That expression can't be simplified. If you know how much the angle (theta) is, you can calculate the sine (do it on a calculator), and then subtract 1.
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).
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.
sin (theta) = [13* sin (32o)]/8 = 13*0.529919264/8 = 0.861118804 [theta] = sin-1 (0.861118804) [theta] = 59.44o
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
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
'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
[]=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)