Michael F.

asked • 01/31/18# Turing Trig Expression into Algebraic Expression

I have no idea how to turn this into an algebraic expression, please help ASAP!

Please give me tips on how to do this with other trig to algebraic expressions, as I have four problems I am quite literally clueless on!

sin (cos

^{-1}√2/2 - sin^{-1}2x)The answer key says the answer is:

-2x√2 + √2-8x

^{2}
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## 2 Answers By Expert Tutors

The number sqrt(2)/2 comes up in several places . To make the typing easier I will replace it with .707 and then restore at the end. I am going to limit my analysis to The standard domain restriction on sin

^{-1}. This restricts x to [-.5 , .5] and generates angles in the first and second quadrant. Further, I will take cos^{-1}(.707) = pi/4 (radian measure) and sin(pi/4) = cos(pi/4) = .707 (first quadrant values)With all of this, the expression is sin( pi/4 -sin

^{-1}(2 x)) This is the sine of a difference, so the trig angle difference formula can be used. The result is sin(pi/4) cos(sin^{-1}(2 x)) - cos(pi/4) sin(sin^{-1}(2 x)) . This evaluates to.707 [ cos(sin

^{-1}(2 x) - sin(sin^{-1}(2x)) ].With the domain restriction mentioned above sin( sin

^{-1}( 2 x)) = 2 x. cos( sin^{-1}(2 x) is a bit more complicated, but a standard triangle analysis show that it is sqrt(1 - 4 x^{2}) So the expression becomes .707[ sqrt( 1 - 4 x

^{2}) - 2 x ] restoring sqrt(2)/2 gives [sqrt(2)/2] [sqrt(1 - 4 x

^{2}) - 2 x] valid for -.5 < x < .5The graph of this expression is interesting. It is clear that it is the top half of a closed curve. Presumable, the rest of the closed curve could be obtained by properly removing the domain restriction.

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Bobosharif S.

Richard D., you are giving very nice, if not the best, explanation.

Report

01/31/18

Bobosharif S. answered • 01/31/18

Tutor

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Mathematics/Statistics Tutor

If it is written correctly, then

sin (cos

^{-1}√2 - sin^{-1}2x)= [use sin(x-y)-sin(x) cos(y)-cos(x) sin(y)]=sin(cos

^{-1}√2)cos(sin^{-1}2x)-cos(cos^{-1}√2)sin(sin^{-1}2x) (⇒)|sin(sin

^{-1}(x)=x and sin(cos^{-1}(x)=√(1-x^{2}), therefore|(⇒)=√(1-(√2)

^{2})√(1-(2x)^{2})-2√(2) x =**√ (-1)**√(1-4x^{2})-2√2 x==-2√2 x+

*√(1-4x2).***i**## Still looking for help? Get the right answer, fast.

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Bobosharif S.

^{2}, then maybe you typed something wrong.01/31/18