In polar coordinates, p = 1 - sin(theta)
csc[]tan[] = sec[]. L: Change csc[] into one over sin[]. Change tan[] into sin[] over cos[]. R: Change sec[] into one over cos[]. 1/sin[] times sin[]/cos[] = 1/cos[]. L: To multiply 2 fractions, multiply the numerators, and multiply the denominators, and put the numerators' product over the denominators' product. R: Nothing more to do. sin[]/sin[]cos[] = 1/cos[]. L: You have a sin[] on both top and bottom. Cross them off to get a one on the top. 1/cos[] = 1/cos[]. Done. [] is theta. L is the left side of the equation. R is the right side.
If X and Y are sides of a right triangle, R is the hypoteneuse, and theta is the angle at the X-R vertex, then sin(theta) is Y / R and cosine(theta) is X / R. It follows, then, that X is R cosine(theta) and Y is R sin(theta)
sin squared
Sin (theta + 180) is equal to -sin (theta) because the sin function is symmetrically opposite every 180 degrees. Proof: Draw a unit circle, radius 1, centered at the origin (x=0, y=0). Pick any point on that circle, and draw a line from that point through the origin and to the opposite edge of the circle. The angle between that line and the x-axis going to the right is theta. It ranges from 0 degrees at (x=1, y=0) to 360 degrees coming back to (x=1, y=0) rotating counter-clockwise. (The angle is called theta to avoid confusion with the question's original use of x.) The x and y coordinates of the first point are symmetrically opposite the x and y coordinates of the second point. (If X1 were 0.35, for instance, then X2 would be -0.35.) The same goes for Y. (There are two right triangles, with the hypotenuses equal and two angles equal; therefore the two triangles are the same, just flipped over.) Sin (theta) in a unit circle is defined in trigonometry as y, so sin (theta + 180) is equal to -y, which is the same as -sin (theta). Sin (theta) is actually y divided by hypotenuse or "opposite over hypotenuse" but, since the hypotenuse is 1, that can be ignored - it does not change the answer.
It is a mathematical expression.
It's 1/2 of sin(2 theta) .
2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))
It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
There can be no significant simplicfication if some of the angles are theta and others are x, so assume that all angles are x. [csc(x) - cot(x)]*[cos(x) + 1] =[1/sin(x) - cos(x)/sin(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos2(x)] =1/sin(x)*[sin2(x)] = sin(x)
Since sin(theta) = 1/cosec(theta) the first two terms simply camcel out and you are left with 1 divided by tan(theta), which is cot(theta).
sin (theta) = [13* sin (32o)]/8 = 13*0.529919264/8 = 0.861118804 [theta] = sin-1 (0.861118804) [theta] = 59.44o
For such simplifications, it is usually convenient to convert any trigonometric function that is not sine or cosine, into sine or cosine. In this case, you have: sin theta / sec theta = sin theta / (1/cos theta) = sin theta cos theta.
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
You will have to bear with the angle being represented by x because this browser will not allow characters from other alphabets!sin^2x + cos^2x = 1=> sin^2x = 1 - cos^x = (1 + cosx)(1 - cosx)Divide both sides by sinx (assuming that sinx is not zero).=> sinx = (1 + cosx)(1 - cosx)/sinxDivide both sides by (1 - cosx)=> sinx/(1 - cosx) = (1 + cosx)/sinx=> sinx/(1 - cosx) - (1 + cosx)/sinx = 0
[]=theta 1. sin[]=0.5sin[] Subtract 0.5sin[] from both sides.2. 0.5sin[]=0. Divide both sides by 0.5.3. Sin[] =0.[]=0 or pi (radians)