How is it possible that the value of cosecant is less than 1 (2/7)?
It helps to convert this kind of equation into one that has only sines and cosines, by using the basic definitions of the other functions in terms of sines and cosines. sin x / (1 - cos x) = csc x + cot x sin x / (1 - cos x) = 1 / sin x + cos x / sin x Now it should be easy to do some simplifications: sin x / (1 - cos x) = (1 + cos x) / sin x Multiply both sides by 1 + cos x: sin x (1 + cos) / ((1 - cos x)(1 + cos x)) = (1 + cos x)2 / sin x sin x (1 + cos) / (1 - cos2x) = (1 + cos x)2 / sin x sin x (1 + cos) / sin2x = (1 + cos x)2 / sin x sin x (1 + cos x) / sin x = (1 + cos x)2 1 + cos x = (1 + cos x)2 1 = 1 + cos x cos x = 0 So, cos x can be pi/2, 3 pi / 2, etc. In some of the simplifications, I divided by a factor that might be equal to zero; this has to be considered separately. For example, what if sin x = 0? Check whether this is a solution to the original equation.
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
No. Cosine, along with sec, is an even function. The odd functions are sin, tan, csc, and cot. The reason for this is because is you take the opposite of the y-value for the cosine function, the overall value of the function is not affected.Take, for example, cos(60 degrees), which equals POSITIVE 1/2.If you flip it over the x-axis, making the y's negative, it becomes cos(-60 degrees), or cos(300 degrees). This equals POSITIVE 1/2.Now let's look at an odd function. For example, sin(30 degrees) equals POSITIVE 1/2. Now take the opposite of this.sin(-30 degrees), or sin(330 degrees), equals NEGATIVE 1/2. This is because it is found in the fourth quadrant, where the y's are negative. Sine of theta, by definition, is y divided by r. If y is negative, sine is negative.
csc(x) = 1/sin(x) = +/- 1/sqrt(1-cos^2(x))
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.
'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
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)
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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.
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
It helps to convert this kind of equation into one that has only sines and cosines, by using the basic definitions of the other functions in terms of sines and cosines. sin x / (1 - cos x) = csc x + cot x sin x / (1 - cos x) = 1 / sin x + cos x / sin x Now it should be easy to do some simplifications: sin x / (1 - cos x) = (1 + cos x) / sin x Multiply both sides by 1 + cos x: sin x (1 + cos) / ((1 - cos x)(1 + cos x)) = (1 + cos x)2 / sin x sin x (1 + cos) / (1 - cos2x) = (1 + cos x)2 / sin x sin x (1 + cos) / sin2x = (1 + cos x)2 / sin x sin x (1 + cos x) / sin x = (1 + cos x)2 1 + cos x = (1 + cos x)2 1 = 1 + cos x cos x = 0 So, cos x can be pi/2, 3 pi / 2, etc. In some of the simplifications, I divided by a factor that might be equal to zero; this has to be considered separately. For example, what if sin x = 0? Check whether this is a solution to the original equation.
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
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.
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tan^2(x) Proof: cos^2(x)+sin^2(x)=1 (Modified Pythagorean theorem) sin^2(x)=1-cos^2(x) (Property of subtraction) cos^2(x)-1/cos^2(x)=? sin^2(x)/cos^2(x)=? (Property of substitution) sin(x)/cos(x) * sin(x)/cos(x) = tan(x) * tan(x) (Definition of tanget) = tan^2(x)
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...).