How to find sin theta and tan theta?
Given that θ is in the first quadrant, we know that the sine, cosine, and tangent are all positive. x = 5cosθ cosθ = x/5 sin²θ + cos²θ = 1 sin²θ = 1 - cos²θ sinθ = √(1 - cos²θ), because sinθ > 0 sinθ = √(1 - x²/25) tanθ = sinθ/cosθ tanθ = √(1 - x²/25) / (x/5) tanθ = (5/x)√(1 - x²/25) tanθ = (1/x)√(25 - x²)
Use the given information to find sin 2 theta, cos 2 theta, and tan 2 theta?
Somewhat formula based. Specifically, I am referring to the following formulas: sin2Θ = 2sinΘcosΘ cos2Θ = (cosΘ)^2 - (sinΘ)^2 tan2Θ = 2tanΘ / (1 - (tanΘ)^2) So, in each case all you got to do is find the value of the triplet (sinΘ, cosΘ, tanΘ). Without actually solving it for you, I will tell you how to find it out. In case 1, both sinΘ and cosΘ are positive. Then cosΘ = [1 - (sinΘ)^2]^(1/2) and tanΘ = sinΘ/cosΘ. In case 2, again, sinΘ, cosΘ > 0. Then sinΘ = [1 - (cosΘ)^2]^(1/2), and tanΘ = sinΘ/cosΘ. In case 3, sinΘ > 0 and cosΘ < 0. Then, let sinΘ = x. This makes cosΘ = -(1 - x^2)^(1/2). Which gives us the equation for tanΘ as x/-(1 - x^2)^(1/2) = -3. Squaring, we get x/(1- x^2) = 9. This gives 9x^2 + x - 1 = 0. You need to use the formula to solve quadratic equations on this one. In case 4, sinΘ < 0 and cosΘ > 0. sinΘ = 1/cscΘ. cosΘ = [1 - (sinΘ)^2]^(1/2) and tanΘ = sinΘ/cosΘ. Read this to find out where the above formulas come from: http://en.wikipedia.org/wiki/List_of_tri... Hope this helps.
Trigonometry: how to find sin(theta) given cos(theta)?
Trig is about the ratio of right triangle side lengths. You have the horizontal side and the vertical side and the "diagonal" side (hypotenuse) from the end of the horizontal side to the end of the vertical side. We know the length of the hypotenuse c is the square root of a^2 plus b^2 where a and b are the lengths of the horizontal and vertical sides respectively. Theta is the angle the hypotenuse makes with the horizontal side. Cosine(theta) is the ratio a/c. Sine(theta) is the ratio b/c. Tangent is the ratio b/a. Cotangent is a/b, secant is c/a, co-secant is c/b. We are told cos(theta) = 1/sqrt(10) = a/c so the horizontal side is 1 and hypotenuse is sqrt(10)=sqrt(a^2+b^2) but a = 1 which means b^2 = 9 and b = 3 Theta = arcos(1/sqrt(10)) = 71.57 degrees in the range 0 to pi (180 degrees). Cos is positive 0 to 90 but negative 90 to 180. 2. If sin(theta) = b/c = 1/sqrt(10) then b = 1, a = 3, c = sqrt(10) Theta = arcsin[1/sqrt(10)] = 18.43 degrees BUT that is between 0 and 90 and we have to find it between 90 and 180 (pi). Imagine the right triangle with vertical side = 1, horizontal side = 3 in the 2nd quadrant (90 to 180 which is pi/2 to pi). The angle is at the origin meaning the horizontal side extends LEFT from the origin 3 units where it makes a right angle with the vertical side of length 1. Thus the angle is 18.43 degrees LESS than 180 = 180-18.43 = 161.57 degrees. Hope this helps
Here, I have taken a small general example to explain you about the doubt let us say tan(theta) being given has value x..Now the respective doubt can be solved by this..
Can you prove that theta= sin-1x=cos-1 in [1-x square=tan-1x/]1-x square?On first inspection, this looked like meaningless gibberish. However, after spending some time in trying to work out what you could possibly mean, I think you have put the ‘]’ in the wrong place: theta= sin-1x=cos-1 in [1-x square]=tan-1x/1-x squareNo, because it isn’t true!Assuming that ‘sin-1x’ is supposed to mean ‘arcsin(x)’ aka ‘sin^-1(x)’, then we have:[math]\theta = \arcsin(x) \Rightarrow \sin(\theta) = x[/math]Using the trigonometric identity [math]\sin^2(\theta) + \cos^2(\theta) = 1[/math], we have:[math]\cos(\theta) = \pm \sqrt{1 - x^2} \Rightarrow \theta = \arccos(\pm \sqrt{1-x^2})[/math][Remember: Except for zero, every number has two square roots]Now [math]\tan(\theta) = \frac {\sin(\theta)}{\cos(\theta)} = \frac {x}{\pm \sqrt{1-x^2}} \Rightarrow \theta = \arctan \left( \frac {\pm x}{\sqrt{1-x^2}} \right)[/math]Assembling this, we have:[math]\theta = \arcsin(x) = \arccos(\sqrt{1-x^2}) = \arctan \left( \frac {\pm x}{\sqrt{1-x^2}} \right)[/math]If I wasn’t able to use LaTeX for formulating maths, then I might write:theta = arcsin(x) = arccos[±√(1 - x^2)] = arctan[±x/(√(1 - x^2))]where x^2 means ‘x squared’ and √(y) means ‘the principal square root of y’[Instead of using ‘arcsin()’ etc, I could use ‘sin^-1() ’ etc, but I dislike this notation as people less familiar with maths might mistake it for being equivalent to ‘1/sin()’ i.e. ‘cosec()’ etc.]If I didn’t know how to write ‘√’ and ‘±’, then:theta = arcsin(x) = arccos[either square root of (1 - x^2)] = arctan[x/(either square root of (1 - x^2))]where the ‘either’ is used to stress that there are two square roots to consider.More long-windedly,theta = arcsin(x) = arccos[either square root of (1 - x squared)] = arctan[x/(either square root of (1 - x squared))]The main difference between this and your question is the presence of square roots.Note how I have used parentheses ‘()’ and brackets ‘[]’ to clearly indicate what I am taking the square roots of, what I am diving ‘x’ by and what I am taking inverse trigonometric functions of.
Given, [math]\cos\theta + \sin\theta = \sqrt{2}[/math]Also, [math]\cos\theta = \sqrt{2} - \sin\theta[/math] ……. (i)Now, we solve [math]\cos\theta + \sin\theta = \sqrt{2},[/math]Squaring both sides,[math](\cos\theta + \sin\theta)^2 = (\sqrt{2})^2[/math][math]\cos^2\theta + \sin^2\theta + 2\cos\theta\sin\theta = 2[/math][math]1 + 2\cos\theta\sin\theta = 2[/math] [Since, [math]\sin^2\theta + \cos^2\theta = 1[/math]][math]1 + 2\cdot(\sqrt{2} - \sin\theta)\cdot\sin\theta = 2[/math] [From (i)][math]1 + 2\sqrt{2}\sin\theta - 2\sin^2\theta = 2[/math]Re-arranging the equation,[math]2\sin^2\theta - 2\sqrt{2}\sin\theta + 1 = 0[/math]([math]\sqrt{2}\sin\theta)^2 - 2\cdot\sqrt{2}\sin\theta\cdot1 + (1)^2 = 0[/math][math](\sqrt{2}\sin\theta - 1)^2 = 0[/math]Square rooting both sides,[math]\sqrt{2}\sin\theta - 1 = 0[/math][math]\sqrt{2}\sin\theta = 1[/math][math]\sin\theta = \frac{1}{\sqrt{2}}[/math][math]\sin\theta = \sin45°[/math] [Since, [math]\sin45° = \frac{1}{\sqrt{2}}][/math]Eliminating [math]\sin[/math] from both sides[math]\theta = 45°[/math]
sinθ=-b±√(b²-4c)/2a=[-(-2)±√{(-2)²-4*1(-1)}]/2={2±√(4+4)}/2={2±√8}/2=(2±2√2)/2=1±√2...reject (+ve) sign ,∵ -1 ≤sinθ ≤ 1 , and (1+√2) ≥ 1∴ sinθ =1–√2≈1–0.414=-0.586<0∵sine of base∠α =sin 24.5°=(0.586)∴ θ is a 3rd or 4th Quadrant angleθ=180+θ or 360-θ=180+24.5 or 360–24.5=204.5° or 335.5° 【IF SPECIFIED 0°≤θ≤360°】〗
If sin theta is 1/2 find cos theta need help with last 4 questions?
1.) sin x = 1/2 where x is in the first quadrant recall sine = opposite / hypotenuse, so opp = 1 and hypotenuse = 2 from pythagorean theorem adjacent side = SQRT[2^2 - 1^2] = SQRT(3) so cos x = adj / hyp = [SQRT(3)] / 2 and is positive because cosine is also positive in the first quadrant 2.) if csc theta is 5/3 find cos theta csc x = 1 / sin x = 5/3 sin x = 3/5, so cos x = plus or minus SQRT[1 - (sin x)^2] = SQRT[1 - (3/5)^2} = 4/5 Since we are in the first quadrant it is positive. 3.) sec theta is 3 find tan theta sec x = 1 / cos x = 3 cos x = 1/3 sin x = plus of minus SQRT[1 - (cos x)^2] = SQRT[1 - (1/3)^2] = (2/3)SQRT(2) Since we are in the first quadrant it is positive. tan x = sinx / cos x = [(2/3)SQRT(2)] / (1/3) = 2 SQRT(2) 4.) sin theta is 4/5 find sec theta sin x = 4/5 cos x = plus or minus SQRT[1 - (sin x)^2] = SQRT[1 - (4/5)^2] = 3/5 sec x = 1 / cos x = 1 / (3/5) = 5/3 Since we are in the first quadrant it is positive.
What is 1/sin^2 theta minus 1/tan^2 theta?
let x = theta 1/sin^2(x) - 1/tan^2(x) = 1/sin^2(x) - 1/(sin^2(x)/cos^2(x)) = 1/sin^2(x) - cos^2(x) / sin^2(x) = (1-cos^2(x))/sin^2(x) = sin^2(x)/sin^2(x) ................... since sin^2(x) + cos^2(x) = 1 = 1