=0,c=-1,d=0 [/math]The discriminant becomes[math]\Delta = 4[/math]So we have that[math]\Delta_0 = 3[/math][math]\Delta_1 = 0[/math][math]C = \sqrt[3]{3i\sqrt{3}}=e^{i\pi/6}\sqrt{3}[/math]where I’ve chosen the plus sign for no particular reason, and written [math]i=e^{i\pi/2}[/math]. I assume that the identity [math]e^{ix}=\cos(x) + i\sin(x)[/math] is familiar - otherwise this gets messy.The first root is therefore[math]x_1 = -\frac{1}{3a}\left(b + C + \frac{\Delta_0}{C}\right)[/math][math] = -\frac{1}{3}\left(\sqrt{3}\cdot e^{i\pi/6} + \sqrt{3}\cdot e^{-i\pi/6}\right) [/math][math] = -\frac{1}{\sqrt{3}}\left(2 \cos(\pi/6)\right) = -1 [/math]The factor [math]\xi = -\frac{1}{2}+i\frac{\sqrt{3}}{2} = e^{2i\pi/3}[/math]. Therefore the next root is[math]x_2 = -\frac{1}{3a}\left(b + \xi C + \frac{\Delta_0}{\xi C}\right) [/math][math] = -\frac{1}{3}\left(\sqrt{3}\cdot e^{2i\pi/3} e^{i\pi/6} + \sqrt{3} e^{-2i\pi/3}e^{-i\pi/6} \right) [/math][math] = -\frac{1}{\sqrt{3}}\left( 2 \cos(5\pi/6)\right) = 1 [/math]And lastly the third root is[math]x_3 = -\frac{1}{3a}\left(b + \xi^2 C + \frac{\Delta_0}{\xi^2 C}\right)[/math][math] = -\frac{1}{3}\left(\sqrt{3}\cdot e^{4i\pi/3} e^{i\pi/6} + \sqrt{3} e^{-4i\pi/3}e^{-i\pi/6} \right) [/math][math] = -\frac{1}{\sqrt{3}}\left( 2 \cos(3\pi/2)\right) = 0 [/math]So the three roots of the equation are -1, 0, and 1 - which of course we would have known immediately by factoring:[math] x^3 - x = x(x^2–1) = x(x+1)(x-1) [/math]Whew. So, this gets messy, and you need to know how to manipulate complex numbers in your sleep, even if all of the roots are real.