find sin x/2 Cos x/2 and tan x/2 in the following
1. Tan x = -4/3 x in quad. 2
2. Sin x =1/4 x in quad.2
3. Cos x = -1/3 in quad.3
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41
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{2}
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[3]
coax = -1/3 .... in the 3rd quadrant
now use this formula...
=> cosx = 2cos²(x/2)-1
=> -1/3 = 2cos²(x/2)-1
=> 2cos²(x/2) = 1-1/3 = 2/3
=> cos²(x/2) = 1/3
=> cos(x/2) = ±root(1/3)
=> cos(x/2) = +root(1/3)
(not a possible value)
=> cos(x/2) = -root(1/3) ...eq(1)
now
=> sin²(x/2) + cos²(x/2) = 1
=> sin²(x/2) = 1 - 1/3 = 2/3
=> sin(x/2) = root(2/3) ...eq(2)
=> tan(x/2) = -root2. ___eq(3)
====================
HOPE IT WILL HELP YOU ☺☺
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DEVIL_KING ▄︻̷̿┻̿═━一
HERE IS YOUR ANSWER ☞
======================
___________________
{2}
__________________
[3]
coax = -1/3 .... in the 3rd quadrant
now use this formula...
=> cosx = 2cos²(x/2)-1
=> -1/3 = 2cos²(x/2)-1
=> 2cos²(x/2) = 1-1/3 = 2/3
=> cos²(x/2) = 1/3
=> cos(x/2) = ±root(1/3)
=> cos(x/2) = +root(1/3)
(not a possible value)
=> cos(x/2) = -root(1/3) ...eq(1)
now
=> sin²(x/2) + cos²(x/2) = 1
=> sin²(x/2) = 1 - 1/3 = 2/3
=> sin(x/2) = root(2/3) ...eq(2)
=> tan(x/2) = -root2. ___eq(3)
====================
HOPE IT WILL HELP YOU ☺☺
====================
DEVIL_KING ▄︻̷̿┻̿═━一
sanya55:
sorry
Answered by
4
Answer:
easy question
Step-by-step explanation:
use formula for tanx=2tanx/1-tan^2x
1+tan^2x=sec^2x
secx+1/cosx
and then sin^2x+cos^2x=1
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