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  1. H

    How to solve this trigonometry equation

    Homework Statement ##\sin a + \cos b## = ##\frac{-1}{2}## ##\cos a + \sin b## = ##\frac{\sqrt 3}{2}## 0 < a < ##\pi/2## ##\pi/2## < b < ##\pi## a + b = ? By calculating sin (a+b) Homework Equations The Attempt at a Solution I tried : ##\sin a + \cos b =...
  2. H

    Find the angle between 2 vectors

    When i examined it carefully. When i realized you substituted the ##\vec a ##. ##\vec {2b}## of #15 into |##\vec a##| |##\vec 2b##|cos##\theta## and resulted the formula in #11. It makes senses...
  3. H

    Find the angle between 2 vectors

    In vector addition for the law of cosines, we define c with a and b. So, it supposed to be ##\vec c = \vec a + \vec b## right? a+ b is head of vec a meets tail of vec b right? Or, when you said that ##\vec c = \vec a - \vec b##. Do you mean that is when head of vec a meets the tail of vec b?
  4. H

    Find the angle between 2 vectors

    Why the law of cosinus is ##c^2 = a^2 + b^2 - 2 ab \cos \theta## How can we get conclusion like that?
  5. H

    Find the angle between 2 vectors

    Yes. But not only negative. My way give wrong value for ##\theta##. It gives cos##\theta## = ##\frac{-12}{8\sqrt 3}##...
  6. H

    Find the angle between 2 vectors

    I meant something like this Can i say that |a+b|^2 = |a|^2 + |b|^2 - 2|a||b|cos##\theta## |a-2b|^2 = |a|^2 + |2b|^2 - 2|a||2b|(-1)cos##\theta##
  7. H

    Find the angle between 2 vectors

    Because i thought finding angle between ##\vec a## and ##\vec { -2b}## i can find angle between a and b. Since 180 - angle(a-2b) = angle(a+b) under the condition, if it's true that ##\vec b## has same degree with regards to x-axis as ##\vec {-2b}## just different direction going left or right...
  8. H

    Find the angle between 2 vectors

    But it still makes more sense to me. Because the |a| |b| and |a-2b| can be seen as triangle right? |a| is a lenght of vector a with magnitude 2 right? Although, i find that expanding (a-2b)^2 also can be the way. Maybe if being elaborated, you mean that : (a-2b)•(a-2b) = |a-2b|.|a-2b|cos0 Thus...
  9. H

    Find the angle between 2 vectors

    Homework Statement |a| = 2 |b| = ## \sqrt3## |a - 2b| = 2 Angle between a and b Homework Equations The Attempt at a Solution ##\theta## is angle between a and b So angle between a and -2b is 180-##\theta## [/B] ##|a-2b|^2## = |a|^2 + |2b|^2 -2|a||2b|cos(180-##\theta##) ##2^2## = 2^2 +...
  10. H

    What is the necessary and sufficient condition?

    Yes. I made mistake. It supposed to be x^2 - 3x - 10 < 0
  11. H

    What is the necessary and sufficient condition?

    Is the necessary condition a>=4 or 0<a<=4 ? Both is possible, i think
  12. H

    What is the necessary and sufficient condition?

    Hm yes. Then, i am confused how to get 0 < a<= 3 itself.. How i get it is when x=-2, a must be 4 When x = 5, a must be 3 When x = 2, a must be 0 The range for a must be between 0 and 4
  13. H

    What is the necessary and sufficient condition?

    If i plug a = 3. I get x = 5. But, the range for x^2-3x-10<0 is x less than 5. Not including 5
  14. H

    What is the necessary and sufficient condition?

    I don't get it why is 0 < a ≤ 3? Why including 3?
  15. H

    What is the necessary and sufficient condition?

    Sorry again.. the value of x for x^2 -3x -10<0 is -2<x<5 When -2<x<2 a value is 0<a<4 When 2<=x<5 0<=a<3 But i cannot decide what is the sufficient and necessary conditions My answer now, sufficient 0<=a<3
  16. H

    What is the necessary and sufficient condition?

    Sorry. I forgot to correct the up side of question. The equation is x^2 - 3x -10 < 0
  17. H

    What is the necessary and sufficient condition?

    Homework Statement |x-2| < a is a necessary condition for x^2 -3x -10 < 10 . What is the range value of a? |x-2| < a is a sufficient condition for x^2 -3x -10 < 10. What is the range value of a?[/B] The options are a>= 4 a>=3 0<a<=2 0<a<=3 0<a<=5 The Attempt at a Solution Range of x for x^2...
  18. H

    Find the limit as h --> 0 for this trigonometery equation

    I did with differentiation right? I want to know different way to solve it, without L Hospital..
  19. H

    Find the limit as h --> 0 for this trigonometery equation

    I want to solve it without derivation..
  20. H

    Find the limit as h --> 0 for this trigonometery equation

    I get your point.. By L Hospital : And applying chain rule ##\lim_{h \to 0} \frac{\cos(x-2h) - \cos (x + h)}{\sin (x + 3h) - \sin(x - h)} ## ##\lim_{h \to 0} \frac{-2 . -\sin(x-2h) + \sin (x + h)}{3. \cos (x + 3h) + \cos(x - h)} ## ##\frac{2\sin(x) + sin(x)}{3\cos (x) + \cos(x)}##...
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