Find the numerical value of sin (40) sin (80) + sin(80) sin (160) + sin (160) sin (320). (All angles are in degrees.)
Here is the Web2.0 Calculator version:
sin (40) sin (80) + sin(80) sin (160) + sin (160) sin (320) = 0.7500000000002045 ~~3/4
Multiply through by 2
2sin(40) sin(80) + 2sin(80)sin(160) + 2sin (160)sin(320)
Using ....[ 2sin AsinB ] = cos(A -B) - cos(A + B)
cos ( 40 - 80) - cos(40+ 80) + cos(80 - 160) - cos(80 + 160) + cos(160 - 320) - cos(160 + 320) =
cos(-40) - cos (120) + cos (-80) - cos(240) + cos (160 - 320) - cos (480) =
[cos(480) = cos(240) = cos (120) = -1/2 ]
cos (-40) - (-1/2) + cos(-80) - (-1/2) + cos (-160) - (-1/2) =
cos (40) + cos(80) + cos (200) + 3/2 =
{Using....cos A + cos B = 2[cos ((A + B)/2)] * [ cos( (A - B)/2) ]
2[ cos ((80 + 40)/2) ] * [ cos ( ((80 - 40)/2) ] + cos(200) + 3/2 =
2cos (60) cos (20) + cos (200) + 3/2 =
2(1/2) cos (20) + cos(200) + 3/2 =
cos (20 ) + cos(200) + 3/2 =
cos(20) + cos (180 + 20) + 3/2 =
cos(20) + [ cos180cos20 - sin180sin20 ] + 3/2 =
cos(20) + (-1)cos(20) - (0)sin(20) + 3/2 =
cos (20) - cos(20) + 3/2 =
3/2
But....since we multplied the original identity by 2.....we must divide the result by 2 to get the correct answer =
3 / 4
Find the numerical value of \(\sin(40^\circ)*\sin(80^\circ) + \sin(80^\circ)*\sin(160^\circ) + \sin(160^\circ)*\sin(320^\circ)\)
\(\begin{array}{|rcll|} \hline && \sin(40^\circ)\sin(80^\circ) + \sin(80^\circ)\sin(160^\circ) + \sin(160^\circ)\mathbf{\sin(320^\circ)} \\\\ && \boxed{ \mathbf{\sin(320^\circ)} = \sin(360^\circ-40^\circ) = -\sin(40^\circ) } \\\\ &=& \sin(40^\circ)\sin(80^\circ) + \sin(80^\circ)\sin(160^\circ) - \sin(160^\circ)\sin(40^\circ) \\ &=& \sin(40^\circ)\sin(120^\circ-40^\circ) + \sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ) - \sin(120^\circ+40^\circ)\sin(40^\circ) \\ &=& \sin(40^\circ)\sin(120^\circ-40^\circ) - \sin(120^\circ+40^\circ)\sin(40^\circ)+ \sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ) \\ &=& \sin(40^\circ)\Big( \underbrace{\sin(120^\circ-40^\circ) - \sin(120^\circ+40^\circ)}_{\boxed{\mathbf{\text{Formula:}~2\cos(x)\sin(x) = \sin(x+y)-\sin(x-y)} \\ \sin(120^\circ-40^\circ) - \sin(120^\circ+40^\circ) = -2\cos(120^\circ)\sin(40^\circ) }} \Big)+ \sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ) \\\\ &=& \sin(40^\circ)\Big( -2\mathbf{\cos(120^\circ)}\sin(40^\circ)\Big)+ \sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ) \\\\ && \boxed{ \mathbf{\cos(120^\circ)} = \cos(180^\circ-60^\circ) = -\cos(60^\circ)=-\dfrac{1}{2} } \\\\ &=& \sin(40^\circ)\Big( -2\left(-\dfrac{1}{2}\right)\sin(40^\circ)\Big)+ \sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ) \\ &=& \sin^2(40^\circ)+\underbrace{\sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ)}_{ \boxed{ \mathbf{\text{Formula:}~-2\sin(x-y)\sin(x+y) = \cos(2x)-\cos(2y)}\\ \sin(120^\circ-40^\circ)\sin(120^\circ+40^\circ)=-\dfrac{1}{2}\Big(\cos(240^\circ)-\cos(80^\circ) \Big) } } \\\\ &=& \sin^2(40^\circ)-\dfrac{1}{2}\Big(\mathbf{\cos(240^\circ)-\cos(80^\circ)} \Big) \\ && \boxed{ \mathbf{\cos(240^\circ)} = \cos(180^\circ+60^\circ) = -\cos(60^\circ)=-\dfrac{1}{2}\\ \cos(80^\circ) = \cos(2*40^\circ) = 2\cos^2(40^\circ)-1 } \\\\ &=& \sin^2(40^\circ)-\dfrac{1}{2} \Bigg( -\dfrac{1}{2} -\Big( 2\cos^2(40^\circ)-1\Big) \Bigg) \\ &=& \sin^2(40^\circ)-\dfrac{1}{2} \Big( -\dfrac{1}{2} - 2\cos^2(40^\circ)+1 \Big) \\ &=& \sin^2(40^\circ)+\dfrac{1}{4} + \cos^2(40^\circ)-\dfrac{1}{2} \\ &=& \underbrace{\sin^2(40^\circ) + \cos^2(40^\circ)}_{=1} +\dfrac{1}{4}-\dfrac{1}{2} \\ &=& 1 +\dfrac{1}{4}-\dfrac{1}{2} \\\\ &=& \dfrac{1}{4}+\dfrac{1}{2} \\\\ &=& \mathbf{ \dfrac{3}{4} } \\ \hline \end{array}\)