Can you help me with this trig. problem?

I assume your Theta is the same as x above.

There's a formula |sin(x/2)| = sqrt((1-cos(x))/2). This gives |sin(x/2)| = sqrt((1-(7/25))/2) = sqrt(9/25) = 3/5.

Now we must only figure out the correct sign. However, from cos x > 0 and tan x < 0, it follows that x is an angle in the lower right quadrant, and so will be the half angle x/2. So its sine must be negative and the answer is -3/5.

P.S. The formula I used is just a better known formula (sin x)^2 = (1-cos(2x))/2 with x/2 in the place of x and square roots taken on both sides. It's easy to find if you know the formula for cos(2x).

Hope this helps!

sin(θ/2)^2=

=[1-cos(θ)]/2=

=(1-7/25)/2=

=9/25=(3/5)^2

hence

sin(θ/2)=+-3/5,

but since tan(θ)<0 and cos(θ)>0, it must be sin(θ)>0, that is θ€(0,TT) and θ/2€(0,TT/2.

So sin(θ/2)>0 and

sin(θ/2)=3/5