# trigonometry (1)

• Aug 9th 2009, 01:26 AM
thereddevils
trigonometry (1)
Without using mathematical tables or calculator , show that
$\displaystyle \theta=\frac{\pi}{10}$ satisfies the equation
$\displaystyle \sin2\theta=\cos3\theta$.

Thanks for ur help .
• Aug 9th 2009, 02:24 AM
Moo
Hello,

Consider : $\displaystyle \cos^2 3\theta-\sin^2 2\theta$
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Remember these identities :
$\displaystyle \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

$\displaystyle \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$

Multiply them :

\displaystyle \begin{aligned} \cos(a-b)\cos(a+b) &=\cos^2(a)\cos^2(b)-\sin^2(a)\sin^2(b) \\ &=\cos^2(b)(1-\sin^2(a))-\sin^2(a)\sin^2(b) \\ &=\cos^2(b)-\sin^2(a)\cos^2(b)-\sin^2(a)\sin^2(b) \\ &=\cos^2(b)-\sin^2(a)(\cos^2(b)+\sin^2(b)) \\ &=\boxed{\cos^2(b)-\sin^2(a)} \end{aligned}
------------------------------------------------------

Hence $\displaystyle \cos^2 3\theta-\sin^2 2\theta=\cos \theta \cos 5\theta$

And if $\displaystyle \theta=\frac{\pi}{10}$, then $\displaystyle \cos 5\theta=0$

So $\displaystyle 0=\cos^2 3\theta-\sin^2 2\theta=(\cos 3\theta-\sin 2\theta)(\cos 3\theta+\sin 2\theta)$ (1)

Now notice that :
$\displaystyle 3\theta=\frac{3\pi}{10}$ and $\displaystyle 2\theta=\frac\pi 5$ are both in the first quadrant (because $\displaystyle <\frac\pi 2$)
So $\displaystyle \cos 3\theta>0$ and $\displaystyle \sin 2\theta>0$

Hence $\displaystyle \cos 3\theta+\sin 2\theta>0$

And it follows from (1) that $\displaystyle \boxed{\cos 3\theta-\sin 2\theta=0}$ if $\displaystyle \theta=\frac{\pi}{10}$

Looks clear to you ?
• Aug 9th 2009, 02:34 AM
mr fantastic
Quote:

Originally Posted by thereddevils
Without using mathematical tables or calculator , show that

$\displaystyle \theta=\frac{\pi}{10}$ satisfies the equation
$\displaystyle \sin2\theta=\cos3\theta$.

Thanks for ur help .

Since it's a 'show question' (rather than a 'solve question') the following is perfectly acceptable:

$\displaystyle \cos \left( \frac{3 \pi}{10} \right) = \sin \left(\frac{\pi}{2} - \frac{3 \pi}{10} \right) = \sin \left( \frac{2 \pi}{10} \right)$
• Aug 9th 2009, 03:57 AM
thereddevils
Quote:

Originally Posted by Moo
Hello,

Consider : $\displaystyle \cos^2 3\theta-\sin^2 2\theta$
------------------------------------------------------
Remember these identities :
$\displaystyle \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

$\displaystyle \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$

Multiply them :

\displaystyle \begin{aligned} \cos(a-b)\cos(a+b) &=\cos^2(a)\cos^2(b)-\sin^2(a)\sin^2(b) \\ &=\cos^2(b)(1-\sin^2(a))-\sin^2(a)\sin^2(b) \\ &=\cos^2(b)-\sin^2(a)\cos^2(b)-\sin^2(a)\sin^2(b) \\ &=\cos^2(b)-\sin^2(a)(\cos^2(b)+\sin^2(b)) \\ &=\boxed{\cos^2(b)-\sin^2(a)} \end{aligned}
------------------------------------------------------

Hence $\displaystyle \cos^2 3\theta-\sin^2 2\theta=\cos \theta \cos 5\theta$

And if $\displaystyle \theta=\frac{\pi}{10}$, then $\displaystyle \cos 5\theta=0$

So $\displaystyle 0=\cos^2 3\theta-\sin^2 2\theta=(\cos 3\theta-\sin 2\theta)(\cos 3\theta+\sin 2\theta)$ (1)

Now notice that :
$\displaystyle 3\theta=\frac{3\pi}{10}$ and $\displaystyle 2\theta=\frac\pi 5$ are both in the first quadrant (because $\displaystyle <\frac\pi 2$)
So $\displaystyle \cos 3\theta>0$ and $\displaystyle \sin 2\theta>0$

Hence $\displaystyle \cos 3\theta+\sin 2\theta>0$

And it follows from (1) that $\displaystyle \boxed{\cos 3\theta-\sin 2\theta=0}$ if $\displaystyle \theta=\frac{\pi}{10}$

Looks clear to you ?

What a beautiful solution ! Thanks Moo . Thanks Mr F too , for that simple and easy approach