Trigonometric Identities for Solving for Exact Values

[pavadrin]

hey i have a problem which I am puzzled over. i I am 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,
Pavadrin.

 

[VietDao29]

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,
Pavadrin.

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 90^o, and 180^o, 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? :)

 

[pavadrin]

hmmmm...i see. i think i understand that now, thanks,
Pavadrin