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# 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.

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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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
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