Andy C. answered 10/01/17
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I will use T to represent the angle Theta.
The first statement is proven to be true as follows:
1 - sin T * cos T * tan T =
1 - sin T * cos T * sin T/ cos T =
1 - sin T* sin T = <--- cosines cancel
1 - sin T^2 =
cos ^2 T <--- trig identity sin^2 + cos^2 = 1 , which means 1 - sin^2 = cos^2
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The second one is false.
Let T=30 = pi/6
sin 30 = 1/2
cos 30 = sqrt3/2
tan 30 = sqrt(3)/3
cos(2T) = cos 60 = 1/2
Left side is 1/2*sqrt(3)/3 = sqrt(3)/6
Just for the fun of it, we can find where
the identity holds for specific values of T.
Using identities, the left side becomes:
cosT^2 * sinT/cosT = cosT*sinT
In order for this identity to hold:
cosT*sinT = sinT
cosT * sin T - sin T = 0
sin T (cos T - 1) = 0
sin T = 0 or cos T - 1 = 0
sin T = 0 ---> T is an integral multple of 2*pi.
That is 2*k*pi where K is an integer.
Specifically ..., -3*pi, -2*pi, -pi, 0, pi, 2*pi, 3*pi, ...
is where the sine function is zero.
cos T - 1 = 0 ---> cos T = 1
The cosine is 1 on EVEN multiples of pi,
namely ..., -4*pi, -2*pi, 0, 2*pi, 4*pi, ...
combining these two solutions produces
integral multiples of pi.