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Q: Why does cos x equal cos -x?

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Cos^2 x = 1 - sin^2 x

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

You can look up "trigonometric identities" in Wikipedia.Cos(2x), among other things, is equal to (cos x)^2 - (sin x)^2 If you meant cos squared x, or (cos x)^2, that is equal to (1 + cos(2x))/2

One solution. (cos x)2 - 2cos x = 3 Factor: (cos x - 3)(cos x + 1)= 0 cos x = {-1, 3} Solve: For cos x = -1, x = 180 deg No solution for cos x = 3

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.

tan x

Start on the left-hand side. cos(x) + tan(x)sin(x) Put tan(x) in terms of sin(x) and cos(x). cos(x) + [sin(x)/cos(x)]sin(x) Multiply. cos(x) + sin2(x)/cos(x) Make the denominators equal. cos2(x)/cos(x) + sin2(x)/cos(x) Add. [cos2(x) + sin2(x)]/cos(x) Use the Pythagorean Theorem to simplify. 1/cos(x) Since 1/cos(x) is the same as sec(x)- the right-hand side- the proof is complete.

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

Cos(2x) = cos^2(x) - sin^2(x) = 2 cos^2(x) - 1 = 1 - 2 sin^2(x).Source: ChaCha.com

sin2x / (1-cos x) = (1-cos2x) / (1-cos x) = (1-cos x)(1+cos x) / (1-cos x) = (1+cos x) sin2x=1-cos2x as sin2x+cos2x=1 1-cos2x = (1-cos x)(1+cos x) as a2-b2=(a-b)(a+b)

Because the slope of the curve of sin(x) is cos(x). Or, equivalently, the limit of sin(x) over x tends to cos(x) as x tends to zero.

cos(x)-cos(x)sin2(x)=[cos(x)][1-sin2(x)]cos(x)-cos(x)sin2(x)=[cos(x)][cos2(x)]cos(x)-cos(x)sin2(x)=cos3(x)

Cos(x) = Sin(2x) Using angle-addition, we have Sin(a+b) = Sin(a)Cos(b) + Sin(b)Cos(a). From that, we see Sin(2x) = Sin(x)Cos(x)+Sin(x)Cos(x) = 2Sin(x)Cos(x) Cos(x) = 2Sin(x)Cos(x) If Cos(x) = 0, then the two sides are equal. This occurs at x= Pi/2 + nPi, where n is an integer and Pi is approximately 3.14. If Cos(x) doesn't equal 0, then we can divide it out. Then, 1 = 2 Sin(x) , or 1/2 = Sin(x) This occurs when x = Pi/6 or 5Pi/6, plus or minus any multiples of 2 Pi.

because sin(2x) = 2sin(x)cos(x)

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

As a first step, I would convert everything to sines and cosines. sin x cot x = sec x - cos x thus becomes: (sin x) (cos x / sin x) = (1 / cos x) - cos x Simplifying: cos x = 1 / cos x - cos x It doesn't look as though they are equal. In fact, if you do the calculations for some specific angle, e.g. 30 degrees, you see that they are not.

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.

Rewrite sec x as 1/cos x. Then, sec x sin x = (1/cos x)(sin x) = sin x/cos x. By definition, this is equal to tan x.

The values depends on what x is. This function sqrt(2)*cos(x - pi/4) [with x in radians], is equivalent to cos(x) + sin(x). or sqrt(2)*cos(x - 90°) [with x in degrees]

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)

Can you please claify if you mean x=y^2/ pi*cos(x) , or x=y^2/cos(pi), since they are very different sums.

The indefinite integral of sin x is equal to -cos x + C.

sec x = 1/cos x sec x cos x = [1/cos x] [cos x] = 1

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

cos 2x = cos2 x - sin2 x = 2 cos2 x - 1; whence, cos 2x / cos x = 2 cos x - (1 / cos x) = 2 cos x - sec x.