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How do you simplify sin theta times csc theta divided by tan theta?
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Answered 2010-11-23 13:32:33
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).
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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 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
How do you simplify csc theta -cot theta cos theta?
For a start, try converting everything to sines and cosines.
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 cos theta multiplied by csc theta?
It is cotangent(theta).
How do you simplify csc theta cot theta cos theta?
cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)
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)
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)
Express csc theta in terms of cot theta theta is in quadrant 3?
It is -sqrt(1 + cot^2 theta)
How do you simplify csc theta tan theta?
With all due respect, you don't really want to know howto solve it.You just want the solution.csc(Θ) = 1/sin(Θ)tan(Θ) = sin(Θ)/cos(Θ)csc(Θ) x tan(Θ) = 1/sin(Θ) x sin(Θ)/cos(Θ) = 1/cos(Θ) = sec(Θ)
What is a reciprocal 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 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)
How do you simplify csc theta cot theta?
There are 6 basic trig functions.sin(x) = 1/csc(x)cos(x) = 1/sec(x)tan(x) = sin(x)/cos(x) or 1/cot(x)csc(x) = 1/sin(x)sec(x) = 1/cos(x)cot(x) = cos(x)/sin(x) or 1/tan(x)---- In your problem csc(x)*cot(x) we can simplify csc(x).csc(x) = 1/sin(x)Similarly, cot(x) = cos(x)/sin(x).csc(x)*cot(x) = (1/sin[x])*(cos[x]/sin[x])= cos(x)/sin2(x) = cos(x) * 1/sin2(x)Either of the above answers should work.In general, try converting your trig functions into sine and cosine to make things simpler.
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
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.
Csc x tan x?
If you want to simplify that, it usually helps to express all the trigonometric functions in terms of sines and cosines.
How do you figure this out Sin theta equals -0.0138 so theta equals what?
Depending on your calculator, you should have an arcsin function, which appears as sin^-1. It's usually a 2nd function of the sin key. If you don't have this function, there are many free calculators you can download... just google scientific calculator downloads.Anyway, this inverse function will give you theta when you plug in the value of sin theta. Here's the algebra written out:sin(theta)=-0.0138arcsin(sin(theta))=arcsin(-0.0138)theta=.......The inverse function applied to both sides of the equation "cancels out" the sin function and yields the value of the angle that was originally plugged into the function, in this case theta. You can use this principle to solve for theta for any of the other trig functions:arccos(cos(theta))=thetaarctan(tan(theta))=thetaand so on, but calculators usually only have these three inverse functions, so if you encounter a problem using sec, csc, or cot, you need to rewrite it as cos, sin, or tan.sec=1/coscsc=1/sincot=1/tan
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)
Values of the 6 trigonometric functions?
Sine Theta (sin θ) = opposite/hypotenuse = a/c Cosine Theta (cos θ) = adjacent/hypotenuse = b/c Tangent Theta (tan θ) = opposite/adjacent = a/b Cotangent Theta (cot θ) = adjacent/opposite = b/a Secant Theta (sec θ) = hypotenuse/adjacent = c/b Cosecant Theta (csc θ) = hypotenuse/opposite = c/a You may need to look on the link below for some sample calculations
What does csc 7 equal?
csc(7 degrees) = 8.2055 csc(7 radians) = 1.5221 csc(7 grads) = 9.1129
What is sec squared x times csc x divided by sec squared x plus csc squared x?
Ah, secant, annoying as always. Why don't we use its definition as 1/cos x and csc as 1/sin x? We will do that Also, please write down the equation, there is at least TWO different equations you are talking about. x^n means x to the power of n 1/(sin x) ^2 is csc squared x, it's actually csc x all squared 1/(cos x) ^2 in the same manner.
What is the derivative of csc x?
The derivative of csc(x) is -cot(x)csc(x).
Any interesting way in teaching right angle trigonometry?
Below link may help.You need to memorize the basic formulas and it will be easy.pythagorean theoremc² = a² + b²Sine Theta (sin θ) = opposite/hypotenuse = a/cCosine Theta (cos θ) = adjacent/hypotenuse = b/cTangent Theta (tan θ) = opposite/adjacent = a/bCotangent Theta (cot θ) = adjacent/opposite = b/aSecant Theta (sec θ) = hypotenuse/adjacent = c/bCosecant Theta (csc θ) = hypotenuse/opposite = c/a
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 you verify 1 divided by cos to the second theta minus tan to the second theta equals cos to the second theta plus 1 divided by csc to the second theta?
Verify the identity:1/(cos θ)^2 - (tan θ)^2 = (cos θ)^2 + 1/(csc θ)^21/(cos θ)^2 - (sin θ)^2/(cos θ)^2 = (cos θ)^2 + (sin θ)^2 ?1 - (sin θ)^2/(cos θ)^2 = (cos θ)^2 + (sin θ)^2 ?(cos θ)^2/(cos θ)^2 = 1 ?1 = 1 TrueMethod 21/(cos θ)2 - (tan θ)2 =? (cos θ)2 + 1/(cscθ)21/(cos θ)2-(sinθ)2/(cos θ)2=? (cosθ)2+ sin(θ)21/(cos θ)2[1-sin(θ)2]=? cos(θ)2+sin(θ)21/cos(θ)2(cos(θ)2)=? 11=1 True
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