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# What does cos squared x - Sin squared x equal?

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###### 2011-04-14 15:39:36

2 x cosine squared x -1 which also equals cos (2x)

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## Related Questions

Sin squared, cos squared...you removed the x in the equation.

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

Yes. Except where sin x = 0, because then you would be dividing by zero so the quotient is undefined.

The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x

(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x

What do you mean sin squared times x? sin ^2 x means one and only one thing, (sin x) ^2 same as sin ^ n x = (sin x) ^n and cos ^n x = (cos x)^n you need to know (and why): sin ^2 x + cos ^2 x = 1 cos 2x = 1 - 2sin ^2 x (= cos ^2 x - sin ^2 x) so sin ^2 x + cos (2x) - cos x = sin ^2 x + 1 - 2 sin ^2 x - cos x = 1 - sin ^2 x - cos x = cos ^2 x - cos x = cos x(cos x - 1)

It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.

No. Cos squared x is not the same as cos x squared. Cos squared x means cos (x) times cos (x) Cos x squared means cos (x squared)

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)

(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)

The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).

Yes. 'sin2x + cos2x = 1' is one of the most basic identities in trigonometry.

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)

Cos2(x-1) is equal to: 1/2 * (1 + Cos(2 - 2x)) (Cos(x) * Cos(1) - Sin(x) * Sin(1))2 1/4 * (2 + e2i - 2ix + e2ix - 2i) where e is the natural log and i is the imaginary unit.

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

if tan x = cos x then sin x / cos x = cos x =&gt; sin x = cos x cos x =&gt; sin x = cos2 x =&gt; sin x = 1 - sin2x =&gt; sin2x + sin x - 1 = 0 Using the quadratic formula =&gt; 1. sin x = 0.61803398874989484820458683436564 =&gt; x = sin-1 (0.61803398874989484820458683436564) or =&gt; 2. sin x = -1.6180339887498948482045868343656 =&gt; x = sin-1 (-1.6180339887498948482045868343656)

1 - 2cos2(x) and also 2sin2(x) - 1 Take your choice. Use whichever one is more convenient.

There are two ways to solve for the double angle formulas in trigonometry. The first is to use the angle addition formulas for sine and cosine. * sin(a + b) = sin(a)cos(b) + cos(a)sin(b) * cos(a + b) = cos(a)cos(b) - sin(a)sin(b) if a = b, then * sin(2a) = sin(a)cos(a) + cos(a)sin(a) = 2sin(a)cos(a) * cos(2a) = cos2(a) - sin2(b) The cooler way to solve for the double angle formulas is to use Euler's identity. eix = cos(x) + i*sin(x). Yes, that is "i" as in imaginary number. we we put 2x in for x, we get * e2ix = cos(2x) + i*sin(2x) This is the same as * (eix)2 = cos(2x) + i*sin(2x) We can substitute our original equation back in for eix. * (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x) We can distribute the squared term. * cos2(x) + i*sin(x)cos(x) + i*sin(x)cos(x) + (i*sin(x))2 = cos(2x) + i*sin(2x) And simplify. Because i is SQRT(-1), the i squared term becomes negative. * cos2(x) + 2i*sin(x)cos(x) - sin2(x) = cos(2x) + i*sin(2x) * cos2(x) - sin2(x) + 2i*sin(x)cos(x) = cos(2x) + i*sin(2x) Now you can plainly see both formulas in the equation arranged quite nicely. I don't yet know how to get rid of the i, but I'm working on it.

cos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan x

(tan x - sin x)/(tan x sin x) = (tan x sin x)/(tan x + sin x)[sin x/cos x) - sin x]/[(sin x/cos x)sin x] =? [(sin x/cos x)sin x]/[sin x/cos x) + sin x][(sin x - sin x cos x)/cos x]/(sin2 x/cos x) =? (sin2 x/cos x)/[(sin x + sin x cos x)/cos x)(sin x - sin x cos x)/sin2 x =? sin2 x/(sin x + sin x cos x)[sin x(1 - cos x)]/sin2 x =? sin2 x/[sin x(1 + cos x)(1 - cos x)/sin x =? sin x/(1 + cos x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[(1 + cos x)(1 - cos x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - cos2 x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - (1 - sin2 x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/sin2 x(1 - cos x)/sin x = (1 - cos x)/sin x True

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;

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

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