# Trigonometric Identities

• Jun 17th 2013, 10:05 PM
brosnan123
Trigonometric Identities

Find the value of sin 15o = sin ( π12 )

• Jun 17th 2013, 10:29 PM
Prove It
Re: Trigonometric Identities
You should know that \displaystyle \begin{align*} \sin{(2\theta)} = 2\sin{(\theta)}\cos{(\theta)} = 2\sin{(\theta)}\,\sqrt{1 - \sin^2{(\theta)} } \end{align*}. So

\displaystyle \begin{align*} \sin{ \left( 30^{\circ} \right) } &= 2\sin{ \left( 15^{\circ} \right) } \, \sqrt{ 1 - \sin^2{ \left( 15^{\circ} \right) } } \\ \frac{1}{2} &= 2\sin{ \left( 15^{\circ} \right) } \, \sqrt{ 1 - \sin^2{ \left( 15^{\circ} \right) } } \\ \frac{1}{4} &= \sin{ \left( 15^{\circ} \right) } \, \sqrt{ 1 - \sin^2{ \left( 15^{\circ} \right) } } \\ \left( \frac{1}{4} \right) ^2 &= \left[ \sin{ \left( 15^{\circ} \right) } \, \sqrt{ 1 - \sin^2{ \left( 15^{\circ} \right) } } \right] ^2 \\ \frac{1}{16} &= \sin^2{ \left( 15^{\circ} \right) } \left[ 1 - \sin^2{ \left( 15^{\circ} \right) } \right] \\ \frac{1}{16} &= \sin^2{ \left( 15^{\circ} \right) } - \sin^4{ \left( 15^{\circ} \right) } \\ \sin^4{ \left( 15^{\circ} \right) } - \sin^2{ \left( 15^{\circ} \right) } + \frac{1}{16} &= 0 \\ x^2 - x + \frac{1}{16} &= 0 \textrm{ if we let } x = \sin^2{ \left( 15^{\circ} \right) } \\ x &= \frac{1 \pm \sqrt{ (-1)^2 - 4(1)\left( \frac{1}{16} \right) } }{2(1)} \end{align*}

\displaystyle \begin{align*} x &= \frac{1 \pm \sqrt{ 1 - \frac{1}{4}}}{2} \\ x &= \frac{1 \pm \sqrt{ \frac{3}{4} } }{2} \\ x &= \frac{1 \pm \frac{\sqrt{3}}{2}}{2} \\ x &= \frac{ \frac{2 \pm \sqrt{3}}{2} }{2} \\ x &= \frac{2 \pm \sqrt{3}}{4} \\ \sin^2{ \left( 15^{\circ} \right) } &= \frac{2 \pm \sqrt{3}}{4} \\ \sin{ \left( 15^{\circ} \right) } &= \pm \frac{\sqrt{2 \pm \sqrt{3}}}{2} \end{align*}

When we note that sine is positive in the first quadrant, and check for extraneous solutions which were obtained through squaring, we find that the solution we want is \displaystyle \begin{align*} \sin{ \left( 15^{\circ} \right) } = \frac{\sqrt{2 - \sqrt{3}}}{2} \end{align*}.
• Jun 18th 2013, 02:26 AM
mpx86
Re: Trigonometric Identities
sin 15° = sin(45°-30°)
= sin 45°cos 30° - cos 45°sin 30°
= $$(1/\sqrt 2 *\sqrt 3 /2) - (1/2*1/\sqrt {2)} = (\sqrt 6 - \sqrt 2 )/4$$
• Jun 18th 2013, 05:48 AM
Prove It
Re: Trigonometric Identities
Quote:

Originally Posted by mpx86
sin 15° = sin(45°-30°)
= sin 45°cos 30° - cos 45°sin 30°
= $$(1/\sqrt 2 *\sqrt 3 /2) - (1/2*1/\sqrt {2)} = (\sqrt 6 - \sqrt 2 )/4$$

How DARE you provide such a simple, concise and beautiful solution after mine took 18 lines? :P hahaha. Well played :)