How to simplify this Cosecant identities?

How to find the value of this equation? I've tried about the multiplication and addition of sine and cosine, but it still useless, and I always ended up with the sin 10 at the end.

#Geometry #HelpMe! #MathProblem

Note by Leonardo Chandra
7 years, 7 months ago

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Comments

Recall the identities

sin(A)+sin(B)=2sin(A+B2) cos(AB2) \sin (A) + \sin (B) = 2 \sin ( \frac {A+B}{2} ) \space \cos ( \frac {A-B}{2} )

cos(A)+cos(B)=2sin(A+B2) sin(AB2) - \cos (A) + \cos (B) = 2 \sin ( \frac {A+B}{2} ) \space \sin ( \frac {A-B}{2} )

cos(A)+cos(B)=2cos(A+B2) cos(AB2) \cos (A) + \cos (B) = 2 \cos ( \frac {A+B}{2} ) \space \cos ( \frac {A-B}{2} )

We have,

1sin(10)+1sin(50)1sin(70) \large \frac {1}{\sin (10^\circ )} + \frac {1}{\sin (50^\circ )} - \frac {1}{\sin (70^\circ )}

=1sin(10)+1sin(50)1cos(20) \large = \frac {1}{\sin (10^\circ )} + \frac {1}{\sin (50^\circ )} - \frac {1}{\cos (20^\circ )}, because sin(A)=cos(90A) \sin (A) = \cos (90^\circ - A)

=2( sin(50)+sin(10) )2sin(50) sin(10)1cos(20) \large = \frac {2 ( \space \sin (50^\circ ) + \sin (10^\circ ) \space ) } { 2 \sin (50^\circ ) \space \sin (10^\circ ) } - \frac {1}{\cos (20^\circ )}

=2(2sin(30)cos(20))cos(60)+cos(40)1cos(20) \large = \frac {2 (2 \sin (30^\circ) \cos (20^\circ) ) } { - \cos (60^\circ) + \cos (40^\circ) } - \frac {1}{\cos (20^\circ )}

=4cos(20)1+2cos(40)1cos(20) \large = \frac {4 \cos (20^\circ) } { - 1 + 2\cos (40^\circ) } - \frac {1}{\cos (20^\circ )}

=4cos2(20)2cos(40)+12cos(40) cos(20)cos(20) \large = \frac {4 \cos^2 (20^\circ) -2 \cos (40^\circ) + 1} { 2 \cos (40^\circ) \space \cos (20^\circ) - \cos (20^\circ) }

=4cos2(20)22cos(40)+3( cos(60)+cos(20))cos(20) \large = \frac {4 \cos^2 (20^\circ) - 2 -2 \cos (40^\circ) + 3} { ( \space \cos (60^\circ) + \cos (20^\circ) ) - \cos (20^\circ) }

=2(2cos2(20)1)2cos(40)+3cos(60) \large = \frac { 2( 2 \cos^2 (20^\circ) - 1 ) -2 \cos (40^\circ) + 3} { \cos (60^\circ) } , because we want to apply cos(2A)=2cos2(A)1 \cos (2A) = 2 \cos^2 (A) - 1

=2cos(40)2cos(40)+312 \large = \frac { 2 \cos (40^\circ) -2 \cos (40^\circ) + 3} { \frac {1}{2} }

=3×2=6 \large = 3 \times 2 = 6

Pi Han Goh - 7 years, 7 months ago
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