What is cot x sin x simplified?

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โˆ™ 2013-12-16 16:42:35

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To simplify such expressions, it helps to express all trigonometric functions in terms of sines and cosines. That is, convert tan, cot, sec or csc to their equivalent in terms of sin and cos.

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A polynomial of degree zero is a constant term

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Q: What is cot x sin x simplified?
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Verify cot x-180 cot x?

cot x = (cos x) / (sin x) cos (x - 180) = cos x cos 180 + sin x sin 180 = - cos x sin (x - 180) = sin x cos 180 - cos x sin 180 = - sin x cot (x - 180) = (cos (x - 180)) / (sin (x - 180)) = (- cos x) / (- sin x) = (cos x) / (sin x) = cot x

Simplify sinx cotx cosx?

== cot(x)== 1/tan(x) = cos(x)/sin(x) Now substitute cos(x)/sin(x) into the expression, in place of cot(x) So now: sin(x) cot(x) cos(x) = sin(x) cos(x) (cos(x)/sin(x) ) sin(x) cos(x) cos(x)/sin(x) The two sin(x) cancel, leaving you with cos(x) cos(x) Which is the same as cos2(x) So: sin(x) cot(x) cos(x) = cos2(x) ===

How do you simplify sec x cot x?

sec(x)*cot(x) = (1/cos(x))*(cos(x)/sin(x)) = (1/sin(x)) = csc(x)

How do you simplify sec x cot x cos x?

y = sec(x)*cot(x)*cos(x)To solve this trigonometric equation, you need to know these identities:sec(x) = 1/(cos(x))cot(x) = 1/(tan(x)) = (cos(x))/(sin(x))Now substitute these identities into the original equation:y = (1/cos(x))*((cos(x))/(sin(x)))*cos(x)Now cancel out the terms that are similar in the numerator and denominator to leave you with:y = (1/(sin(x)))*cos(x)y = (cos(x))/(sin(x))From the aforementioned known identity, the final simplified trigonometric equation becomes:y = cot(x)

How do you solve csc x sin x equals cos x cot x plus?

Suppose csc(x)*sin(x) = cos(x)*cot(x) + y then, ince csc(x) = 1/sin(x), and cot(x) = cos(x)/sin(x), 1 = cos(x)*cos(x)/sin(x) + y so y = 1 - cos2(x)/sin(x) = 1 - [1 - sin2(x)]/sin(x) = [sin2(x) + sin(x) - 1]/sin(x)

Csc squared divided by cot equals csc x sec. can someone make them equal?

cot(x)=1/tan(x)=1/(sin(x)/cos(x))=cos(x)/sin(x) csc(x)=1/sin(x) sec(x)=1/cos(x) Therefore, (csc(x))2/cot(x)=(1/(sin(x))2)/cot(x)=(1/(sin(x))2)/(cos(x)/sin(x))=(1/(sin(x))2)(sin(x)/cos(x))=(1/sin(x))*(1/cos(x))=csc(x)*sec(x)

Is 2 cot x sin x cos x equals 2 - 2 sin 2 x an identity?

The easiest way to approach this problem is by rewriting the left hand side entirely in terms of sin and cos and then simplifying. To do so, use the fact that cot(x)=cos(x)/sin(x) to get that 2*cot(x)*sin(x)*cos(x)=2*cos(x)/sin(x)*sin(x)*cos(x)=2*cos(x)² Next, we will try to simplify the right hand side by factoring and utilizing the formula cos(x)²+sin(x)²=1 which implies that 1-sin(x)²=cos(x)² 2-2sin(x)²=2*(1-sin(x)²)=2*cos(x)² Since both sides can be simplified to equal the same thing, both sides must always be equal, and the equation 2*cot(x)*sin(x)*cos(x)=2-2sin(x)² must be an identity

What is cot x?

Cot x is 1/tan x or cos x / sin x or +- sqrt cosec^2 x -1

2 cot x plus 1 equals -1?

2 cot(x) + 1 = -1 2 cot(x) = -2 cot(x) = -1 cos(x)/sin(x) = -1 cos(x) = - sin(x) x = 135°, 315°, 495°, ... another one every 180 degrees

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 cot of a 68 degree angle?

The trig identaty of cot(x) is cos(x)/sin(x) so then if we want to evaluate cot (68) deg. we just plug into the identady. so cos(68)/sin(68)=.404

Solution for tan x plus cot x divided by sec x csc x?

(tan x + cot x)/sec x . csc x The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form. Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x The solution is: [(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x) then sin x. sin x + cos x . cos x sin2x+cos2x =1 The answer is 1.

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