240 Followers Problem - 1

Geometry Level 3

cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 \large \cos \theta _{1} + \cos \theta _{2} + \cos \theta_{3}

Given that cos ⁡ ( θ 1 − θ 2 ) + cos ⁡ ( θ 2 − θ 3 ) + cos ⁡ ( θ 3 − θ 1 ) = − 3 2 \cos \left( \theta_{1} -\theta_{2} \right) + \cos \left( \theta_{2} -\theta_{3} \right) + \cos \left( \theta_{3} -\theta_{1} \right) = -\dfrac{3}{2} , find the value of the expression above.


The answer is 0.

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Surya Prakash by Surya Prakash , 22, India

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3 solutions

P = [ cos ⁡ θ 1 cos ⁡ θ 2 + sin ⁡ θ 1 sin ⁡ θ 2 ] + [ cos ⁡ θ 2 cos ⁡ θ 3 + sin ⁡ θ 2 sin ⁡ θ 3 ] + [ cos ⁡ θ 3 cos ⁡ θ 1 + sin ⁡ θ 3 sin ⁡ θ 1 ] P=[\cos{\theta_1}\cos{\theta_2}+\sin{\theta_1}\sin{\theta_2}]+[\cos{\theta_2}\cos{\theta_3}+\sin{\theta_2}\sin{\theta_3}]+[\cos{\theta_3}\cos{\theta_1}+\sin{\theta_3}\sin{\theta_1}] P = [ cos ⁡ θ 1 cos ⁡ θ 2 + cos ⁡ θ 2 cos ⁡ θ 3 + cos ⁡ θ 3 cos ⁡ θ 1 ] + [ sin ⁡ θ 1 sin ⁡ θ 2 + sin ⁡ θ 2 sin ⁡ θ 3 + sin ⁡ θ 3 sin ⁡ θ 1 ] P=\displaystyle[\cos{\theta_1}\cos{\theta_2}+\cos{\theta_2}\cos{\theta_3}+\cos{\theta_3}\cos{\theta_1}]+[\sin{\theta_1}\sin{\theta_2}+\sin{\theta_2}\sin{\theta_3}+\sin{\theta_3}\sin{\theta_1}] P = ( cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 ) 2 − cos ⁡ 2 θ 1 − cos ⁡ 2 θ 2 − cos ⁡ 2 θ 3 2 + ( sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 ) 2 − sin ⁡ 2 θ 1 − sin ⁡ 2 θ 2 − sin ⁡ 2 θ 3 2 P=\displaystyle\frac{(\cos \theta _{1} + \cos \theta _{2} + \cos \theta_{3})^2-\cos^2 \theta _{1} - \cos^2 \theta _{2} - \cos^2 \theta_{3}}{2}+\frac{(\sin \theta _{1} + \sin \theta _{2} + \sin \theta_{3})^2-\sin^2 \theta _{1} - \sin^2 \theta _{2} - \sin^2 \theta_{3}}{2} P = ( cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 ) 2 + ( sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 ) 2 − 3 2 = − 3 2 P=\displaystyle\frac{(\cos \theta _{1} + \cos \theta _{2} + \cos \theta_{3})^2+(\sin \theta _{1} + \sin \theta _{2} + \sin \theta_{3})^2-3}{2}=-\frac{3}{2}

( cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 ) 2 + ( sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 ) 2 = 0 (\cos \theta _{1} + \cos \theta _{2} + \cos \theta_{3})^2+(\sin \theta _{1} + \sin \theta _{2} + \sin \theta_{3})^2=0

cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 = 0 \cos \theta _{1} + \cos \theta _{2} + \cos \theta_{3}=0

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Nice solution. You can even do it another way.

Surya Prakash - 5 years, 7 months ago

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@Surya Prakash I think it isn't good to repeat essentially the same problem, your 3rd problem is a repeat for this.

Venkata Karthik Bandaru - 5 years, 7 months ago

Let a , b , c a,b,c be three complex numbers such that a = e θ 1 i a=e^{\theta_1 i} , b = e θ 2 i b=e^{\theta_2 i} and c = e θ 3 i c=e^{\theta_3 i} . Then let's find the magnitude of their sum, i.e., ∣ a + b + c ∣ |a+b+c| :

∣ a + b + c ∣ = ∣ cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 + i ( sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 ) ∣ a + b + c ∣ = ( cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 ) 2 + ( sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 ) 2 ∣ a + b + c ∣ = 3 + 2 ( cos ⁡ θ 1 cos ⁡ θ 2 + sin ⁡ θ 1 sin ⁡ θ 2 + cos ⁡ θ 2 cos ⁡ θ 3 + sin ⁡ θ 2 sin ⁡ θ 3 + cos ⁡ θ 3 cos ⁡ θ 1 + sin ⁡ θ 3 sin ⁡ θ 1 ) ∣ a + b + c ∣ = 3 + 2 ( cos ⁡ ( θ 1 − θ 2 ) + cos ⁡ ( θ 2 − θ 3 ) + cos ⁡ ( θ 3 − θ 1 ) ) ∣ a + b + c ∣ = 3 + 2 ( − 3 2 ) ∣ a + b + c ∣ = 0 ⟹ a + b + c = 0 |a+b+c|=|\cos \theta_1+\cos \theta_2+\cos \theta_3+i(\sin \theta_1+\sin \theta_2+\sin \theta_3) \\ |a+b+c|=\sqrt{(\cos \theta_1+\cos \theta_2+\cos \theta_3)^2+(\sin \theta_1+\sin \theta_2+\sin \theta_3)^2} \\ |a+b+c|=\sqrt{3+2(\cos \theta_1 \cos \theta_2+\sin \theta_1 \sin \theta_2+\cos \theta_2 \cos \theta_3+\sin \theta_2 \sin \theta_3+\cos \theta_3 \cos \theta_1+\sin \theta_3 \sin \theta_1)} \\ |a+b+c|=\sqrt{3+2(\cos(\theta_1-\theta_2)+\cos(\theta_2-\theta_3)+\cos(\theta_3-\theta_1))} \\ |a+b+c|=\sqrt{3+2\left(-\dfrac{3}{2}\right)} \\ |a+b+c|=0 \implies a+b+c=0

Hence, cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 + i ( sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 ) = 0 \cos \theta_1+\cos \theta_2+\cos \theta_3+i(\sin \theta_1+\sin \theta_2+\sin \theta_3)=0 . Comparing the real part we get 0 \boxed{0} as our answer.

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Nice solution!

Miloje Đukanović - 5 years, 7 months ago

cos ⁡ ( θ 1 − θ 2 ) + cos ⁡ ( θ 2 − θ 3 ) + cos ⁡ ( θ 3 − θ 1 ) = − 3 2 \small \cos(\theta_{1}-\theta_{2})+\cos(\theta_{2} -\theta_{3})+\cos(\theta_{3} -\theta_{1})=-\dfrac{3}{2}

⟹ 3 + 2 [ cos ⁡ ( θ 1 − θ 2 ) + cos ⁡ ( θ 2 − θ 3 ) + cos ⁡ ( θ 3 − θ 1 ) ] = 0 \small \implies 3+2[\cos(\theta_{1}-\theta_{2})+\cos(\theta_{2}-\theta_{3})+\cos(\theta_{3}-\theta_{1})]=0

⟹ ( sin ⁡ 2 ( θ 1 ) + cos ⁡ 2 ( θ 1 ) ) + ( sin ⁡ 2 ( θ 2 ) + cos ⁡ 2 ( θ 2 ) ) + ( sin ⁡ 2 ( θ 3 ) + cos ⁡ 2 ( θ 3 ) ) + 2 cos ⁡ ( θ 1 ) cos ⁡ ( θ 2 ) + 2 sin ⁡ ( θ 1 ) sin ⁡ ( θ 2 ) + 2 cos ⁡ ( θ 2 ) cos ⁡ ( θ 3 ) + 2 sin ⁡ ( θ 2 ) sin ⁡ ( θ 3 ) + 2 cos ⁡ ( θ 3 ) cos ⁡ ( θ 1 ) + 2 sin ⁡ ( θ 3 ) sin ⁡ ( θ 1 ) = 0 \small \implies (\sin^{2}(\theta_{1})+\cos^{2}(\theta_{1}))+(\sin^{2}(\theta_{2})+\cos^{2}(\theta_{2}))+(\sin^{2}(\theta_{3})+\cos^{2}(\theta_{3}))+2\cos(\theta_{1})\cos(\theta_{2})+2\sin(\theta_{1})\sin(\theta_{2})+2\cos(\theta_{2})\cos(\theta_{3})+2\sin(\theta_{2})\sin(\theta_{3})+2\cos(\theta_{3})\cos(\theta_{1})+2\sin(\theta_{3})\sin(\theta_{1})=0

⟹ ( sin ⁡ 2 ( θ 1 ) + sin ⁡ 2 ( θ 2 ) + sin ⁡ 2 ( θ 3 ) + 2 sin ⁡ ( θ 1 ) sin ⁡ ( θ 2 ) + 2 sin ⁡ ( θ 2 ) sin ⁡ ( θ 3 ) + 2 sin ⁡ ( θ 3 ) sin ⁡ ( θ 1 ) ) + ( cos ⁡ 2 ( θ 1 ) + cos ⁡ 2 ( θ 2 ) + cos ⁡ 2 ( θ 3 ) + 2 cos ⁡ ( θ 1 ) cos ⁡ ( θ 2 ) + 2 cos ⁡ ( θ 2 ) cos ⁡ ( θ 3 ) + 2 cos ⁡ ( θ 3 ) cos ⁡ ( θ 1 ) ) = 0 \small \implies (\sin^{2}(\theta_{1})+\sin^{2}(\theta_{2})+\sin^{2}(\theta_{3})+2\sin(\theta_{1})\sin(\theta_{2})+2\sin(\theta_{2})\sin(\theta_{3})+2\sin(\theta_{3})\sin(\theta_{1}))+(\cos^{2}(\theta_{1})+\cos^{2}(\theta_{2})+\cos^{2}(\theta_{3})+2\cos(\theta_{1})\cos(\theta_{2})+2\cos(\theta_{2})\cos(\theta_{3})+2\cos(\theta_{3})\cos(\theta_{1}))=0

⟹ ( sin ⁡ ( θ 1 ) + sin ⁡ ( θ 2 ) + sin ⁡ ( θ 3 ) ) 2 + ( cos ⁡ ( θ 1 ) + cos ⁡ ( θ 2 ) + cos ⁡ ( θ 3 ) ) 2 = 0 \small \implies (\sin(\theta_{1})+\sin(\theta_{2})+\sin(\theta_{3}))^{2}+(\cos(\theta_{1})+\cos(\theta_{2})+\cos(\theta_{3}))^{2}=0

⟹ cos ⁡ ( θ 1 ) + cos ⁡ ( θ 2 ) + cos ⁡ ( θ 3 ) = sin ⁡ ( θ 1 ) + sin ⁡ ( θ 2 ) + sin ⁡ ( θ 3 ) = 0 \small \implies \cos(\theta_{1})+\cos(\theta_{2})+\cos(\theta_{3})=\sin(\theta_{1})+\sin(\theta_{2})+\sin(\theta_{3})=\boxed{0}

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