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How do you find vertical asymptotes for trig functions?
Only the cofunctions have asymptotes. Because csc x = 1/sin x, csc x has vertical asymptotes whenever the denominator is equal to 0, or whenever sin x = 0, which are the multiples of pi (0,1,2,3,4,...). For sec x, it's 1/cos x, thus cos x = 0, x = pi/2 + pi*n, where n is a counting number (0,1,2,etc...). cot x = cos x/sin x, thus its vertical asymptotes are the same as those of csc x. If the function is transformed, look at the number in front of x (for example, csc (2x), that number would be 2)), and divide the fundamental asymptotes (above) by that number. The vertical asymptotes of csc (2x) would be (pi/2, 2pi/2, 3pi/2, etc...).
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.
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.
How do you simplify sec x cot x?
sec(x)*cot(x) = (1/cos(x))*(cos(x)/sin(x)) = (1/sin(x)) = csc(x)
What is the derivative of csc x?
The derivative of csc(x) is -cot(x)csc(x).
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
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)
What is the csc of a 74 degree angle?
csc = 1/sin csc (74o) = 1/sin(74o) = 1/0.9613 = 1.0403
What is co-tangent squared?
Start with the identity (sin a)2 + (cos a)2 = 1. Divide both sides by (sin a)2 to get1 + (cot a)2 = (csc a)2. Then subtract 1 from both sides. (cot a)2 = (csc a)2 - 1.
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.
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.
What is csc A if cos A equals 0.6?
cos A=3/5 sin=square root of (1-cos2) sin=square root of (1-3/52) sin=square root of (1-9/25) sin=square root of (16/25) sin=4/5 csc=1/sin csc=1/(4/5) csc=5/4
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.
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.
How do you simplify this expression... cot2x over csc2x-cscx... please note that cot2x and csc2x are really just raised to the second power?
I assume the expression is cot^2 x / ( csc^2 x - csc x) express it in terms of sin x and cos x: =(cos^2 x / sin^2 x) / (1/sin^2 x - 1/sin x) =(cos^2 x / sin^2 x) / [(1 - sin x)/sin^2 x] =cos^2 x / (1 - sin x) = (1 - sin^2 x) / (1 - sin x) = (1 + sin x)(1 - sin x) / (1 - sin x) = 1 + sin x
What is sin cos tan csc sec cot of 30 45 60 degrees?
Sin(30) = 1/2 Sin(45) = root(2)/2 Sin(60) = root(3)/2 Cos(30) = root(3)/2 Cos(45) = root(2)/2 Cos(60) = 1/2 Tan(30) = root(3)/3 Tan(45) = 1 Tan(60) = root(3) Csc(30) = 2 Csc(45) = root(2) Csc(60) = 2root(3)/3 Sec(30) = 2root(3)/3 Sec(45) = root(2) Sec(60) = 2 Cot(30) = root(3) Cot(45) = 1 Cot(60) = root(3)/3
Write the expression in terms of sine and cosine and simplify so that no quotients appear in the final expression. cscx(sinx plus cosx)?
csc(x)*{sin(x) + cos(x)} = csc(x)*sin(x) + csc(x)*cos(x) =1/sin*(x)*sin(x) + 1/sin(x)*cos(x) = 1 + cot(x)
