Nov 6, 2014 · Best Answer. cos(0) = 1. So [cos(0)]^2 = [cos(0)] * [cos(0)] = 1 * 1 = 1 ....!!! CPhill Nov 6, 2014. Post New Answer

Jan 9, 2018 · cos 2^@ ~~ 0.99939083 cos 2^@ ~~ 0.99939083 is an irrational number which is not expressible in terms of real nth roots. Note that the trigonometric functions of any integer multiple of 3^@ are expressible in terms of radicals, but other whole number of degrees do not have any such expressions. We can effectively calculate approximations for the sin or cos of …

Apr 4, 2018 · Using the Pythagorean identity (sin^2theta+cos^2theta=1), the value of sin^2 0+cos^2 0 is 1.

theta = 0 or text{Arc}text{cos}(-1/4) or 2pi - text{Arc}text{cos}(-1/4) Sure you didn't mean 4cos^3 theta \\ ... ? 4 cos ^2 theta - 3 cos theta = 1 That's a quadratic equation in unknown cos theta. 4 cos ^2 theta - 3 cos theta - 1 = 0 It's easily factored, (4 cos theta + 1)(cos theta - 1) = 0 cos theta = -1/4 or cos theta =1 The latter is of course theta = 0 in the range. The former is ...

May 6, 2015 · int_0^ pi sqrt cos^2 xdx= int_0^pi cos x dx This definite integral means the area bounded by the curve y= cos x between x=0 to x= pi. To evaluate this area correctly, divide the interval in two parts 0 to pi/2 and pi/2 to pi = [sinx]_0^(pi/2) + [sin x]_(pi/2)^(pi) The area above x axis would come to be +1 and that below x axis would come to be -1. The sum of the two …

Nov 15, 2014 · 2cos 2 Θ + cos(Θ) - 1 = 0 factor (2cos(Θ) - 1) (cos(Θ) +1 ) = 0. So. setting both factors to 0, we have (2cos(Θ) - 1) = 0 add 1 to both sides. 2cos(Θ) = 1 divide both sides by 2. cos(Θ) = 1/2 and this happens at 60 degrees and at 300 degrees on [0, 360] And setiing the other factor to 0, we have

Sep 19, 2016 · At x=0, f(x) takes the value f(0)(=cos0+cos^2 0=1+1^2=2 Hence, we are seeking a tangent at point (0,2) on the curve. The slope of tangent at x=0 is given by f'(0) As f(x)=cosx+cos^2x, f'(x)=-sinx+2cosx xx (-sinx) and f'(0)=-sin0-2cos0xxsin0=0-2xx1xx0=0 Hence slope of tangent is 0 and it passes through (0,2) Hence, equation of tangent is y-2=0xx ...

Nov 28, 2018 · Best Answer. Find the area enclosed by the graph of the parametric equations \(\begin{align*} x &= 6 \cos t \sin t, \\ y &= 6 \cos^2 t. \end{align*}\)

Mar 29, 2020 · cos 2 (0) + cos 2 (1) + cos 2 (2)..... cos 2 (44) + sin 2 (45) + sin 2 (44) ..... sin 2 (0). Next we make use of a basic trig identity: \(\cos^2\theta + \sin^2\theta = 1\) Given that the two angles are the same. We realize we have 0-44 = 45 total …

Mar 20, 2018 · Approximately 0.44 There are two ways I can think of to solve this. sin^2+cos^2=1 rarr It is like the Pythagorean theorem; they will add up to the hypotenuse squared (think of sin 90 (sin of hypotenuse), which is 1, sin x plus cos x is the sum of the sines of each of the non-right angles). Plug the sin (0.9) in and solve for cosine 0.9^2+cos^2=1 0.81+cos^2=1 cos^2=0.19 …