Too Complex! Part- III

Geometry Level 4

sin ⁡ 5 θ = 5 sin ⁡ θ ( 1 − sin ⁡ 2 θ sin ⁡ 2 p ) ( 1 − sin ⁡ 2 θ sin ⁡ 2 q ) \sin 5\theta = 5 \sin \theta \left ( 1 - \frac{\sin^{2} \theta}{\sin ^{2} p} \right) \left( 1 - \frac{\sin^{2} \theta}{\sin ^{2} q} \right)

If the equation above is a trigonometric identities for constants p p and q q such that 0 ≤ p , q ≤ π 2 0 \leq p,q\leq \frac \pi2 , find the value of p + q p + q .

Clue : Use the equation z 10 − 1 = 0 z^{10} - 1 = 0


The answer is 1.885.

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Avineil Jain by Avineil Jain , 24, Netherlands

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2 solutions

Ayush Verma
Oct 29, 2014

Chew-Seong Cheong's solution is universal but my solution only for this question.

s i n 5 θ = 0 & s i n θ ≠ 0 i f s i n 2 θ = s i n 2 p o r , s i n 2 θ = s i n 2 q sin5\theta =0\quad \& \quad sin\theta \neq 0\quad if\\ \\ { sin }^{ 2 }\theta ={ sin }^{ 2 }p\\ \\ or,\quad { sin }^{ 2 }\theta ={ sin }^{ 2 }q

If we only consider θ ∈ ( 0 , π 2 ) \theta \in \left( 0,\cfrac { \pi }{ 2 } \right)

f o r s i n 5 θ = 0 & s i n θ ≠ 0 , 5 θ = π , 2 π ( o n l y t w o s o l n ) ⇒ θ = π 5 , 2 π 5 for\quad sin5\theta =0\quad \& \quad sin\theta \neq 0,\\ \\ 5\theta =\pi ,2\pi (only\quad two\quad { sol }^{ n })\Rightarrow \theta =\cfrac { \pi }{ 5 } ,\cfrac { 2\pi }{ 5 }

one of them is p & other is q,so

p + q = π 5 + 2 π 5 = 3 π 5 = 1.885 p+q=\cfrac { \pi }{ 5 } +\cfrac { 2\pi }{ 5 } =\cfrac { 3\pi }{ 5 } =1.885

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice. I liked that you figured out what the exact values of p p and q q are.

Can you generalize this identity to

( sin ⁡ ( 2 n + 1 ) θ ) = ( 2 n + 1 ) ( sin ⁡ θ ) ∏ ( 1 − sin ⁡ 2 θ sin ⁡ 2 p i ) ? \left( \sin ( 2n + 1 ) \theta \right) = (2n+1)( \sin \theta ) \prod ( 1 - \frac{ \sin ^2 \theta } { \sin ^ 2 p_i } )?

Calvin Lin Staff - 6 years, 7 months ago

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Hello Sir. This can be generalized using chebyshev polynomial of the second kind. I have posted it here . I f you have a method other than mine, I'd like to see it, please.

Ishan Singh - 5 years, 8 months ago

I used the same method!!

Kunal Gupta - 6 years, 7 months ago

Did the same

Aishwary Omkar - 6 years, 5 months ago
Chew-Seong Cheong
Oct 29, 2014

Using the identity e i θ ≡ cos ⁡ θ + i sin ⁡ θ e^{i\theta} \equiv \cos{\theta} + i\sin{\theta} , where i = − 1 i = \sqrt {-1} :

We have, e i 5 θ = ( e i θ ) 5 e^{i5\theta} = (e^{i\theta})^5

⇒ cos ⁡ 5 θ + i sin ⁡ 5 θ = ( cos ⁡ θ + i sin ⁡ θ ) 5 \Rightarrow \cos{5\theta} + i\sin{5\theta} = (\cos{\theta} + i\sin{\theta})^5 = cos ⁡ 5 θ + i 5 cos ⁡ 4 θ sin ⁡ θ − 10 cos ⁡ 3 θ sin ⁡ 2 θ − i 10 cos ⁡ 2 θ sin ⁡ 3 θ + 5 cos ⁡ θ sin ⁡ 4 θ + i sin ⁡ 5 θ = \cos^5{\theta} + i 5 \cos ^4 {\theta} \sin {\theta} - 10 \cos ^3 {\theta} \sin ^2 {\theta} - i 10 \cos ^2 {\theta} \sin ^3 {\theta} + 5 \cos {\theta} \sin ^4 {\theta} + i \sin ^5 {\theta}

Equating the imaginary part:

sin ⁡ 5 θ = 5 cos ⁡ 4 θ sin ⁡ θ − 10 cos ⁡ 2 θ sin ⁡ 3 θ + sin ⁡ 5 θ \sin {5\theta} = 5 \cos ^4 {\theta} \sin {\theta} - 10 \cos ^2 {\theta} \sin ^3 {\theta} + \sin ^5 {\theta}

= 5 sin ⁡ θ ( cos ⁡ 4 θ − 2 cos ⁡ 2 θ sin ⁡ 2 θ + 1 5 sin ⁡ 4 θ ) \quad \quad = 5\sin {\theta} (\cos ^4 {\theta} - 2 \cos ^2 {\theta} \sin ^2 {\theta} + \frac {1}{5} \sin ^4 {\theta})

= 5 sin ⁡ θ ( cos ⁡ 4 θ − 2 cos ⁡ 2 θ sin ⁡ 2 θ + sin ⁡ 4 θ − 4 5 sin ⁡ 4 θ ) \quad \quad = 5\sin {\theta} (\cos ^4 {\theta} - 2 \cos ^2 {\theta} \sin ^2 {\theta} + \sin^4{\theta} - \frac {4}{5} \sin ^4 {\theta})

= 5 sin ⁡ θ ( [ cos ⁡ 2 θ − sin ⁡ 2 θ ] 2 − 4 5 sin ⁡ 4 θ ) \quad \quad = 5\sin {\theta} ( [\cos ^2 {\theta} - \sin ^2 {\theta} ] ^2 - \frac {4}{5} \sin ^4 {\theta})

= 5 sin ⁡ θ ( [ 1 − 2 sin ⁡ 2 θ ] 2 − 4 5 sin ⁡ 4 θ ) \quad \quad = 5\sin {\theta} ([1 - 2 \sin ^2 {\theta} ] ^2 - \frac {4}{5} \sin ^4 {\theta})

= 5 sin ⁡ θ ( 1 − 4 sin ⁡ 2 θ + 16 5 sin ⁡ 4 θ ) \quad \quad = 5\sin {\theta} (1 - 4 \sin ^2 {\theta} + \frac {16}{5} \sin ^4 {\theta})

= 5 sin ⁡ θ ( 1 − [ 2 + 2 5 ] sin ⁡ 2 θ ) ( 1 − [ 2 − 2 5 ] sin ⁡ 2 θ ) \quad \quad = 5\sin {\theta} (1 - [2+\frac {2}{\sqrt{5}}] \sin ^2 {\theta} ) (1 - [2-\frac {2}{\sqrt{5}}] \sin ^2 {\theta} )

⇒ sin ⁡ 2 p = 1 2 + 2 5 sin ⁡ 2 q = 1 2 − 2 5 \Rightarrow \sin^2 {p} = \dfrac {1} {2+\frac {2}{\sqrt{5}}} \quad \quad \sin^2 {q} = \dfrac {1} {2-\frac {2}{\sqrt{5}}}

⇒ p = 0.6283 , q = 1.2566 ⇒ p + q = 1.885 \Rightarrow p = 0.6283, \quad q = 1.2566 \quad \Rightarrow p + q = \boxed {1.885}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Could you please explain how it came?

( cos ⁡ θ + i sin ⁡ θ ) 5 (\cos{\theta} + i\sin{\theta})^5 = cos ⁡ 5 θ + i 5 cos ⁡ 4 θ sin ⁡ θ − 10 cos ⁡ 3 θ sin ⁡ 2 θ − i 10 cos ⁡ 2 θ sin ⁡ 3 θ + 5 cos ⁡ θ sin ⁡ 4 θ + i sin ⁡ 5 θ = \cos^5{\theta} + i 5 \cos ^4 {\theta} \sin {\theta} - 10 \cos ^3 {\theta} \sin ^2 {\theta} - i 10 \cos ^2 {\theta} \sin ^3 {\theta} + 5 \cos {\theta} \sin ^4 {\theta} + i \sin ^5 {\theta}

Anandhu Raj - 6 years, 3 months ago

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Hey, It is just the binomial expansion of ( cos ⁡ θ + ι sin ⁡ θ ) 5 (\cos\theta + \iota \sin\theta)^{5} .

I hope you are aware of binomial expansion of type ( a + b ) n (a+b) ^{n} .

Prakhar Gupta - 6 years, 1 month ago

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Oh! didn't noticed that. Thank You. :)

Anandhu Raj - 5 years, 11 months ago

( a + b ) 5 = a 5 + 5 a 4 b + 10 a 3 b 2 + 10 a 2 b 3 + 5 a b 4 + b 5 (a+b)^5 = a^5 + 5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5

( cos ⁡ θ + i sin ⁡ 2 θ ) 5 = cos ⁡ 5 θ + 5 cos ⁡ 4 θ ( i sin ⁡ θ ) + 10 cos ⁡ 3 θ ( i 2 sin ⁡ 2 θ ) + 10 cos ⁡ 2 θ ( i 3 sin ⁡ 3 θ ) + 5 cos ⁡ θ ( i 4 sin ⁡ 4 θ ) + i 5 sin ⁡ 5 θ = cos ⁡ 5 θ + i 5 cos ⁡ 4 θ sin ⁡ θ − 10 cos ⁡ 3 θ sin ⁡ 2 θ − i 10 cos ⁡ 2 θ sin ⁡ 3 θ + 5 cos ⁡ θ sin ⁡ 4 θ + i sin ⁡ 5 θ (\cos{\theta} + i\sin^2{\theta})^5 \\ = \cos^5{\theta} + 5\cos^4{\theta} (\color{#D61F06}{i}\sin{\theta}) + 10\cos^3{\theta} (\color{#D61F06}{i^2} \sin^2{\theta}) + 10\cos^2{\theta} (\color{#D61F06}{i^3} \sin^3{\theta}) + 5\cos{\theta} (\color{#D61F06}{i^4} \sin^4{\theta}) + \color{#D61F06}{i^5} \sin^5{\theta} \\ = \cos^5{\theta} \color{#D61F06}{+i}5\cos^4{\theta} \sin{\theta} \color{#D61F06}{-} 10\cos^3{\theta} \sin^2{\theta} \color{#D61F06}{-i}10\cos^2{\theta}\sin^3{\theta} \color{#D61F06}{+} 5\cos{\theta}\sin^4{\theta} \color{#D61F06}{+i} \sin^5{\theta}

Chew-Seong Cheong - 5 years, 11 months ago

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