CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

$$sin 36^{\circ} sin72^osin 108^{\circ} sin 144^{\circ}$$=


A
1/4
loader
B
1/16
loader
C
3/4
loader
D
5/16
loader

Solution

The correct option is D 5/16
$$\sin 144^{\circ}=\sin (180^{\circ}-36^{\circ})=\sin 36^{\circ}$$

$$\sin 108^{\circ}\sin 72^{\circ}\\=\sin (90^{\circ}+18^{\circ})\sin (90^{\circ}-18^{\circ})=\sin^2 90^{\circ}-\sin ^2 18^{\circ}=1-\dfrac{(\sqrt{5}-1)^2}{16}=\dfrac{16-5-1+2\sqrt{5}}{16}=\dfrac{5+\sqrt{5}}{8}$$


$$\sin 36^{\circ}\sin 72^{\circ}\sin 108^{\circ}\sin 144^{\circ}=\sin 36^{\circ}\times (\sin 108^{\circ}\sin 72^{\circ})\times \sin 36^{\circ}$$
                                                         $$=\sin^2 36^{\circ}\times \dfrac{5+\sqrt{5}}{8}=\dfrac{10-2\sqrt{5}}{16}\times \dfrac{5+\sqrt{5}}{8}=\dfrac{5^2-5}{8^2}=\dfrac{5}{16}$$

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image