# Trigonometry Identities.

hey i have a problem which im puzzled over. i im given that:
$$\sin A = \frac{3}{5}$$
and i am asked to find exact values for:
$$\sin 2A$$
$$\cos 2A$$
$$\tan 2A$$
where $$90^0 \leq A \leq 180^0$$ (the power to zero is the degree sign)
i have gone about solving this by the use of a pythagorien (?) triple, therefore
$$\cos A = \frac{4}{5}$$ and
$$\tan A = \frac{3}{4}$$.
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,

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#### VietDao29

Homework Helper
i have gone about solving this by the use of a pythagorien (?) triple, therefore
$$\cos A = \frac{4}{5}$$ and
$$\tan A = \frac{3}{4}$$.
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,
The Pythagorean Identity states that: sin2x + cos2x = 1, where x is some angle.
So in this case, we have:
sin2A + cos2A = 1
$$\Leftrightarrow \cos ^ 2 A = 1 - \sin ^ 2 A = 1 - \left( \frac{3}{5} \right) ^ 2 = \frac{16}{25}$$
So there will be 2 possible value for cosA:
$$\Leftrightarrow \cos A = \pm \frac{4}{5}$$, 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,

"Trigonometry Identities."

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