Question

Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π).
tan θ/2 − sin θ = 0

Answer 1

tan(u/2)=((sin u)/(1+cos u))

Answer 2

\[\large \tan \frac{ \theta }{ 2 }=\frac{ \sin \theta }{ 1+\cos \theta }=\frac{ 1-\cos \theta }{ \sin \theta }\]. the second form is more convenient for this problem.

Answer 3

\[\tan \frac{ \theta }{ 2 }-\sin \theta =0\]\[\tan \frac{ \theta }{ 2 }-\sin 2\left( \frac{ \theta }{ 2 } \right)=0\]\[\frac{ \sin \frac{ \theta }{ 2 } }{ \cos \frac{ \theta }{ 2 } }-2\sin \frac{ \theta }{ 2 }\cos \frac{ \theta }{ 2 }=0\]\[\frac{ \sin \frac{ \theta }{ 2 }-2\sin \frac{ \theta }{ 2 }\cos ^{2}\frac{ \theta }{ 2 } }{ \cos \frac{ \theta }{ 2 } }=0\]\[\frac{ \sin \frac{ \theta }{ 2 }(1-2\cos ^{2}\frac{ \theta }{ 2 }) }{ \cos \frac{ \theta }{ 2 } }=0\]\[\frac{ \sin \frac{ \theta }{ 2 }\cos 2(\frac{ \theta }{ 2 }) }{ \cos \frac{ \theta }{ 2 } }=0\]\[\frac{ \sin \frac{ \theta }{ 2 }\cos \theta }{ \cos \frac{ \theta }{ 2 } }=0\]A fraction equals zero when the numerator is zero, thus\[\sin \frac{ \theta }{ 2 }\cos \theta =0\]I've done more than I should have. Now I think you should finish it.

Answer 4

confused?! :(

Answer 5

Where?

Answer 6

Where to go from there... Trig not my strong suit

Answer 7

Your interval was\[0 \le \theta < 2\pi\]We use this to solve for\[\cos \theta =0\]
You can see from the graph that the cosine graph has zeros at π/2 and 3π/2.

Now divide each side of the given interval by 2 to obtain\[0 \le \frac{ \theta }{ 2 } < \pi \]and use this interval to solve for\[\sin \frac{ \theta }{ 2 }=0\]From the sine graph you will see that this happens at 0 and π. However since π is not included, we only use the 0.

∴ θ = 0, π/2, 3π/2.