Solve the inequality: sin ⁡ θ − tan ⁡ θ &lt; .01

Question

Solve the inequality:  sin  ⁡  θ  −  tan  ⁡  θ  &amp;lt;  .01

kitskjeja · Accepted Answer

The easiest thing to do is use a converging numerical approachlike Newton's Method.

x_{n + 1} = x_{n} - \frac{f (x_{n})}{f * (x_{n})} w h e r e f (x) = \sin (x) - \tan (x) - .01 a n d f^{'} (x) = \cos (x) - \sec^{2} (x)

Just plug an initial guess say "3" into the right hand side ofNewtons formula get the answer on the left and then plug that answer back into the right side. I readthe other answers to this question and I agree that the two locations between zero and 2 Pi can increase by adding multiples of 2Pi all the way to infinity. If we restrict thesolution to the interval

[0, 2 π)

though and use an initial guess of 6you will find your answer. The initial guess of 3 gives the other answer but its not the largest.

Mauricio Mathis · Accepted Answer

sin  ⁡  (  x  )  =  tan  ⁡  (  x  )  sin  ⁡  (  x  )  =            sin      ⁡      (      x      )              cos      ⁡      (      x      )        cos  ⁡  (  x  )  =            sin      ⁡      (      x      )              sin      ⁡      (      x      )        cos  ⁡  (  x  )  =  1  x  =      cos          −      1        ⁡  (  1  )  x  =  (  0  ,  2  π  )  ±  2  π  k (where &#039;k&#039; is any positive integer)The integer &#039;k&#039; can go up to infinity, so your angle can getas high as you want. Regardless, the sine and cosine of theangle (as long as it is a multiple   2  π ) will always be the same, 0 (zero). Therefore, you can have an infinitely large angle, and the valuesof sine and tangent with agree exactly.

Solve the inequality: sin theta - tan theta < .01

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