st

Wednesday, July 15, 2020

trigonometry miscellaneous question 2 ncert cbse 11th

trigonometry miscellaneous question 2 ncert cbse 11th

Prove that ( sin3x + sinx ) sinx + (cos3x - cosx) cosx = 0

Using trigonometry formula trigonometry identities
sinx + siny =2sin[(x+y)/2] cos[(x-y)/2]
cosx - cosy = -2sin[(x+y)/2] sin[(x-y)/2]

LHS = {2sin[(3x+x)/2] cos[(3x-x)/2]}sinx +{-2sin[(3x+x)/2] sin[(3x-x)/2]} cosx

=2sin2xcosxsinx - 2sin2xsinxcosx = 0 = RHS


1.Prove that 2cos(pi/13)cos(9pi/13)+cos (3pi/13)+cos(5pi/13) = 0
solution 

2. Prove that ( sin3x + sinx ) sinx + (cos3x - cosx) cosx = 0

solution

3. Prove that (cosx +cosy)^2 + ( sinx - siny )^2 = 4 { cos[(x+y)/2] }^2

solution

4. Prove that (cosx - cosy)^2 + ( sinx - siny )^2 = 4 { sin[(x-y)/2] }^2

solution 



5.Show that sinx +sin3x+ sin5x +sin7x = 4cosx cos2x sin 4x

solution



disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work

No comments:

Post a Comment

please leave your comments

SAT Math Probability: Conditional Probability, Two-Way Tables, Formulas and Practice

  SAT Math Probability: Formulas, Conditional Probability, Two-Way Tables and Practice Probability on the SAT Math section is often less abo...