Can I please have someone help me with some trigonometry? I would really appreciate it!!?
tan = sin/cos arctan just means that they have given you the sin/cos value of tan already, and they want to know the corresponding radian value. so first, figure out what sin and cos values divide to give you √3. the answer is when sin=√3/2 and cos=1/2 (you can figure that out by looking at a unit circle.) next, figure out which radian value corresponds to these sin and cos values. The answer is that π/3 = sin√3/2 and cos1/2 so, arctan(√3)=π/3. However, we're looking for -√3, so we need the value on the other side of the y axis, which, if you look at a unit circle, you'll see is 2π/3. Therefore, arctan(-√3)=2π/3. same drill for tan θ: what values of sin and cos divide to equal -1? Well, if you remember reciprocals, for instance 1/2*2/1=1, this should give you a hint... (√2/2)*(2/√2)=1 So you know that the sin and cos values will be a variation of √2/2. However, since it's a negative 1, one of the values must be negative. So looking at a unit circle, the answer can be either 3π/4 or 7π/4, because for both those terms we have one negative √2/2 and one positive √2/2, which = -1. Last one: (cos(x)-(√2/2))((sec(x)-1)=0 (in picture) as you can see, many values would work for x, as long as at least one item in parentheses equals 0, since anything*0=zero. Sorry, trig is really hard to explain in typing! I hope I helped a bit, so sorry!
Trigonometry HELP!!! need it ASAP ?
Only one problem per question please. Let x = distance of observer from wall α = angle of elevation to from observer's eye to base of picture θ = angle subtended by observer's eye of picture α + θ = angle of elevation to from observer's eye to top of picture Solve for θ. ___________ We have: tanα = 9/x tan(α+θ) = (9+7)/x = 16/x Then by the tangent addition formula we have: tan(α+θ) = (tanα + tanθ) / [1 - (tanα)(tanθ)] tan(α+θ) [1 - (tanα)(tanθ)] = tanα + tanθ tan(α+θ) - tan(α+θ)(tanα)(tanθ) = tanα + tanθ tan(α+θ) + tanα = tanθ + tan(α+θ)(tanα)(tanθ) tan(α+θ) + tanα = tanθ[1 + tan(α+θ)(tanα)] (tan(α+θ) + tanα) / [1 + tan(α+θ)(tanα)] = tanθ tanθ = (tan(α+θ) + tanα) / [1 + tan(α+θ)(tanα)] Plug in values. tanθ = (16/x + 9/x) / [1 + (16/x)(9/x)] = (16x + 9x) / (x² + 16*9) tanθ = 25x / (x² + 144) θ = arctan[25x / (x² + 144)] You can plug in the values for x and solve with a calculator.
The value for what trigonometric ratios (sin, cos etc.) of acute angles is always less than 1?
Sine and Cosine are the trignomatric ratios, whose values are less that 1 for an acute angle. An acute angle is angle greater than 0 degree but less than 90 degrees. Since both Sine and Cosine has the value 1 at angles 90 degrees and 0 degree respectively. But these angles aren't acute angles. Tangent and Cotangent has value 1 at angle 45 degrees. And Secant and Cosecant don't have their values less than 1 at any angle.
Trigonometry help-Give the ordered pair for the points pi/3 and pi/6 on the unit circle.?
Draw a horizontal line (x-axis) and vertical line (y-axis) so the origin (0,0) coincides with the center of the circle. Draw a circle of unit radius. Take any point P on the unit circle Suppose the coordinates of P are (x,y) Suppose (x,y) are the coordinates of the point P. Take the point Q=(1,0) on the circle. Find the angle that made as you move the ray OQ counter clockwise so it coincides with OP. with the x-axis, Suppose this angle is theta. This is considered the positive angle. You could also reach OP from OQ by going clockwise, the corresponding angle is considered negative. For example going from (1.0) to (0.1) anticlockwise, the angle is 90 degrees or pi/2 radians. Going from (1.0) to (0,1) clockwise, the angle would be -270 degrees or -3pi/2 In fact you can go around the circle a few times (clockwise or counterclockwise) before you land at (0,1), in which case the angle is n(360)+90 or n(-360)-270, 2npi+pi/2 or -2npi-3pi/2 Once you find the theta (positive or negative,with or without multiple rotations of circle), the relation between theta and (x,y) is x=cos(theta) y=sin(theta) So in your case, if theta=pi/3=60 degrees, find cos(60 deg)=1/2 and sin(60 deg)=sqrt(3)/2 So coordinate is (1/2,sqrt(3)/2) if theta=pi/6=30 degrees then cos(30 deg)=sqrt(3)/2, sin(30 deg)=1/2 so corodinate is (sqrt(3)/2,1/2) if theta=(-pi/6)=(-30 deg), i.e., you go clockwise and end up in the fourth quadrant, then x=cos(-30deg)=cos(30 deg)=sqrt(3)/2 but y=sin(-30 deg)=(-sin 30 deg)=-(1/2) Thus the point is (sqrt(3,2),(-1/2)) which does fall in the fourth quadrant Hope this helps. Then the
Give the exact values of the six trigonometric functions of θ?
I'm not entirely sure but if I had to guess I would say put the equation in slope intercept form giving you: y=-3x. This equation is the line going through the origin with a slope of -3. I assume the restriction means you should only look at the portion of the line in the fourth quadrant since that is where x is positive. The slope gives you two side lengths in your triangle and using the pythagorean theorem you can find the third. The only reason x is restricted is to give the quadrant allowing you to determine the sign of each trigonometric function. Using the side lenghts and quadrant you should be able to find all six funcitons.
How do I find the two smallest positive values of sin (30-X) =1/2 and cos (90-X) =-0.571 (separately) using trigonometric properties?
I take it that these are separate problems since they don’t have any values of [math]x[/math] in common.For [math]\displaystyle \sin(30^o - x) = \frac12[/math], [math]\sin[/math] is positive in the first and second quadrants.So [math]30^o - x = 30^o[/math] or [math]30^o - x = 180^o - 30^o = 150^o[/math]Thus, [math]x = 0^o[/math] or [math]x = -120^o[/math] but we want positive values for [math]x[/math] so we have to add [math]360^o[/math] to the second answer. ([math]\displaystyle \sin(-210^o) = \frac12[/math])Hence [math]x = 0^o[/math] or [math]x = 240^o[/math].The second problem is easier since [math]\cos(90^o - x) = sin x = -0.571[/math].Since [math]\sin[/math] is negative in the third and fourth quadrants, [math]x = 180^o + 34.82^o[/math] or [math]x = 360^o = 34.82^o[/math]So [math]x = 214.82^o[/math] or [math]x = 325.18^o[/math]
Precalculus/ Trig Help!!! PLEASE!!?
Asin(6 x) cos(10 x) - cos(6 x) sin(10 x) = -0.15 sin(6x-10x) = -0.15 sin -4x = -0.15 -sin 4x = -0.15 sin 4x = 0.15 4x = sin^-1 0.15 x = (1/4)(sin^-1 0.15) B) A sin(x + φ) = - 6 sin x + 3 cos x A (sin x cos φ + cos x sin φ) = - 6 sin x + 3 cos x A cos φ = -6 A sin φ = 3 A^2 sin^2 φ + cos φ = (-6)^2 + 3^2 A^2 = 45 A = sqrt 45 = 3 sqrt 5 cos φ = - 2 / sqrt 5 φ = cos^-1(-2/sqrt 5) what quadrant is φ in? cos φ < 0 sin φ >0 φ is in QII the standard definition of cos^-1 will work. C) you can either use the steps shown above to find φ. max will be when x + φ = pi/2 x = pi/2 - φ an alternate way to find φ first find: -6.7 sin(x) + 5.1 cos(x) = A sin(x+φ) -6.7 sin(-φ) + 5.1 cos(-φ) = A sin(0) = 0 5.1 cos -φ = -6.7 sin -φ tan -φ = 5.1/6.7 -φ = tan^-1 5.1/6.7 x = pi/2 + tan^-1 5.1/6.7 D) same as above, except... A sin x + φ = 1 sin x + φ = 1/A x + φ = sin^-1 (1/A) x = sin^-1 (1/A) - φ
What is the least value of 2^sin^2x+2^cos^2 x?
The least value of the expression[math]2^{\sin^2x} + 2^{\cos^2x}[/math]is to be found.First, the trick answer. The expression does not favour [math]\sin{x}[/math] or [math]\cos{x}[/math] (the coefficients of both are unity, and both are squared). Further, the two are basically the same functions (one is a shifted version of the other): if one achieves an extreme value, the other is zero. Hence, the function is maximum when the two are equal.[math]\sin{x} = \cos{x}[/math]When this happens, they are equal to [math]\pm\frac{1}{\sqrt{2}}[/math]. Hence, the maximum value of the expression is easily found.[math]M = 2^{\left(\frac{1}{\sqrt{2}}\right)^2} + 2^{\left(\frac{1}{\sqrt{2}}\right)^2}[/math][math]M = 2^{\frac{1}{2}} + 2^{\frac{1}{2}}[/math][math]M = 2\times2^{\frac{1}{2}}[/math][math]M = 2\sqrt{2}[/math]A more rigorous method would require the inequality involving the arithmetic and geometric means of two positive quantities. You may guess that a trigonometric identity may also have to be used somewhere.Specifically, the inequality is as follows.[math]a,b\in\mathbb{R}^{+}\implies\cfrac{a + b}{2}\geq\sqrt{ab}[/math]Apply the above inequality on [math]a = 2^{\sin^2x}[/math] and [math]b = 2^{\cos^2x}[/math].[math]\cfrac{2^{\sin^2x} + 2^{\cos^2x}}{2}\geq\sqrt{2^{\sin^2x}2^{\cos^2x}}[/math][math]\implies2^{\sin^2x} + 2^{\cos^2x}\geq2\sqrt{2^{\sin^2x}2^{\cos^2x}}[/math]Hence, the maximum value can be found.[math]M = 2\sqrt{2^{\sin^2x}2^{\cos^2x}}[/math][math]M = 2\sqrt{2^{\sin^2x + \cos^2x}}[/math][math]M = 2\sqrt{2^{1}}[/math][math]M = 2\sqrt{2}[/math]