Given sin alpha and cosine beta...?
Since sin alpha is negative and alpha is in 4th quadrand according to 3pi/2 < alpha < 2pi, cos alpha is going to be positive because cosine in the fourth quadrant is positive..Now.. sin = y/r where 'r' is the hypotenuse sin alpha = -4/5 cos = x/r cos alpha = 3/5 tan = y/x tan alpha = -4/3 Anyways, now lets take the arccos or arcsin of either one. Lets use arccos.. cos alpha = 3/5 arccos = 53.13 alpha = 53.13 Now lets find beta.. Cos is negative in the second and third quadrant, according to pi/2 < beta < pi, beta will be in second quadrant.. cos beta = -5/13 sin beta = 12/13 tan beta = -12/5 Now, we take the arcsin or arccos of either one.. cos beta = -5/13 arccos = 112.61 - 90 = 22.33 beta = 22.33 Now we plug in these numbers.. a) sin ( 53.13 - 22.33 ) = .51204 b) cos ( 53.13 + 22.33 ) = .25106 c) tan ( 53.13 + 22.33 ) = 3.85561
Given [math]\sin (\alpha)+\sin (\beta)=1 , \cos (\alpha)+\cos (\beta)=1[/math], how do you find the value of [math]\sin (\alpha)-\cos (\beta)[/math]?Let [math]x=\sin\alpha[/math] and [math]y=\cos\beta[/math]. We want to find [math]x-y[/math] given[math]x+\sqrt{1-y^2}=1\tag{1}[/math][math]y+\sqrt{1-x^2}=1\tag{2}[/math](Note that while we would normally write, e.g., [math]\sin\beta = \pm\sqrt{1-y^2}[/math], we know that [math]\sin\beta\ge0[/math] because [math]\sin\alpha\le 1[/math].)The symmetry of (1) and (2) makes us suspect that any solution must have [math]x=y[/math], and if that is in fact the case, then [math]x-y=0[/math]. Indeed, two obvious solution pairs are [math]x=y=0[/math] and [math]x=y=1[/math]; we want to confirm that any other solutions (if they exist) also satisfy [math]x=y[/math].Equation (1) leads to[math]\sqrt{1-y^2}=1-x\implies 1-y^2=1-2x+x^2 \implies x^2+y^2=2x\tag{3}[/math]while equation (2) similarly leads to [math]x^2+y^2=2y[/math], confirming that [math]x=y[/math]. Note that we are actually done at this point, because we have confirmed that [math]x-y=\sin\alpha - \cos\beta = 0[/math].For the sake of closure, we’ll also confirm that the only solutions are [math]x=y=0[/math] or [math]x=y=1[/math]: Equation (3) becomes[math]x^2+x^2=2x\implies x^2=x\implies x\in\{0,1\}\tag*{}[/math]
Given that tan(alpha)=3/4 with alpha in quadrant 3 and that cos(beta)=1/4 with beta in quadrant 4. Find the?
Use: cos(A + B) = cos(A)cos(B) - sin(A)sin(B) sin(A - B) = sin(A)cos(B) - cos(A)sin(B). If we construct a 3-4-5 triangle in Quadrant III such that tan(A) = 3/4, we can see that: sin(A) = -3/5 cos(A) = -4/5. Similarly, if we construct a right triangle in Quadrant IV such that cos(B) = 1/4, we see that the other leg has length √(4^2 - 1^2) = √15 units. Thus: sin(B) = √15/4. Finally, using the above formulas: cos(A + B) => cos(A)cos(B) - sin(A)sin(B) = (-4/5)(1/4) - (-3/5)(√15/4) = (-4 + 3√15)/20. sin(A - B) => sin(A)cos(B) - cos(A)sin(B) = (-3/5)(1/4) - (-4/5)(√15/4) = (-3 + 4√15)/20. I hope this helps!
(pi/4) = 45 deg so simplify the equation astan(x-45) = (tan x - tan 45)/(tan x + tan 45)= (tan x -1)/(tan x + 1) = 1/6, or6(tan x -1) = (tan x + 1), or6 tan x - 6 = tan x + 1, or5 tan x = 7, ortan x = 7/5 = 1.4, orx - tan (inverse) 1.4 = 54.46232221 deg.Check: tan (54.46232221 -45) = tan 9.46232221 = 0.166666666 = 1/6x = 54.46232221 deg. and tan x = 1.4.
Given cos(alpha)=5/9 and alpha is in quadrant IV, find the exact value of cos(alpha/2)?
i hope you will be able to see the solution. it wasn't so easy sending this image and yahoo only allows one image. hope it also helped
If you know sin. alpha , how do you find cos. alpha and tan. apha?
You could easily find this using the pythagorean identity: 1. sin^2(a) + cos^2(a) = 1 so cosine alpha is the square root of 1-sin^2(a) Since you are given angle restrictions in part one, the answer is the positive square root of 1-sin^2(a) The tangent of alpha is the sine of alpha divided by the cosine of alpha, both of which you now know. 2. Use the same logic to find cosine, except in this situation if you could visualize the unit circle, the angle position indicates that the cosine would be the negative square root of 1-sin^2(a) for the cosine of alpha. Tangent is found similarly: sine of alpha divided by cosine of alpha.
Given Sin(Theta)= -1/5 and Tan(Theta)<0, find the exact value of Cos(Theta)?
Given: sin θ = -1/5 tan θ < 0 Imagine a right triangle with hypotenuse length 5 and vertical leg length of 1 going downwards. θ would be the angle between the x-axis and the hypotenuse. Using the Pythagorean Theorem, we can find out that the length of the horizontal leg is √((5)² − (-1)²) = √24. Thus, cos θ = adjacent / hypotenuse = √(24)/5. This also satisfies the condition tan θ = (sin θ)/(cos θ) = (-1/5)/(√(24)/5)) = -1/√24 < 0.
Finding tan(2 alpha)?
sin(alpha) = y / r -3/8 = y/r that means y = -3 and r = 8. r^2 = x^2 + y^2 8^2 = x^2 + (-3)^2 64 - 9 = x^2 x^2 = 55 x = sqrt 55 or x = -sqrt 55 alpha is in the third quadrant so x must be negative. Therefore, x = -sqrt55. tan(alpha) = y/x = -3/-sqrt55 = 3sqrt55 / 55. tan(2alpha) = 2tan(alpha) / (1 - tan^2(alpha)) You can finish it...
Hope it helps…
Hi, I can see that people have come up with many different methods like using trigonometric identities like [math]sin^2 ({\theta})+ cos^2 ({\theta})= 1[/math] and then finding out the value of [math]tan {\theta}[/math]I will be explaining this question in a method which I think is the easiest and makes sense. I am going to be using 2 concepts, namely the pythagoras’ theorem and the signs of the 4 quadrants in the cartesian plane.Firstly, according to your question, [math]sin {\theta}= \dfrac{-3}{5}[/math] and [math]{\pi}< {\theta} < \dfrac{3{\pi}} {2}[/math]Now, we can consider a right angled triangle say [math]{\Delta}ABC [/math]where angle[math] ABC= {\theta} [/math]We can ignore the negative sign of [math]sin {\theta}[/math] and find out the value of your desired trigonometric function which is [math]tan {\theta}[/math]By applying pythagoras’ theorem, the unknown side is 4 units which is the side adjacent to angle [math]{\theta}[/math]Now we know that [math]tan {\theta}= \dfrac {opposite side}{adjacent side} [/math]We get [math]tan {\theta}= \dfrac{\pm {3}}{4}[/math]Here, the value of [math]tan {\theta} [/math]can be [math]{\pm}[/math] depending upon which quadrant it is present in.Now we can consider the second part of the question, where [math]{\pi}<{\theta} < \dfrac{3{\pi}}{2}[/math]This just implies that theta lies in the 3rd quadrant, henceIn the third quadrant, tan and cot are postive, you can remember these signs by knowing the acronym All students take chocolates where the initial letters give the positive sign of all functions, sin and its reciprocal, you get the idea…..Hence, your value of [math]tan {\theta}= \dfrac{+3}{4}[/math]If you just use the pythagoras theorem and the signs, you can practically get the answer in 2 lines.