Brilliant New Identity

Geometry Level 1

sin ⁡ 10 5 ∘ × sin ⁡ 1 5 ∘ = ? \large \sin 105^{\circ} \times \sin 15^{\circ} = \, ?

6 2 \frac{\sqrt{6}}{2} None of the choices 1 4 \frac{1}{4} 1 3 \frac{1}{3}

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Paul Ryan Longhas by Paul Ryan Longhas , 24, Philippines

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5 solutions

sin ⁡ ( 105 ) sin ⁡ ( 15 ) = cos ⁡ ( 15 ) sin ⁡ ( 15 ) = sin ⁡ ( 30 ) 2 = 1 4 \sin(105)\sin(15)=\cos(15)\sin(15)=\dfrac{\sin(30)}{2}=\dfrac{1}{4}

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Can you explain why can't I do like

sin ⁡ 10 5 ∘ sin ⁡ 1 5 ∘ = cos ⁡ ( 105 − 15 ) ∘ − cos ⁡ ( 105 + 15 ) ∘ 2 \sin 105^\circ \sin 15^\circ = \frac{\cos (105 - 15)^\circ - \cos (105 + 15)^\circ}{2}

cos ⁡ 9 0 ∘ − cos ⁡ 12 0 ∘ 2 = 0 − cos ⁡ 12 0 ∘ 2 \frac{\cos 90^\circ - \cos 120^\circ}{2} = \frac{0 - \cos 120^\circ}{2}

(we know it will be a negative result so we don't need to calculate cos ⁡ 12 0 ∘ \cos 120^\circ to tell it's "None of the choices")

João Arruda - 5 years, 5 months ago

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Why do you think your answer is negative? I can see clearly that it is positive and equals 1 4 \frac{1}{4}

Hjalmar Orellana Soto - 5 years, 5 months ago

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Oh right, I forgot that cos ⁡ 12 0 ∘ = − 1 2 \cos 120^\circ = -\frac{1}{2}

João Arruda - 5 years, 5 months ago
Shivam Jadhav
Jan 3, 2016

sin ⁡ ( 105 ) sin ⁡ ( 15 ) \sin(105)\sin(15) = sin ⁡ ( 90 + 15 ) sin ⁡ ( 15 ) =\sin(90+15)\sin(15) = cos ⁡ ( 15 ) sin ⁡ ( 15 ) =\cos(15)\sin(15) since sin ⁡ ( 90 + ω ) = cos ⁡ ( ω ) \sin(90+ω)=\cos(ω) = 2 cos ⁡ ( 15 ) sin ⁡ ( 15 ) 2 =\frac{2\cos(15)\sin(15)}{2} = sin ⁡ ( 30 ) 2 =\frac{\sin(30)}{2} Since sin ⁡ ( 2 ω ) = 2 sin ⁡ ( ω ) cos ⁡ ( ω ) \sin(2ω)=2\sin(ω)\cos(ω) = 1 4 =\boxed{\frac{1}{4}}

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Hobart Pao
Jan 4, 2016

sin ⁡ 10 5 ∘ = sin ⁡ 7 5 ∘ \sin 105^{\circ} = \sin 75^{\circ} sin ⁡ a sin ⁡ b = 1 2 [ cos ⁡ ( a − b ) − cos ⁡ ( a + b ) ] \sin a \sin b = \dfrac{1}{2}\left[\cos(a-b) - \cos(a+b) \right] = 1 2 ( cos ⁡ 6 0 ∘ + cos ⁡ 9 0 ∘ ) = \dfrac{1}{2} \left( \cos 60^{\circ} + \cos 90^{\circ}\right) = 1 4 = \boxed{ \dfrac{1}{4} }

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sin ⁡ 105 sin ⁡ 15 = sin ⁡ ( 60 + 45 ) × sin ⁡ ( 60 − 45 ) \sin 105 \sin 15 = \sin(60+45) \times \sin(60 - 45)

By identity

sin ⁡ ( 60 + 45 ) × sin ⁡ ( 60 − 45 ) = sin ⁡ 2 60 − sin ⁡ 2 45 = 1 4 \sin(60+45) \times \sin(60 - 45) = \sin^2 60 - \sin^2 45 = \frac{1}{4}

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Ram R
Jan 9, 2016

sin ⁡ ( 105 ) × sin ⁡ ( 15 ) = sin ⁡ ( 90 + 15 ) × sin ⁡ ( 15 ) = cos ⁡ ( 15 ) × sin ⁡ ( 15 ) = 2 × cos ⁡ ( 15 ) × sin ⁡ ( 15 ) 2 = sin ⁡ ( 30 ) 2 = 1 4 \sin (105) \times \sin (15)=\sin (90+15) \times \sin (15)= \cos (15) \times \sin (15)=\dfrac{2 \times \cos (15) \times \sin (15)}{2}=\dfrac{\sin (30)}{2}=\dfrac{1}{4}

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