Prove without using of calculator, that sin ⁡ ( 40 ∘ ) &lt; 3 7 My attempt.Since sin ⁡ ( 40 ∘ ) = 2 sin ⁡ ( 20 ∘ ) cos ⁡ ( 20 ∘ ) &lt; 2 sin ⁡ ( 20 ∘ ) = 2 sin ⁡ ( 60 ∘ − 40 ∘ ) = 3 cos ⁡ ( 40 ∘ ) − sin ⁡ ( 40 ∘ ) , 2 sin ⁡ ( 40 ∘ ) &lt; 3 cos ⁡ ( 40 ∘ ) .Hence, 4 sin 2 ⁡ ( 40 ∘ ) &lt; 3 cos 2 ⁡ ( 40 ∘ ) = 3 ( 1 − sin 2 ⁡ ( 40 ∘ ) ) 7 sin 2 ⁡ ( 40 ∘ ) &lt; 3 sin ⁡ ( 40 ∘ ) &lt; 3 7

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Prove without using of calculator, that   sin  ⁡  (      40    ∘    )  &amp;lt;            3      7      My attempt.Since  sin  ⁡  (      40    ∘    )  =  2  sin  ⁡  (      20    ∘    )  cos  ⁡  (      20    ∘    )  &amp;lt;  2  sin  ⁡  (      20    ∘    )  =  2  sin  ⁡  (      60    ∘    −      40    ∘    )  =      3    cos  ⁡  (      40    ∘    )  −  sin  ⁡  (      40    ∘    )  ,  2  sin  ⁡  (      40    ∘    )  &amp;lt;      3    cos  ⁡  (      40    ∘    )  .Hence,  4      sin    2    ⁡  (      40    ∘    )  &amp;lt;  3      cos    2    ⁡  (      40    ∘    )  =  3  (  1  −      sin    2    ⁡  (      40    ∘    )  )  7      sin    2    ⁡  (      40    ∘    )  &amp;lt;  3  sin  ⁡  (      40    ∘    )  &amp;lt;            3      7

cerfweddrq · Accepted Answer

We need to prove that            1      −      cos      ⁡              80                  ∘                      2    &amp;lt;      3    7  or  sin  ⁡      10          ∘        &amp;gt;      1    7    .Let   sin  ⁡      10          ∘        =  xThus,  3  x  −  4      x    3    =      1    2  or   f  (  x  )  =  0, where  f  (  x  )  =      x    3    −      3    4    x  +      1    8  and since  f      (          1      7        )    =      1    343    −      3    28    +      1    8    =      57    2744    &amp;gt;  0  ,we are done!Indeed,       f    ′    (  x  )  =  3      x    2    −      3    4    =  3      (    x    −          1      2        )        (    x    +          1      2        )  ,which says that   sin  ⁡      10          ∘       is an unique root of the equation on       (    0    ,          1      2        ]    .

Prove without using of calculator, that sin(40^o)<sqrt(3/7)

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