JEE Novice - (3)

Geometry Level 3

sin ⁡ 8 7 5 ∘ − cos ⁡ 8 7 5 ∘ = ? \Large \sin^8 75^\circ - \cos^8 75^\circ = \ ?

  • Give your answer to correct to 3 decimal places.
This question is a part of JEE Novices .


The answer is 0.757.

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Nihar Mahajan by Nihar Mahajan , 20, India

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2 solutions

Sandeep Bhardwaj
May 26, 2015

sin ⁡ 8 7 5 ∘ − cos ⁡ 8 7 5 ∘ \sin^8 75^\circ - \cos^8 75^\circ

= ( sin ⁡ 4 7 5 ∘ − cos ⁡ 4 7 5 ∘ ) ( sin ⁡ 4 7 5 ∘ + cos ⁡ 4 7 5 ∘ ) =\left( \sin^4 75^\circ-\cos^475^\circ \right) \left( \sin^475^\circ+\cos^475^\circ \right)

= ( sin ⁡ 2 7 5 ∘ − cos ⁡ 2 7 5 ∘ ) ( sin ⁡ 2 7 5 ∘ + cos ⁡ 2 7 5 ∘ ) ( sin ⁡ 4 7 5 ∘ + cos ⁡ 4 7 5 ∘ ) =\left( \sin^275^\circ-\cos^275^\circ \right) \left( \sin^275^\circ+\cos^275^\circ \right) \left( \sin^475^\circ+\cos^475^\circ \right)

= ( sin ⁡ 2 7 5 ∘ − cos ⁡ 2 7 5 ∘ ) ( ( sin ⁡ 2 7 5 ∘ + cos ⁡ 2 7 5 ∘ ) 2 − 2 sin ⁡ 2 7 5 ∘ c o s 2 7 5 ∘ ) =\left( \sin^275^\circ-\cos^275^\circ \right) \left( (\sin^275^\circ+\cos^275^\circ)^2-2\sin^275^\circ cos^275^\circ \right)

= ( − cos ⁡ ( 2 × 7 5 ∘ ) ) ( 1 − 2 sin ⁡ 2 7 5 ∘ c o s 2 7 5 ∘ ) =\left( -\cos(2 \times 75^\circ) \right) \left( 1-2\sin^275^\circ cos^275^\circ\right)

= − cos ⁡ 15 0 ∘ × ( 1 − sin ⁡ 2 15 0 ∘ 2 ) =-\cos150^\circ \times \left( 1-\dfrac{\sin^2150^\circ}{2} \right)

= − ( 3 2 ) × ( 1 − 1 8 ) =- \left( \dfrac{\sqrt{3}}{2} \right) \times \left(1-\dfrac{1}{8} \right)

= 7 3 16 = 0.758 (rounding off up to 3 decimal places) =\dfrac{7 \sqrt{3}}{16}=\boxed{0.758 \text{(rounding off up to 3 decimal places)}}

Identities used : \color{#D61F06}{\text{Identities used : }}

A 2 − B 2 = ( A + B ) ( A − B ) sin ⁡ 2 λ + cos ⁡ 2 λ = 1 sin ⁡ 2 λ − cos ⁡ 2 λ = − cos ⁡ ( 2 ⋅ λ ) 2 ⋅ sin ⁡ λ ⋅ cos ⁡ λ = sin ⁡ ( 2 ⋅ λ ) \boxed{A^2-B^2=(A+B)(A-B) \\ \sin^2\lambda+\cos^2\lambda=1 \\ \sin^2\lambda-\cos^2\lambda=- \cos(2 \cdot \lambda) \\ 2 \cdot \sin\lambda \cdot \cos\lambda=\sin(2 \cdot \lambda)}

enjoy !

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Moderator note:

Bonus question: Can you find the exact form of sin ⁡ ( 7 5 ∘ ) \sin(75^\circ) ?

Bonus question: Method 1:

cos ⁡ ( 15 0 ∘ ) = 1 − 2 sin ⁡ 2 ( 7 5 ∘ ) \cos(150^{\circ}) = 1 - 2\sin^{2}(75^{\circ})

⟹ sin ⁡ 2 ( 7 5 ∘ ) = 1 2 ( 1 − cos ⁡ ( 15 0 ∘ ) ) = 1 2 ( 1 + 3 2 ) = 1 4 ( 2 + 3 ) \Longrightarrow \sin^{2}(75^{\circ}) = \dfrac{1}{2}(1 - \cos(150^{\circ})) = \dfrac{1}{2}\left(1 + \dfrac{\sqrt{3}}{2}\right) = \dfrac{1}{4}(2 + \sqrt{3})

⟹ sin ⁡ ( 7 5 ∘ ) = 2 + 3 2 , \Longrightarrow \sin(75^{\circ}) = \dfrac{\sqrt{2 + \sqrt{3}}}{2},

where we have taken the positive root as we know that sin ⁡ ( 7 5 ∘ ) > 0. \sin(75^{\circ}) \gt 0.

Further, cos ⁡ ( 7 5 ∘ ) = 1 − sin ⁡ 2 ( 7 5 ∘ ) = 2 − 3 2 . \cos(75^{\circ}) = \sqrt{1 - \sin^{2}(75^{\circ})} = \dfrac{\sqrt{2 - \sqrt{3}}}{2}.

Method 2:

sin ⁡ ( 7 5 ∘ ) = sin ⁡ ( 3 0 ∘ ) + 4 5 ∘ ) = sin ⁡ ( 3 0 ∘ ) cos ⁡ ( 4 5 ∘ ) + cos ⁡ ( 3 0 ∘ ) sin ⁡ ( 4 5 ∘ ) = \sin(75^{\circ}) = \sin(30^{\circ}) + 45^{\circ}) = \sin(30^{\circ})\cos(45^{\circ}) + \cos(30^{\circ})\sin(45^{\circ}) =

1 2 × 2 2 + 3 2 × 2 2 = 2 + 6 4 . \dfrac{1}{2} \times \dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{2} + \sqrt{6}}{4}.

Note that ( 2 + 6 ) 2 = 8 + 2 12 = 8 + 4 3 = 2 2 + 3 , \sqrt{(\sqrt{2} + \sqrt{6})^{2}} = \sqrt{8 + 2\sqrt{12}} = \sqrt{8 + 4\sqrt{3}} = 2\sqrt{2 + \sqrt{3}},

so the two expressions found for sin ⁡ ( 7 5 ∘ ) \sin(75^{\circ}) are in fact identical.

Brian Charlesworth - 6 years ago

Great efforts! Nice.

Nihar Mahajan - 6 years ago

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thank you. ⌣ ¨ \ddot \smile

Sandeep Bhardwaj - 6 years ago
Michael Fuller
May 27, 2015

sin ⁡ 8 75 ° − cos ⁡ 8 75 ° = ( sin ⁡ ( 30 ° + 45 ° ) ) 8 − ( cos ⁡ ( 30 ° + 45 ° ) ) 8 = ( sin ⁡ 30 cos ⁡ 45 + sin ⁡ 45 cos ⁡ 30 ) 8 − ( cos ⁡ 30 cos ⁡ 45 − sin ⁡ 30 sin ⁡ 45 ) 8 = ( 1 2 ⋅ 2 2 + 2 2 ⋅ 3 2 ) 8 − ( 3 2 ⋅ 2 2 − 1 2 ⋅ 2 2 ) 8 = ( 2 + 6 4 ) 8 − ( 6 − 2 4 ) 8 = ( 8 + 4 3 16 ) 4 − ( 8 − 4 3 16 ) 4 = ( 2 + 3 4 ) 4 − ( 2 − 3 4 ) 4 = ( 7 + 4 3 16 ) 2 − ( 7 − 4 3 16 ) 2 = ( 97 + 56 3 256 ) − ( 97 − 56 3 256 ) = 112 3 256 = 7 3 16 = 0.758 ( 3 d p ) \sin ^{ 8 }{ 75° } -\cos ^{ 8 }{ 75° } \\ ={ \left( \sin { \left( 30°+45° \right) } \right) }^{ 8 }-{ \left( \cos { \left( 30°+45° \right) } \right) }^{ 8 }\\ ={ \left( \sin { 30 } \cos { 45 } +\sin { 45 } \cos { 30 } \right) }^{ 8 }-{ \left( \cos { 30 } \cos { 45 } -\sin { 30 } \sin { 45 } \right) }^{ 8 }\\ ={ \left( \frac { 1 }{ 2 } \cdot \frac { \sqrt { 2 } }{ 2 } +\frac { \sqrt { 2 } }{ 2 } \cdot \frac { \sqrt { 3 } }{ 2 } \right) }^{ 8 }-{ \left( \frac { \sqrt { 3 } }{ 2 } \cdot \frac { \sqrt { 2 } }{ 2 } -\frac { 1 }{ 2 } \cdot \frac { \sqrt { 2 } }{ 2 } \right) }^{ 8 }\\ ={ \left( \frac { \sqrt { 2 } +\sqrt { 6 } }{ 4 } \right) }^{ 8 }-{ \left( \frac { \sqrt { 6 } -\sqrt { 2 } }{ 4 } \right) }^{ 8 }\\ ={ \left( \frac { 8+4\sqrt { 3 } }{ 16 } \right) }^{ 4 }-{ \left( \frac { 8-4\sqrt { 3 } }{ 16 } \right) }^{ 4 }={ \left( \frac { 2+\sqrt { 3 } }{ 4 } \right) }^{ 4 }-{ \left( \frac { 2-\sqrt { 3 } }{ 4 } \right) }^{ 4 }\\ ={ \left( \frac { 7+4\sqrt { 3 } }{ 16 } \right) }^{ 2 }-{ \left( \frac { 7-4\sqrt { 3 } }{ 16 } \right) }^{ 2 }\\ ={ \left( \frac { 97+56\sqrt { 3 } }{ 256 } \right) }-{ \left( \frac { 97-56\sqrt { 3 } }{ 256 } \right) }\\ =\frac { 112\sqrt { 3 } }{ 256 } \\ =\frac { 7\sqrt { 3 } }{ 16 } \\ =\boxed { 0.758 } (3dp)

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