# Thread: trigonometry (2)

1. ## trigonometry (2)

Show that $\theta=\pi/10$ satisfies the equation $sin2\theta=cos3\theta$ (I know how to do this ). Express this equation in terms of $sin \theta$ and $cos \theta$. (I can do this too ) .

Show that $4sin^2\frac{\pi}{10}+2sin\frac{\pi}{10}-1=0$

2. Originally Posted by thereddevils
Show that $\theta=\pi/10$ satisfies the equation $sin2\theta=cos3\theta$ (I know how to do this ). Express this equation in terms of $sin \theta$ and $cos \theta$. (I can do this too ) .

Show that $4sin^2\frac{\pi}{10}+2sin\frac{\pi}{10}-1=0$
Haha I've got this equation while trying to do your other question

What did you get for $\sin 2\theta=\cos 3\theta$ in terms of sin and cos ?

Something like $2\cos\theta\sin\theta=4\cos^3\theta-3\cos\theta$ (1) isn't it ? (next time, it's better you write the result you got, so that we can check it )

Now, let $\theta=\frac{\pi}{10}$, which is correct, since it satisfies the initial equation.

$\cos\theta\neq 0$, so you can divide both sides of (1) by $\cos\theta$ to get $2\sin \frac{\pi}{10}=4\cos^2\frac{\pi}{10}-3$

Then just use the well-known identity $\cos^2 a+\sin^2 a=1$ to transform $\cos$ into $\sin$

That's all folks !

3. Originally Posted by Moo
Haha I've got this equation while trying to do your other question

What did you get for $\sin 2\theta=\cos 3\theta$ in terms of sin and cos ?

Something like $2\cos\theta\sin\theta=4\cos^3\theta-3\cos\theta$ (1) isn't it ? (next time, it's better you write the result you got, so that we can check it )

Now, let $\theta=\frac{\pi}{10}$, which is correct, since it satisfies the initial equation.

$\cos\theta\neq 0$, so you can divide both sides of (1) by $\cos\theta$ to get $2\sin \frac{\pi}{10}=4\cos^2\frac{\pi}{10}-3$

Then just use the well-known identity $\cos^2 a+\sin^2 a=1$ to transform $\cos$ into $\sin$

That's all folks !

Thanks thank thanks thanks Moo , you are just so nice .. Thanks again

ok i think i understand dy