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Q: What is the solution to cos tan-sin over cot equals 0?

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Prove that tan(x)sin(x) = sec(x)-cos(x) tan(x)sin(x) = [sin(x) / cos (x)] sin(x) = sin2(x) / cos(x) = [1-cos2(x)] / cos(x) = 1/cos(x) - cos2(x)/ cos(x) = sec(x)-cos(x) Q.E.D

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sec(x)=1/cos(x), by definition of secant.

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.

cos x - 0.5 = 0 ⇒ cos x = 0.5 ⇒ x = 2nπ ± π/3

Sin[x] = Cos[x] + (1/3)

A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.

cos-1(0.2874) = 73.29763833

I'm not really sure what you mean by "the solution", but that equation cos = sec - sintan does simplify down to sin^2 + cos^2 = 1 which so happens to be an identity. I'm not sure if that's what you're looking for, but if it is, here are the steps in simplifying it. 1. Convert sec to 1/cos 2. Convert tan into sin/cos and multiply it by sin sintan = sin(sin/cos) = (sin^2)/cos You then have cos = 1/cos - (sin^2/cos) 3. Multiply everything by cos cos^2 = 1 - sin^2 4. And finally, send the sin^2 over to the left side by adding it (since it is being subracted on the right) You should see this sin^2 + cos^2 = 1 which is an identity.

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

cos pi over four equals the square root of 2 over 2 This value can be found by looking at a unit circle. Cos indicates it is the x value of the point pi/4 which is (square root 2 over 2, square root 2 over 2)

No, but cos(-x) = cos(x), because the cosine function is an even function.

It would be 1 over square root 5.

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

7.66

0.766

1. Anything divided by itself always equals 1.

cos x equals cos -x because cos is an even function. An even function f is such that f(x) = f(-x).

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If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²

The fisrt thing to note is that there is no variable in the question and so there cannot be a solution set. The only possibilities are the statement is true, false or indeterminate. Now, 2 cos 0 - 1 = 0 is equivalent to 2*1 - 1 = 0 [since cos(0) = 1] or 2 - 1 = 0 which is false.

To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED

y = 2 sin 3x y' = 2(sin 3x)'(3x)' y' = 2(cos 3x)(3) y' = 6 cos 3x

sec(x) = 2 so cos(x) = 1/2 and so x = pi/3

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)