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Cos^-1 t = sin^-1 t Solve please!? Trigonometry?

Cos^-1 t = sin^-1 t Solve please!? Trigonometry? Topic: Cos^-1 t = sin^-1 t Solve please!? Trigonometry?
June 16, 2019 / By Clarice
Question: 1. Solve cos-1 t = sin-1 t My book says to take the cosine of both sides, but it's not working out. 2. 2tan-1(x)=tan-1(1/4x) The negative ones mean the inverse function. Please explain all steps. 1 hour ago - 4 days left to answer.
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Best Answers: Cos^-1 t = sin^-1 t Solve please!? Trigonometry?

Beatie Beatie | 3 days ago
When dealing with inverses, it is best to draw an accompanying triangle. cos^-1(t) = sin^-1(t) cos(cos^-1(t)) = cos(sin^-1(t)) t = cos(sin^-1(t)) Now focus on sin^-1(t), this means that sin(θ) = t. Draw a right triangle. Label one of the angles θ. The opposite side of that angle is t and the hypotenuse is 1. Then by the pythagorean theorem, you get the adjacent side to be √(1 - t^2). Since you drew a triangle and sin^-1(t) refers to θ in that triangle, then cos(sin^-1(t)) = cos(θ) = √(1 - t^2). Now you get: t = √(1 - t^2) Square both sides to get: t^2 = 1 - t^2 2t^2 = 1 t^2 = 1/2 t = ±1/√2 Because funny things can happen when when you square square roots (it has to do with range/domain restrictions on functions) you need to check your answers for extraneous solutions. You'll see that the only solution that works is 1/√2, that is the correct answer. 2tan^-1(x) = tan^-1(x/4) tan(2tan^-1(x)) = tan(tan^-1(x/4)) x/4 = tan(2tan^-1(x)) Just like the previous problem, draw a triangle. Remember that tan^-1(x) is the same as tan(θ) = x. Draw your right triangle, the opposite side will be x, the adjacent side will be 1, and the hypotenuse will be √(x^2 + 1). Since tan^-1(x) is the same as θ, you can think of tan(2tan^-1(x)) as tan(2θ). The formula for tan(2θ): tan(2θ) = 2tan(θ)/(1 - tan^2(θ)) Based upon the triangle we have, tan(θ) = x, now just replace tan(θ) with x to get: tan(2θ) = 2x/(1 - x^2) Since tan(2θ) is the same as tan(2tan^-1(x)) and tan(tan^-1(x)) = x/4, you get: x/4 = 2x/(1 - x^2) Now just solve for x: x(1 - x^2) = 8x x - x^3 = 8x x^3 - 7x = 0 x(x^2 - 7) = 0 x = 0, ±√7 Again you need to check your answers for extraneous solutions. You'll find that x = 0 is the only solution that works. I'm assuming that 1/4x meant 1/4*x or x/4 and not 1/(4x) or 1 over 4x. Now if you meant 1 over 4x you'd get: 1/(4x) = 2x/(1 - x^2) 1 - x^2 = 8x^2 9x^2 - 1 = 0 (3x - 1)(3x + 1) = 0 x = ±1/3 In this example both 1/3 and -1/3 both work. So decide which case is yours and go with that example.
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Beatie Originally Answered: How do you solve the trigonometry problem?
Assuming degrees for the measure, rather than radians... A+B+C = 180 A+B = 90 if A = x and B = x^2 then x + x^2 = 90 so solve for x x^2 + x - 90 = 0 x = (-1 ± √(361))/2 x = 9 or - 10 Ignore the -10 so A = 9 degrees and B = 81 degrees
Beatie Originally Answered: How do you solve the trigonometry problem?
Since all triangles must equal 180 degrees x + x^2 + 90 = 180 x^2 + x -90 = 0 (x + 10)(x - 9) = 0 x = -10, 9 -10 can be thrown out as impossible x = 9 x^2 = 81

Adelphie Adelphie
acos(t) = asin(t) ; Immediately I think 45° but we'll solve it real way. cos(acos(x)) = x, similar to how e^(ln(x)) = x. cos • acos(t) = cos • asin(t) ; function chaining both sides t = cos • asin(t); # to change the arcsine into an arccosine use √(1-t^2) # this gives alternate position on unit circle t = cos • acos[√(1-t^2)]; asin turned into an acos, now compatible for inversing t = √(1-t^2); t^2 = 1-t^2; 2(t^2) = 1; t^2 = (1/2); t = √2/2; which matches 45° =) ***************** That 2 in front of atan(x) confounds applying tan directly, doesn't it... We will need to use identities to transform that left hand side to something we can take tan of. These ARE tricky problems, 3 steps with pattern recognition needed, but we can do it! ;) Let me look up the tangent half/double-angle formula lol. http://www.sosmath.com/trig/douangl/douangl.html Okay. 2atan(x) is twice the angle of 1atan(x) so... 2atan(x) = 1atan[2x/(1-x^2)]. atan[2x/(1-x^2)] = atan(1/4x); <-- now we can inverse both sides! tan • atan[2x/(1-x^2)] = tan • atan(1/4x) ; [2x/(1-x^2)] = (1/4x) ; [8x^2/(1-x^2)] = 1; 8x^2 = (1-x^2); 9x^2 = 1; x^2 = (1/9); x = ±1/3.
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Teige Teige
question - one million: sin^2(x) = one million - cos^2(x) [via identification] = {one million + cos(x)}*{one million - cos(x)} [using a^2 - b^2 = (a+b)*(a-b)] So, left area = one million - [{one million + cos(x)}*{one million - cos(x)}]/{one million - cos(x)} = one million - one million - cos(x) [{one million - cos(x)} cancels with one yet another] = - cos(x) = precise area subsequently one million - ((sin^2 x)/(one million - cos x)) = - cos x (Proved) question - 2: Left area: [tan(x) + tan(y)]/[cot(x) + cot(y)] = Multiplying numerator and denominator via {tan(x)*tan(y)}, [tan(x) + tan(y)]/[cot(x) + cot(y)] = = {tan(x)*tan(y)}[tan(x) + tan(y)]/{tan(x)*tan(y)}[cot(x) + cot(y)] = = {tan(x)*tan(y)}[tan(x) + tan(y)]/[tan(x) + tan(y)] {in view that tan(x)*cot(x) = tan(y)*cot(y)=one million} This simplifies to tan(x)*tan(y) = precise area [Proved]
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Teige Originally Answered: Solve the trigonometry problem?
4 sin^2 A = 1 + 6 cos^2 A 10sin²A=1+6sin²A+6cos²A 10 sin²A=1+6 sin²A=7/10 sinA=±√(7/10) A=asin(±√7/10) A=56.789° A=-56.789°

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