Question

xiii) cos 7º cos 14º cos28° cos 56°=sin68°​

Answer 1

Step-by-step explanation:

How can you show that cos 7° cos 14° cos 28° cos 56° = (sin 68°) / (16 cos 83°)?

[math]\text{Let }x = 7°[/math]

[math]S = \cos x \cos 2x \cos 4x \cos 8x = \dfrac{1}{2\sin x}. [\underbrace{(2 \sin x \cos x)}_{=\sin 2x} \cos 2x \cos 4x \cos 8x ][/math]

[math]= \dfrac{1}{4\sin x}. [\underbrace{(2 \sin 2x \cos 2x)}_{=\sin 4x} \cos 4x \cos 8x ][/math]

[math]= \dfrac{1}{8\sin x}. [\underbrace{(2 \sin 4x \cos 4x)}_{\sin 8x} \cos 8x ][/math]

[math]= \dfrac{1}{16\sin x}. [2 \sin 8x \cos 8x ] = \dfrac{\sin 16x}{16\sin x}[/math]

[math]\text{Substitute }x = 7°[/math]

[math]S = \dfrac{\sin (16 \times 7)°}{16\sin 7°} = \dfrac{\sin 112°}{16\sin 7°}[/math]

[math]\sin 7° = \sin (90-83)° = \cos 83°[/math]

[math]\sin 112° = \sin (180 - 68)° = \sin 68°[/math]

[math]\therefore S = \dfrac{\sin 68°}{16\cos 83°} = RHS [/math]

proved .