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Find the numerical value of sin (40) sin (80) + sin(80) sin (160) + sin (160) sin (320).  (All angles are in degrees.)

Dec 16, 2019

#1
+23562
0

Here is the Web2.0 Calculator version:

sin (40) sin (80) + sin(80) sin (160) + sin (160) sin (320) = 0.7500000000002045   ~~3/4

Dec 16, 2019
#2
+111321
+1

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

Dec 16, 2019
edited by CPhill  Dec 16, 2019
#3
+24949
+2

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}$$

Dec 16, 2019