Find the value of { \sin ^6 15^{\circ}+ \sin ^6 75^{\circ}+6 \sin ^2 15^{\circ} \sin ^2 75^{\circ}}/{ \sin ^4 15^{\circ}+ \sin ^4 75^{\circ}+5 \sin ^2 15^{\circ} \sin ^2 75^{\circ}}

\sin 15^{\circ}+ \sin 75^{\circ}

\sin 15^{\circ} \cos 15^{\circ}

Xavier Aptitude Test 2023 Paper

Question : 55 of 102

Marks: +1, -0

Find the value of

\;{ \sin \;^6 15^{∘}+ \sin \;^6 75^{∘}+6 \sin \;^2 15^{∘} \sin \;^2 75^{∘}}/{ \sin \;^4 15^{∘}+ \sin \;^4 75^{∘}+5 \sin \;^2 15^{∘} \sin \;^2 75^{∘}}

Solution:

Let

\sin \;^2 15^{∘}=a

and

\sin \;^2 75^{∘}=b

Then the given equation becomes,

\;{a^3+b^3+6 a b}/{a^2+b^2+5 a b}

=\;{(a+b)^3-3 a b(a+b)+6 a b}/{(a+b)^2-2 a b+5 a b} \;\; ∴ a^3+b^3=(a+b)^3-3 a b(a+b), ∴ a^2+b^2=(a+b)^2-2 a b

Using

\sin \;(θ)=\cos (90-θ)

we get,

sin \;^2 75^{∘}=\cos ^2 15^{∘}

now,

{\a}+ {\b}= \sin \;^2 15^{∘}+ \sin \;^2 75^{∘}= \sin \;^2 15^{∘}+\cos ^2 15^{∘}=1

The equation becomes,

=\;{1-3 a b+6 a b}/{1-2 a b+5 a b}=\;{1+3 a b}/{1+3 a b}=1

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