Trignometry - 7

Geometry Level 4

3 sin ⁡ 2 θ + 7 cos ⁡ θ sin ⁡ 2 θ + 91 cos ⁡ 2 θ sin ⁡ 3 θ 3 \sin^2\theta + 7 \cos\theta \sin2\theta + 91\cos^2\theta \sin3\theta is expressed as a polynomial in terms of sin ⁡ θ \sin\theta . Then the product of the numerical value of the greatest coefficient of the polynomial and the highest degree of polynomial is equal to?


The answer is 3255.

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Kunal Joshi by Kunal Joshi , 24, India

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2 solutions

Chew-Seong Cheong
Oct 14, 2014

3 sin ⁡ 2 θ + 7 cos ⁡ θ sin ⁡ 2 θ + 91 cos ⁡ 2 θ sin ⁡ 3 θ 3\sin^2{\theta}+7\cos{\theta}\sin{2\theta}+91\cos^2{\theta} \sin{3\theta}

= 3 sin ⁡ 2 θ + 14 sin ⁡ θ cos ⁡ 2 θ + 91 ( 1 − sin ⁡ 2 θ ) ( sin ⁡ θ cos ⁡ 2 θ + sin ⁡ 2 θ cos ⁡ θ ) =3\sin^2{\theta}+14\sin {\theta}\cos^2 {\theta}+91 (1 - \sin ^2 {\theta} )( \sin {\theta}\cos {2\theta} + \sin {2\theta}\cos {\theta})

= 3 sin ⁡ 2 θ + 14 sin ⁡ θ ( 1 − sin ⁡ 2 θ ) + 91 [ ( 1 − sin ⁡ 2 θ ) ( sin ⁡ θ ( 1 − 2 sin ⁡ 2 θ ) + 2 sin ⁡ θ cos ⁡ 2 θ ] =3\sin^2{\theta}+14\sin {\theta} (1 - \sin^2 {\theta})+91 [(1 - \sin ^2 {\theta} )( \sin {\theta} (1 - 2\sin^2 {\theta}) + 2\sin {\theta}\cos^2 {\theta}]

= 3 sin ⁡ 2 θ + 14 sin ⁡ θ − 14 sin ⁡ 3 θ + 91 [ ( 1 − sin ⁡ 2 θ ) ( sin ⁡ θ ( 1 − 2 sin ⁡ 2 θ ) + 2 sin ⁡ θ cos ⁡ 2 θ ] =3\sin^2{\theta}+14\sin {\theta} - 14 \sin^3 {\theta}+91 [(1 - \sin ^2 {\theta} )( \sin {\theta} (1 - 2\sin^2 {\theta}) + 2\sin {\theta}\cos^2 {\theta}]

= 14 sin ⁡ θ + 3 sin ⁡ 2 θ − 14 sin ⁡ 3 θ + 91 [ ( 1 − sin ⁡ 2 θ ) ( sin ⁡ θ − 2 sin ⁡ 3 θ + 2 sin ⁡ θ ( 1 − sin ⁡ 2 θ ) ] =14\sin {\theta} + 3 \sin^2 {\theta} - 14 \sin^3 {\theta}+91 [(1 - \sin ^2 {\theta} )( \sin {\theta} - 2\sin^3 {\theta} + 2\sin {\theta} ( 1 - \sin ^2 {\theta})]

= 14 sin ⁡ θ + 3 sin ⁡ 2 θ − 14 sin ⁡ 3 θ + 91 [ ( 1 − sin ⁡ 2 θ ) ( 3 sin ⁡ θ − 4 sin ⁡ 3 θ ) ] = 14\sin {\theta} + 3 \sin^2 {\theta} - 14 \sin^3 {\theta}+91 [(1 - \sin ^2 {\theta} )( 3\sin {\theta} - 4\sin^3 {\theta} )]

= 14 sin ⁡ θ + 3 sin ⁡ 2 θ − 14 sin ⁡ 3 θ + 91 ( 3 sin ⁡ θ − 7 sin ⁡ 3 θ + 4 sin ⁡ 5 θ ) = 14\sin {\theta} + 3 \sin^2 {\theta}- 14 \sin^3 {\theta}+91 ( 3 \sin {\theta} - 7 \sin ^3 {\theta} + 4\sin^5 {\theta} )

= 14 sin ⁡ θ + 3 sin ⁡ 2 θ − 14 sin ⁡ 3 θ + 273 sin ⁡ θ − 637 sin ⁡ 3 θ + 364 sin ⁡ 5 θ = 14\sin {\theta} + 3 \sin^2 {\theta} - 14 \sin^3 {\theta} +273 \sin {\theta} - 637 \sin ^3 {\theta} + 364 \sin^5 {\theta}

= 287 sin ⁡ θ + 3 sin ⁡ 2 θ − 651 sin ⁡ 3 θ + 364 sin ⁡ 5 θ = 287 \sin {\theta} + 3 \sin^2 {\theta} - 651 \sin ^3 {\theta} + 364 \sin^5 {\theta}

The required answer = 651 × 5 = 3255 = 651 \times 5 = \boxed {3255}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

This question has so less solvers and such a high level as nobody wanted to expand the hell-of-a-big thing!!

Adarsh Kumar - 6 years, 8 months ago

Problem is simple but both coefficients were multiplied wrongly instead of coefficient and highest degree of x

Kolli Venkateswarlu - 6 years, 5 months ago

You really should reword your problem statement to "the absolute value of the greatest coefficient value" or similar.

tom engelsman - 1 year, 4 months ago

solving and converting all the cos in sin, we get 364(sin(x))^5-651(sin(x))^3+3(sin(x))^2+287sin(x) highest numeric value=651, highest coefficient=5 their product=3255

1 Helpful 0 Interesting 0 Brilliant 0 Confused

somebody pls convert this in LATEX, i dont know how to do that!! thnks

A Former Brilliant Member - 6 years, 8 months ago

We can straight away see that last terms would give highest degree as 5. It will also be included in the term with biggest coefficient.
So simplifying last terms first,
N o t e : C o s 2 ( θ ) = 1 − S i n 2 ( θ ) , . . . . . ( 1 ) . . . . . . a n d . . . . . . . S i n ( 3 θ ) = 3 ∗ S i n ( θ ) − 4 ∗ S i n 3 ( θ ) . . . . . . ( 2 ) S i n ( 2 θ ) = 2 ∗ S i n ( θ ) ∗ C o s ( θ ) . . . . . . . ( 3 ) C o n s i d e r i n g o n l y u s e f u l t e r m s C o s 2 ( θ ) ∗ S i n ( 3 θ ) = 3 ∗ S i n ( θ ) − 4 ∗ S i n 3 ( θ ) − S i n 2 ( θ ) ∗ 3 ∗ S i n ( θ ) + S i n 2 ( θ ) ∗ 4 ∗ S i n 3 ( θ ) = . . . . . . . . . . . . . . . . . . . − ( 4 + 3 ) ∗ S i n 3 ( θ ) + . . . . . . . . . . . . . S o − 91 ∗ 7 ∗ S i n 3 ( θ ) = − 637 ∗ S i n 3 ( θ ) , b u t t h e r e i s S i n 3 ( θ ) i n t h e s e c o n d t e r m a l s o . B y ( 1 ) a n d ( 3 ) 7 ∗ C o s ( θ ) ∗ S i n ( 2 θ ) = 14 ∗ S i n ( θ ) ∗ ( 1 − S i n 2 ( θ ) ) = . . . . . . . . . . . . . − 14 ∗ S i n 3 ( θ ) . ∴ − 637 − 14 = − 651 o f S i n 3 ( θ ) a n d 5 t h d e g r e e . 651 ∗ 5 = 3255. Note:Cos^2(\theta)=1 - Sin^2(\theta),.....(1)......and.......Sin(3\theta)=3*Sin(\theta) - 4*Sin^3(\theta)......(2)\\ Sin(2\theta)=2*Sin(\theta)*Cos(\theta).......(3)~~~~~Considering~only~useful~terms\\ Cos^2(\theta)*Sin(3\theta)\\ =3*Sin(\theta)-{\color{#3D99F6}{4*Sin^3(\theta)~-~~Sin^2(\theta)*3*Sin(\theta)}}+Sin^2(\theta)* 4*Sin^3(\theta)\\ =................... -(4+3)*Sin^3(\theta)+............. \\ So~- 91*7 *Sin^3(\theta) = - 637*Sin^3(\theta),~but~there~is~Sin^3(\theta)~in~the~second~term~also.\\ By~(1)~and~(3)~~7*Cos(\theta)*Sin(2\theta) =14*Sin(\theta)*(1 - Sin^2(\theta))=.............-14*Sin^3(\theta).\\ \therefore~-637-14=-651~of~Sin^3(\theta)~~and~~5th~degree.~~~~651*5=3255.

In the last correct try I forgot the "5", while first two wrong ones I did not !

Niranjan Khanderia - 3 years, 3 months ago

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