Trigonometry Identities.

  • Thread starter pavadrin
  • Start date
hey i have a problem which im puzzled over. i im given that:
[tex]\sin A = \frac{3}{5}[/tex]
and i am asked to find exact values for:
[tex]\sin 2A[/tex]
[tex]\cos 2A[/tex]
[tex]\tan 2A[/tex]
where [tex]90^0 \leq A \leq 180^0[/tex] (the power to zero is the degree sign)
i have gone about solving this by the use of a pythagorien (?) triple, therefore
[tex]\cos A = \frac{4}{5}[/tex] and
[tex]\tan A = \frac{3}{4}[/tex].
however i do not understand how A can be greator than 90 since this is in a right angle triangle. thanks in advance for those who help,
Pavadrin.
 

VietDao29

Homework Helper
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pavadrin said:
i have gone about solving this by the use of a pythagorien (?) triple, therefore
[tex]\cos A = \frac{4}{5}[/tex] and
[tex]\tan A = \frac{3}{4}[/tex].
however i do not understand how A can be greator than 90 since this is in a right angle triangle. thanks in advance for those who help,
Pavadrin.
The Pythagorean Identity states that: sin2x + cos2x = 1, where x is some angle.
So in this case, we have:
sin2A + cos2A = 1
[tex]\Leftrightarrow \cos ^ 2 A = 1 - \sin ^ 2 A = 1 - \left( \frac{3}{5} \right) ^ 2 = \frac{16}{25}[/tex]
So there will be 2 possible value for cosA:
[tex]\Leftrightarrow \cos A = \pm \frac{4}{5}[/tex], right?
Since (4 / 5)2, and (-4 / 5)2 both return 16 / 25.
So how can we know what the value of cos A is? The problem gives us more information, that the angle A is between 90o, and 180o, i.e in the second quadrant. By looking at the unit circle, can you see what sign cos A takes? Is it positive or negative?
After having cos A, one can use the Double-Angle Formulae to finish the problem:
sin(2A) = 2 sin(A) cos(A)
cos(2A) = cos2(A) - sin2(A) = 1 - 2sin2(A) = 2cos2(A) - 1.
Ok, can you go from here? :)
 
hmmmm.....i see. i think i understand that now, thanks,
Pavadrin
 

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