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What is the cardiod in mathematics?
In polar coordinates, p = 1 - sin(theta)
H ow do you verify that csc theta tan theta sec 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.
How do you solve for x and y in an angle?
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)
Why sin x plus 180 equals - sin x?
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.
2 sin squared x plus sin x minus 1 is equal to 0 solve for x?
2 sin2(x) + sin(x) - 1 = 0(2 sin + 1) (sin - 1) = 0Either 2 sin(x) + 1 = 02sin(x) = -1sin(x) = -0.5x = 210°, 330°or sin(x) - 1 = 0sin(x) = 1x = 90°
What is sin theta minus 2?
It is a mathematical expression.
If cos and theta 0.65 what is the value of sin and theta?
Remember use the Pythagorean Trig/ Identity. Sin^(2)(Theta) + Cos^(2)(Theta) = 1 Algebraically rearrange Sin^(2)(Theta) = 1 - Cos^(2)(Theta) Substitute Sin^(2)(Theta) = 1 - 0.65^(2) Factor Sin^(2)(Theta) = ( 1- 0.65 )( 1 + 0.65) Sin^(2)(Theta) = (0.35)(1.65) Sin^(2)(Theta) = 0.5775 Sin(Theta) = sqrt(0.5775) Sin(Theta) = 0.759934207.... Theta = Sun^(-1)(0.759934207...) Theta = 49.45839813 degrees.
What is sin theta cos theta?
It's 1/2 of sin(2 theta) .
How do you solve 2 sin squared theta equals one fourth?
2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))
How can you verify 1 plus tan theta divided by 1 minus tan theta equals cot theta plus 1 divided by cot theta minus 1?
It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
How do you simplify csc theta minus cot x theta times cos theta plus 1?
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)
What is angle Theta if sine theta 0.03125?
Sin(theta) = 0.03125 Hence theta = ArcSin(0.03125) theta = 1.790784659... degrees.
How do you simplify sin theta times csc theta divided by tan theta?
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).
What does negative sine squared plus cosine squared equal?
-Sin^(2)(Theta) + Cos^(2)Theta => Cos^(2)Theta - Sin^(2)Theta Factor (Cos(Theta) - Sin(Theta))( Cos(Theta) + Sin(Theta)) #Is the Pythagorean factors . Or -Sin^(2)Theta = -(1 - Cos^(2)Theta) = Cos(2)Theta - 1 Substitute Cos^(2)Thetqa - 1 + Cos^(2) Theta = 2Cos^(2)Theta - 1
How do you prove the following equation the quantity of sin theta divided by 1 minus cos theta minus the quantity 1 plus cos theta divided by sin theta equals 0?
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
What is the value of theta if 4sin theta is equal to 2?
4Sin(theta) = 2 Sin(Theta) = 2/4 = 1/2 - 0.5 Theta = Sin^(-1) [0.5] Theta = 30 degrees.
How do you calculate Sin theta equals 13 times sin 32 degrees divided by 8?
sin (theta) = [13* sin (32o)]/8 = 13*0.529919264/8 = 0.861118804 [theta] = sin-1 (0.861118804) [theta] = 59.44o
