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What is csc theta when sin theta equals two-third with theta in quadrant one?
Wiki User
∙ 12y ago
Your question is CscΘ=? when SinΘ=2/3 in Q1
well bases on the fundamental identities.. Sin Θ= 1/CscΘ and CscΘ= 1/SinΘ
So when your in is 2/3
CscΘ=1/sinΘ
CscΘ =1/(2/3)
CscΘ = 3/2 -The value of cscΘ and it is positive because all functions in quadrant 1 are positive..
If you have more questions, please comment..:))
Wiki User
∙ 12y ago
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How do you simplify csc theta -cot theta cos theta?
For a start, try converting everything to sines and cosines.
How do you simplify cos theta times csc theta divided by tan theta?
'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
By using trigonometric identities find the value of sin A if tan A equals a half?
If tan A = 1/2, then sin A = ? We use the Pythagorean identity 1 + cot2 A = csc2 A to find csc A, and then the reciprocal identity sin A = 1/csc A to find sin A. tan A = 1/2 (since tan A is positive, A is in the first or the third quadrant) cot A = 1/tan A = 1/(1/2) = 2 1 + cot2 A = csc2 A 1 + (2)2 = csc2 A 5 = csc2 A √5 = csc A (when A is in the first quadrant) 1/√5 = sin A √5/5 = sin A If A is in the third quadrant, then sin A = -√5/5.
How do you get the second derivative of g of x equals xcscx where x equals theta?
T=theta so that it will not look so messy. g(T)=TcscT To find the first derivative, you must use the product rule. Product rule is derivative of the first times the second, plus the first times the derivative of the second, which will give you: g'(T)=0xcscT + Tx-cscTcotT, which simplifies: g'(T)= -cscTxcotT Now, take the derivative of that to get the second derivatice. In order to do that, you have to do the product rule again. g"(T)=(cscTcotT)cotT + -cscT(-csc^2T) {that's csc squared} which simplifies: g"(T)= cscTcot^2(T) + csc^3 (T)
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)
Express csc theta in terms of cot theta theta is in quadrant 3?
It is -sqrt(1 + cot^2 theta)
How do you get the csc theta given tan theta in quadrant 1?
If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)
What is csc theta?
That depends on the value of the angle, theta. csc is short for "cosecans", and is the reciprocal of the sine. That is, csc theta = 1 / sin theta.
What is the exact value of cos theta if csc theta -4 with theta in quadrant III?
csc θ = 1/sin θ → sin θ = -1/4 cos² θ + sin² θ = 1 → cos θ = ± √(1 - sin² θ) = ± √(1 - ¼²) = ± √(1- 1/16) = ± √(15/16) = ± (√15)/4 In Quadrant III both cos and sin are negative → cos θ= -(√15)/4
Verify that Cos theta cot theta plus sin theta equals csc theta?
It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)
What is cos theta multiplied by csc theta?
It is cotangent(theta).
Why does Sine Theta equal Sine 180 minus Theta?
When you subtract theta from 180 ( if theta is between 90 degrees and 180 degrees) you will get the reference angle of theta; the results of sine theta and sine of its reference angle will be the same and only the sign will be different depends on which quadrant the angle is located. Ex. 150 degrees' reference angle will be 30 degrees (180-150) sin150=1/2 (2nd quadrant); sin30=1/2 (1st quadrant) 1st quadrant: all trig functions are positive 2nd: sine and csc are positive 3rd: tangent and cot are positive 4th: cosine and secant are positive
How do i simplify csc theta divided by sec theta?
By converting cosecants and secants to the equivalent sine and cosine functions. For example, csc theta is the same as 1 / sin thetha.
How do you simplify csc theta -cot theta cos theta?
For a start, try converting everything to sines and cosines.
How do you simplify cos theta times csc theta divided by tan theta?
'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
What is a trig function?
sin theta and csc theta are reciprocal functions because sin = y/r and csc = r/y you use the same 2 sides of a triangle, but you use the reciprocal.
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).
