Respuesta :
We are given with the equation
1/sin 18 - 1/sin 54 = 2
We are meant to verify if the equation is true
So,
(sin 54 - sin 18)/ sin 18 sin 54 = 2
sin 54 - sin 18 = 2 sin 18 sin 54
sin 18*3 - sin 18 = 2 sin 18 sin 18*3
Use trigonometric identities to verify the equation.
The identities to use are the double and product of angles.
1/sin 18 - 1/sin 54 = 2
We are meant to verify if the equation is true
So,
(sin 54 - sin 18)/ sin 18 sin 54 = 2
sin 54 - sin 18 = 2 sin 18 sin 54
sin 18*3 - sin 18 = 2 sin 18 sin 18*3
Use trigonometric identities to verify the equation.
The identities to use are the double and product of angles.
Answer with explanation:
We have to prove that
[tex]\frac{1}{sin18}-\frac{1}{sin54}=2[/tex]
sin54°=sin3×18°=3 sin18°-4sin³18°
LHS
[tex]\frac{1}{sin18^{\circ}}-\frac{1}{3sin18^{\circ}-4sin^318^{\circ}}[/tex]
[tex]=\frac{1}{sin18^{\circ}}\times(1-\frac{1}{3-4 sin^218^{\circ}})\\\\=\frac{1}{sin18^{\circ}}\times(\frac{3-4sin^218^{\circ}-1}{3-4 sin^2 18^{\circ}})\\\\=\frac{1}{sin18^{\circ}}\times(\frac{2-4sin^218^{\circ}}{3-4 sin^2 18^{\circ}})\\\\=\frac{2}{sin18^{\circ}}\times(\frac{1-2sin^218^{\circ}}{3-4 sin^2 18^{\circ}})\\\\=\frac{2cos36^{\circ}}{sin54^{\circ}}[/tex]
[tex]=2\frac{sin54^{\circ}}{sin54^{\circ}}\\\\=2[/tex]
Hence Proved.
Used the following trigonometric Identity to solve the problem
Sin3A=3 SinA-4 sin³A
Sin(90°-A)=Cos A &Cos(90°-A)=SinA
Cos 2A=2Cos²A-1=1-2Sin²A=Cos²A-Sin²A