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estimation for sin function


Gaussian approximationnon linear equations solving methods?Inequality for $fracsin xsin y$How find this maximum of the $sin^2theta_1+sin^2theta_2+cdots+sin^2theta_n$Show $fracsin(x)x>cos(x)$ for $0<x<pi$ using the Mean Value TheoremProve inequality $arccos left( fracsin 1-sin x1-x right) leq sqrtfrac1+x+x^23$Impossible inequality : Dottie NumberOn branches of Lambert W functionThe limit of $2dfracsinBig(dfracMtheta_k2Big) cosBig(dfrac12(1+M)theta_kBig)sinBig(dfractheta_k2Big) $Prove that the sum of squares of sin and cos equals 1













0












$begingroup$


We have that $$-a_nsin(sqrtlambda_nt)+b_ncos(sqrtlambda_nt)=sqrta_n^2+b_n^2sin(sqrtlambda_nt+theta_n)$$ for some $theta_nin R$



Then $$|-a_nsin(sqrtlambda_nt)+b_ncos(sqrtlambda_nt)|leqlambda_nsqrta_n^2+b_n^2 |t-t_0|$$

for every $t_0$ belonging to $t_0mid sin(lambda_nt_0+theta_n)=0$.



The question is: how we can get this inequality,and if there is some rules on the difference between $sin$ function that can give this. (I mean if $|sin(lambda_nt+theta_n)-sin(lambda_nt_0+theta_n)|leq lambda_n|t-t_0|$).










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    We have that $$-a_nsin(sqrtlambda_nt)+b_ncos(sqrtlambda_nt)=sqrta_n^2+b_n^2sin(sqrtlambda_nt+theta_n)$$ for some $theta_nin R$



    Then $$|-a_nsin(sqrtlambda_nt)+b_ncos(sqrtlambda_nt)|leqlambda_nsqrta_n^2+b_n^2 |t-t_0|$$

    for every $t_0$ belonging to $t_0mid sin(lambda_nt_0+theta_n)=0$.



    The question is: how we can get this inequality,and if there is some rules on the difference between $sin$ function that can give this. (I mean if $|sin(lambda_nt+theta_n)-sin(lambda_nt_0+theta_n)|leq lambda_n|t-t_0|$).










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      We have that $$-a_nsin(sqrtlambda_nt)+b_ncos(sqrtlambda_nt)=sqrta_n^2+b_n^2sin(sqrtlambda_nt+theta_n)$$ for some $theta_nin R$



      Then $$|-a_nsin(sqrtlambda_nt)+b_ncos(sqrtlambda_nt)|leqlambda_nsqrta_n^2+b_n^2 |t-t_0|$$

      for every $t_0$ belonging to $t_0mid sin(lambda_nt_0+theta_n)=0$.



      The question is: how we can get this inequality,and if there is some rules on the difference between $sin$ function that can give this. (I mean if $|sin(lambda_nt+theta_n)-sin(lambda_nt_0+theta_n)|leq lambda_n|t-t_0|$).










      share|cite|improve this question











      $endgroup$




      We have that $$-a_nsin(sqrtlambda_nt)+b_ncos(sqrtlambda_nt)=sqrta_n^2+b_n^2sin(sqrtlambda_nt+theta_n)$$ for some $theta_nin R$



      Then $$|-a_nsin(sqrtlambda_nt)+b_ncos(sqrtlambda_nt)|leqlambda_nsqrta_n^2+b_n^2 |t-t_0|$$

      for every $t_0$ belonging to $t_0mid sin(lambda_nt_0+theta_n)=0$.



      The question is: how we can get this inequality,and if there is some rules on the difference between $sin$ function that can give this. (I mean if $|sin(lambda_nt+theta_n)-sin(lambda_nt_0+theta_n)|leq lambda_n|t-t_0|$).







      calculus algebra-precalculus trigonometry inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 29 at 9:41









      Dylan

      14.2k31127




      14.2k31127










      asked Mar 28 at 19:42









      i.issai.issa

      143




      143




















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