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The above expression cannot be expressed in an algebraic form.

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Q: What is this expression as an algebraic expression cos parenthesis arccosx-arcsinx parenthesis?
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How do you solve sin parenthesis cos to the negative 1 parenthesis 2 over 5 closed parenthesis and another closed parenthesis?

Let y = sin(cos-1(2/5)) Suppose x = cos-1(2/5): that is, cos(x) = 2/5 then sin2(x) = 1 - cos2(x) = 1 - 4/25 = 21/25 so that sin(x) = sqrt(21)/5 which gives x = sin-1[sqrt(21)/5] Then y = sin(cos-1(2/5)) = sin(x) : since x = cos-1(2/5) =sin{sin-1[sqrt(21)/5]} = sqrt(21)/5 There will be other solutions that are cyclically related to this one but no range has been given for the solutions.


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What is this expression as the cosine of an angle cos30cos55 plus sin30sin55?

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What is cos theta minus cos theta times sin squared theta?

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What is sin23A minus sin7A upon sin2A plus sin14A if A equals pi upon 21?

Using the identity, sin(X)+sin(Y) = 2*sin[(x+y)/2]*cos[(x-y)/2] the expression becomes {2*sin[(23A-7A)/2]*cos[(23A+7A)/2]}/{2*sin[(2A+14A)/2]*cos[(2A-14A)/2]} = {2*sin(8A)*cos(15A)}/{2*sin(8A)*cos(-6A)} = cos(15A)/cos(-6A)} = cos(15A)/cos(6A)} since cos(-x) = cos(x) When A = pi/21, 15A = 15*pi/21 and 6A = 6*pi/21 = pi - 15pi/21 Therefore, cos(6A) = - cos(15A) and hence the expression = -1.


How do you solve tan parenthesis sec to the negative 1 parenthesis 5 over 2 closed parenthesis and another closed parenthesis?

tan(sec-1(5/2))Start with sec-1(5/2), which is the same as cos-1(2/5). So there is a right triangle, where the side adjacent the angle is 2, and the hypotenuse is 5. Solve for the opposite side: sqrt(5² - 2²) = sqrt(21).Tangent is opposite over adjacent, so the answer is sqrt(21)/2