2018年4月27日 · When you see a trigonometric function such as sine (or sin(#theta#)) or cosine (or cos(#theta#)), it refers the point on the circumference of the circle that intersects the line coming from the origin at a given angle (#theta#) counter-clockwise from the axis between Quadrant I and Quadrant IV of the coordinate plane.

2016年4月11日 · Apply the trig identity: #cos 2a = 1 - 2sin^2 a# #cos ((2pi)/12) = cos (pi/6) = sqrt3/2 = 1 - 2sin^2 (pi/12)# ...

2015年4月15日 · sin(pi-x) = sin pi cosx -cos pi sinx color(white)"sssssssssss" =(0) cos x - (-1)sinx color(white)"sssssssssss" =sinx Yes.

2018年1月11日 · sin(pi/6) = 1/2 Start with an equilateral triangle of side 2. The interior angle at each vertex must be pi/3 since 6 such angles make up a complete 2pi circle. Then bisect the triangle through a vertex and the middle of the opposite side, dividing it into two right angled triangles. These will have sides of length 2, 1 and sqrt(2^2-1^2) = sqrt(3). The interior angles of each right angled ...

2016年2月11日 · sqrt(2)/2 In the trigonometric circle pi/4 is the bisectrix between 0 and pi/2, where x=y. By the Pythogoras theorem we know that x^2+y^2=1. If you don't know the trigonometric circle, you can see that if one of the small angles of a rectangle triangle is pi/4, the other will be also pi/4, which implies the catets are equal. So if x=y, you will have x^2+y^2=1 2y^2=1 y^2=1/2 y=sqrt(1/2)=sqrt(2 ...

2016年7月2日 · sin(pi/5)=sqrt(10-2sqrt5)/4 Let theta=pi/5, then 5theta=pi and 3theta=pi-2theta. Note theta) is an acute angle. Hence sin3theta=sin(pi-2theta) but as sin(pi-A)=sinA This can be written as sin3theta=sin2theta expanding them or 3sintheta-4sin^3theta=2sinthetacostheta as theta=pi/5 we have sintheta!=0 and dividing by it we get 3-4sin^2theta=2costheta or 3-4(1-cos^2theta)=2costheta or 4cos^2theta ...

2016年7月31日 · How do you evaluate #sin(45)cos(15)+cos(45)sin(15)#? How do you write #cos75cos35+sin75sin 35# as a single trigonometric function? How do you prove that #cos(x-y) = cosxcosy + sinxsiny#?

2016年8月16日 · #sin(x- pi/2)# is just #sinx# translated by #(pi/2,0)# We see that #sinx#: graph{sinx [-4.006, 4.006, -2.003, 2.003]} #therefore sin(x-pi/2) #: graph{sin(x- pi/2) [-4.006, 4.006, -2.003, 2.003]} By observing this new graph, we see that this is just #cosx# reflected in the #x# axis. Hence just: #=> -cosx #

2017年2月27日 · sin (pi/8) = sqrt(2 - sqrt2)/2 Use trig identity: 2sin^2 a = 1 - cos 2a In this case, a = pi/8 and 2a = pi ...

2015年8月8日 · Sin #(5pi)/6# = sin #(pi- pi/6)# = sin #pi/6# = sin 30 = #1/2# Another way to think about it is to draw the angle in a Unit circle and create the "new" triangle in Quadrant II. Drop a perpendicular to the x-axis and you will have the correct triangle to use.