Solve for theta sin θ/2 − cos θ = 0
help please
Check me on this, but one possible answer is 2pi/3.
I don't know the best pedagogical approach to helping, except that I recommend you review your unit circle....
\[\large \sin \frac{\theta}{2}-\cos \theta=\sin \frac{\theta}{2}-\cos 2(\frac{\theta}{2})\] Using your double angle results. Expand/Distribute cos2(theta/2). \[\large \cos 2t=\cos^2 t-\sin^2 t\] Apply that rule to your given question. \[\large \sin\frac{\theta}{2}-(\cos^2 \frac{\theta}{2}-\sin^2 \frac{\theta}{2})\]
Using trigonometric identities: Make cos^2(theta/2) in terms of sine. \[\huge \sin \frac{\theta}{2}-(1-\sin^2 \frac{\theta}{2}-\sin^2 \frac{\theta}{2})=0\] \[\huge \sin \frac{\theta}{2}-1+2\sin^2 \frac{\theta}{2}=0\] Solve it as a quadratic by factorising. \[\huge 2\sin^2 \frac{\theta}{2}+\sin \frac{\theta}{2}-1=0\]
How can I factor if it's still theta/2?
You look at it as if it were x. \frac{\theta}{2} is simply x. And you get to the end before finding theta.
\[\huge (2\sin \frac{\theta}{2}-1)(\sin \frac{\theta}{2}+1)=0\]
That's the factored form.
I think you're afraid of theta/2 because of you think when you multiply, something is going to happen inside the sine. Well fortunately, it doesn't. So theta/2 is not different to theta. You just find theta when you solve for theta/2. \[\sin\frac{\theta}{2}\times \sin \frac{\theta}{2}=\sin^2\frac{\theta}{2}\]
because you think*
Yeah lol! Thank you so much though! Very in depth and helpful explanation ! Can't thank you enough! I get it!! :)
Okay. Now let's continue on: There are two solutions to the above quadratic as you can see. \[\huge \sin \frac{\theta}{2}=\frac{1}{2}\] or \[\huge \sin \frac{\theta}{2}=-1\]
But I chose my angles as 60 180 and 300 and it's still wrong
We haven't finished yet. Using your calculator we can find what theta/2 is equal to. \[\frac{\theta}{2}=30, 150, 270\] Now we find theta. And may I ask, do you have a range to what theta can become?
because theta can just get bigger and bigger without any range. is theta in between the range 0<theta<360?
Oh yeah 0<theta<360
Nice! I got it right! THANKS SO MUCH!
No worries mate.
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