Trigonometric powers?

Geometry Level 1

Evaluate ( 1 2 sin ⁡ 2 4 ∘ ) sin ⁡ 2 4 ∘ × ( 1 2 sin ⁡ 6 6 ∘ ) sin ⁡ 6 6 ∘ × ( 1 2 cos ⁡ 3 6 ∘ ) cos ⁡ 3 6 ∘ × ( 1 2 cos ⁡ 5 4 ∘ ) cos ⁡ 5 4 ∘ . \large { \begin{aligned} & \left(12 ^{\sin 24 ^\circ}\right)^{\sin 24 ^\circ} \times \left(12 ^{\sin 66 ^\circ}\right)^{\sin 66 ^\circ} \\ & \times \left(12 ^{\cos 36 ^\circ}\right)^{\cos 36 ^\circ} \times \left(12 ^{\cos 54 ^\circ}\right)^{\cos 54 ^\circ}. \end{aligned}}

144 216 72 288

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Mahindra Jain by Mahindra Jain

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5 solutions

Shabarish Ch
Mar 7, 2014

( 1 2 sin ⁡ 2 4 ∘ ) sin ⁡ 2 4 ∘ × ( 1 2 sin ⁡ 6 6 ∘ ) sin ⁡ 6 6 ∘ × ( 1 2 cos ⁡ 3 6 ∘ ) cos ⁡ 3 6 ∘ × ( 1 2 cos ⁡ 5 4 ∘ ) cos ⁡ 5 4 ∘ (12^{ \sin 24^{ \circ } } )^{ \sin 24^{ \circ }} \times (12^{ \sin 66^{ \circ } } )^{ \sin 66^{ \circ }} \times (12^{ \cos 36^{ \circ } } )^{ \cos 36^{ \circ }} \times (12^{ \cos 54^{ \circ } } )^{ \cos 54^{ \circ }}

First we remove the brackets to get,

1 2 sin ⁡ 2 2 4 ∘ × 1 2 sin ⁡ 2 6 6 ∘ × 1 2 cos ⁡ 2 3 6 ∘ × 1 2 cos ⁡ 2 5 4 ∘ 12^{ \sin^2 24^{\circ} } \times 12^{ \sin^2 66^{\circ} } \times 12^{ \cos^2 36^{\circ} } \times 12^{ \cos^2 54^{\circ} }

We know that sin ⁡ ( 90 − θ ) = cos ⁡ θ \sin (90 - \theta) = \cos \theta and cos ⁡ ( 90 − θ ) = sin ⁡ θ \cos (90 - \theta) = \sin \theta , when angles are expressed in degrees.

So, sin ⁡ 6 6 ∘ = cos ⁡ ( 9 0 ∘ − 6 6 ∘ ) = cos ⁡ 2 4 ∘ \sin 66^{\circ} = \cos ( 90^{\circ} - 66^{\circ}) = \cos 24^{\circ} and cos ⁡ 5 4 ∘ = sin ⁡ ( 9 0 ∘ − 5 4 ∘ ) = sin ⁡ 3 6 ∘ \cos 54^{\circ} = \sin ( 90^{\circ} - 54^{\circ}) = \sin 36^{\circ}

Substituting these values, we get,

1 2 sin ⁡ 2 2 4 ∘ × 1 2 cos ⁡ 2 2 4 ∘ × 1 2 cos ⁡ 2 3 6 ∘ × 1 2 sin ⁡ 2 3 6 ∘ 12^{ \sin^2 24^{\circ} } \times 12^{ \cos^2 24^{\circ} } \times 12^{ \cos^2 36^{\circ} } \times 12^{ \sin^2 36^{\circ} }

or,

1 2 sin ⁡ 2 2 4 ∘ + cos ⁡ 2 2 4 ∘ × 1 2 sin ⁡ 2 3 6 ∘ + cos ⁡ 2 3 6 ∘ 12^{ \sin^2 24^{\circ} + \cos^2 24^{\circ} } \times 12^{ \sin^2 36^{\circ} + \cos^2 36^{\circ} }

We know that sin ⁡ 2 θ + cos ⁡ 2 θ = 1 \sin^{2} \theta + \cos^{2} \theta = 1

Substituting this in the above expression, we get

1 2 1 × 1 2 1 = 144 12^1 \times 12^1 = \boxed{144}

20 Helpful 1 Interesting 3 Brilliant 0 Confused
Prasun Biswas
Mar 7, 2014

We should first know the formula that sin ⁡ 2 θ + cos ⁡ 2 θ = 1 \sin^2 \theta + \cos^2 \theta = 1 and the laws of indices like a m × a n = a m + n a^m\times a^n = a^{m+n} and ( a m ) n = a m n (a^m)^n = a^{mn} . Also, we must know that sin ⁡ θ = cos ⁡ ( 9 0 ∘ − θ ) \sin \theta = \cos (90^{\circ}- \theta) from which on squaring both sides of it, we get sin ⁡ 2 θ = cos ⁡ 2 ( 9 0 ∘ − θ ) \sin^2 \theta = \cos^2 (90^{\circ}- \theta) Now, lets get back to the given problem----

( 1 2 sin ⁡ 2 4 ∘ ) sin ⁡ 2 4 ∘ × ( 1 2 sin ⁡ 6 6 ∘ ) sin ⁡ 6 6 ∘ + ( 1 2 cos ⁡ 3 6 ∘ ) cos ⁡ 3 6 ∘ + ( 1 2 cos ⁡ 5 4 ∘ ) cos ⁡ 5 4 ∘ \large (12^{\sin 24^{\circ}})^{\sin 24^{\circ}}\times (12^{\sin 66^{\circ}})^{\sin 66^{\circ}}+(12^{\cos 36^{\circ}})^{\cos 36^{\circ}}+(12^{\cos 54^{\circ}})^{\cos 54^{\circ}}

= 1 2 sin ⁡ 2 4 ∘ × sin ⁡ 2 4 ∘ × 1 2 sin ⁡ 6 6 ∘ × sin ⁡ 6 6 ∘ × 1 2 cos ⁡ 3 6 ∘ × cos ⁡ 3 6 ∘ × 1 2 cos ⁡ 5 4 ∘ × cos ⁡ 5 4 ∘ \large =12^{\sin 24^{\circ}\times \sin 24^{\circ}}\times 12^{\sin 66^{\circ}\times \sin 66^{\circ}}\times 12^{\cos 36^{\circ}\times \cos 36^{\circ}}\times 12^{\cos 54^{\circ}\times \cos 54^{\circ}}

= 1 2 sin ⁡ 2 2 4 ∘ × 1 2 sin ⁡ 2 6 6 ∘ × 1 2 cos ⁡ 2 3 6 ∘ × 1 2 cos ⁡ 2 5 4 ∘ \large =12^{\sin^2 24^{\circ}}\times 12^{\sin^2 66^{\circ}}\times 12^{\cos^2 36^{\circ}}\times 12^{\cos^2 54^{\circ}}

= 1 2 ( sin ⁡ 2 2 4 ∘ + sin ⁡ 2 6 6 ∘ + cos ⁡ 2 3 6 ∘ + cos ⁡ 2 5 4 ∘ ) \large =12^{(\sin^2 24^{\circ}+\sin^2 66^{\circ}+\cos^2 36^{\circ}+\cos^2 54^{\circ}})

= 1 2 ( sin ⁡ 2 ( 90 − 66 ) ∘ + sin ⁡ 2 6 6 ∘ + sin ⁡ 2 ( 90 − 54 ) ∘ + cos ⁡ 2 5 4 ∘ ) \large =12^{(\sin^2 (90-66)^{\circ}+\sin^2 66^{\circ}+\sin^2 (90-54)^{\circ}+\cos^2 54^{\circ}})

= 1 2 ( cos ⁡ 2 6 6 ∘ + sin ⁡ 2 6 6 ∘ + sin ⁡ 2 5 4 ∘ + cos ⁡ 2 5 4 ∘ ) \large =12^{(\cos^2 66^{\circ}+\sin^2 66^{\circ}+\sin^2 54^{\circ}+\cos^2 54^{\circ}})

= 1 2 1 + 1 = 1 2 2 = 144 \large =12^{1+1}=12^2=\boxed{144}

6 Helpful 0 Interesting 0 Brilliant 0 Confused
Alec Camhi
May 23, 2015

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Ritesh Surve
Apr 17, 2014

12*12=144

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Very good question...

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Yes indeed, I agree

Mahdi Raza - 1 year, 4 months ago

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