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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...).
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How do you find the horizontal asymptote of a trig function?
The only trig functions i can think of with horizontal assymptotes are the inverse trig functions. and they go assymptotic for everytime the non-inverse function is equal to zero.
How do you subtract trig functions?
basic algebra, group like terms Sin30-Sin1=Sin(30-2)=Sin(28)
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It is used in hyperbolic functions; it's the rule to change a normal trig function into hyperbolic trig function. Example: cos(x-y) = cosx cosy + sinx siny Cosh(x-y) = coshx coshy - sinhx sinhy Whenever you have a multiplication of sin, you write the hyperbolic version as sinh but change the sign. also applied when: tanxsinx (sinx)^2 etc... Hope this helps you
Sine divided by cosine?
Trig identity... sin/cos = tangent
