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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

If ((1)/(4) 36^ 54^ )^2-( 12^ 36^ 48^ 72^ )^2= a -b c where a, c...

If (14cos⁡36∘sin⁡54∘)2−(sin⁡12∘sin⁡36∘sin⁡48∘sin⁡72∘)2=a−bc\left(\frac{1}{4} \cos 36^{\circ} \sin 54^{\circ}\right)^2-\left(\sin 12^{\circ} \sin 36^{\circ} \sin 48^{\circ} \sin 72^{\circ}\right)^2=\frac{\sqrt{a}-b}{c} where a,ca, c are relatively prime numbers, then find the value of (a+b+c)(a+b+c).

Correct answer

70

Step-by-step solution

116(cos⁡36∘sin⁡54∘)2−(14sin⁡36∘sin⁡36∘)2116(cos⁡236∘−sin⁡236∘)=116cos⁡72∘=5−164≡a−bc∴(a+b+c)=5+1+64=70\begin{aligned} & \frac{1}{16}\left(\cos 36^{\circ} \sin 54^{\circ}\right)^2-\left(\frac{1}{4} \sin 36^{\circ} \sin 36^{\circ}\right)^2 \\ & \frac{1}{16}\left(\cos ^2 36^{\circ}-\sin ^2 36^{\circ}\right)=\frac{1}{16} \cos 72^{\circ}=\frac{\sqrt{5}-1}{64} \equiv \frac{\sqrt{a}-b}{c} \\ & \therefore(a+b+c)=5+1+64=70\end{aligned}
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