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Author Topic: Trigonometry Problem  (Read 2658 times)

Offline ashwinkandel

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Trigonometry Problem
« on: September 27, 2011, 05:06:33 am »
Hey guys, How to solve this :
[Right Click the image and click view image to see the enlarged version]
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Offline astarmathsandphysics

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Re: Trigonometry Problem
« Reply #1 on: September 27, 2011, 09:25:36 am »
5 mins
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Offline astarmathsandphysics

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Re: Trigonometry Problem
« Reply #2 on: September 27, 2011, 09:32:38 am »
Here
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Offline ashwinkandel

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Re: Trigonometry Problem
« Reply #3 on: September 27, 2011, 03:26:15 pm »
Thanks  :)
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Offline astarmathsandphysics

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Re: Trigonometry Problem
« Reply #4 on: September 27, 2011, 03:34:05 pm »
There is a page for these problems on my website
All the C2 notes are here
http://astarmathsandphysics.com/a_level_maths_notes/a_level_maths_notes_c2_menu.html
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Offline angel_ak

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Re: Trigonometry Problem
« Reply #5 on: November 13, 2011, 05:52:19 am »
Express cos x + (under root 3) sin x in the form R cos ( x - a),
where R > 0 and 0 < a < 1/2 pie ,
giving the exact values of R and a.

m soryy i din knw how to insert that under root or pie symbol  :(
 reply sooon pleeeeeeeeeeease :(
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Offline angel_ak

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Re: Trigonometry Problem
« Reply #6 on: November 13, 2011, 05:54:05 am »
its  from paper june 2007, p03 question 5
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Offline Arthur Bon Zavi

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Re: Trigonometry Problem
« Reply #7 on: November 13, 2011, 07:25:05 am »
Doing it.
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Offline Arthur Bon Zavi

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Re: Trigonometry Problem
« Reply #8 on: November 13, 2011, 07:59:59 am »
cos x + \sqrt 3 sin x \equiv Rcos (x - a)
= cos x + \sqrt 3 sin x \equiv [(R) (cos x) (cos a)] + [(R) (sin x) (sin a)]

Keeping x = 0, we get :
1 \equiv R cos a

Keeping x = \frac{1}{2}\pi, we get :
\sqrt3 \equiv R sin a

R2 = (\sqrt 3)2 + (1)2
R = 2

There are two equations now :

1 = 2cos a AND \sqrt 3 = 2sin a

Take any one of them and you will get the value of a, which is 600 or \frac {1}{3} \pi.

Therefore the equation is cos x + \sqrt 3 sin x \equiv 2 cos (x - 60)
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