Math and Arithmetic
Algebra
Calculus

# What equals 1 cos squared x divided by cos squared x?

234 ###### 2012-05-01 00:32:52

1. Anything divided by itself always equals 1.

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## Related Questions 2 x cosine squared x -1 which also equals cos (2x) Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos. Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.) (This is diff of 2 squares because 1 is the square of 1.) And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos squared) - 1. Then from basic trig we know that (sin squared) + (cos squared) = 1, so this is 0.  If x = sin &theta; and y = cos &theta; then: sin&sup2; &theta; + cos&sup2; &theta; = 1 &rarr; x&sup2; + y&sup2; = 1 &rarr; x&sup2; = 1 - y&sup2; 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) [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, Sin squared is equal to 1 - cos squared. sin2 x = (1/2)(1 - cos 2x) cos2 x = (1/2)(1 + cos 2x) Multiplying both you get (1/4) (1 - cos2 2x) Which is equal to (1/4) (1 - (1/2) (1 + cos 4x) = (1/8) (2 - 1 - cos 4x) = (1/8) (1 - cos 4x) Or If it is the trigonomic function, sin squared x and cosine squared x is equal to one  sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos) cos2(theta) = 1 so cos(theta) = &Acirc;&plusmn;1 cos(theta) = -1 =&gt; theta = pi cos(theta) = 1 =&gt; theta = 0 The answer is 1. sin^2 x cos^2/sin^2 x 1/cos^2 cos^2 will be cancelled =1 sin^2 also will be cancelled=1 1/1 = 1 (1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)  2cos2x - cosx -1 = 0 Factor: (2cosx + 1)(cosx - 1) = 0 cosx = {-.5, 1} x = {...0, 120, 240, 360,...} degrees There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos &theta; + b sin &theta; = 8a sin &theta; - b cos &theta; = 5then square both sides of each to get:a&sup2; cos&sup2; &theta; + 2ab cos &theta; sin &theta; + b&sup2; sin&sup2; &theta; = 64a&sup2; sin&sup2; &theta; - 2ab sin &theta; cos &theta; + b&sup2; cos&sup2; &theta; = 25Now add the two together:a&sup2; cos&sup2; &theta; + a&sup2; sin&sup2; &theta; + b&sup2; sin&sup2; &theta; + b&sup2; cos&sup2; &theta; = 89&rarr; a&sup2;(cos&sup2; &theta; + sin&sup2; &theta;) + b&sup2; (sin&sup2; &theta; + cos&sup2; &theta;) = 89using cos&sup2; &theta; + sin&sup2; &theta; = 1&rarr; a&sup2; + b&sup2; = 89 Yes and the answer is 1 squared equals 1  How is it possible that the value of cosecant is less than 1 (2/7)? Rewrite 1/cos x as (cos x)-1 and use chain rule.   This is geometry/trigonometry, not calculus. That said, cos 2a = .3 The property you need: cos 2u = 1 - 2sin2u = 2cos2u - 1 Using the first equality: cos(2a) = 1 - 2sin2a Let u = a (.3) = 1 - 2sin2a Substitute cos(2a) = .3 .7 = sin2a Subtract .3 from both sides, get rid of negative sign sin a = .71/2 Square root both sides a = sin-1 .71/2= .991156 Not a special angle, use calculator.  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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