Long trigo

Geometry Level 4

8 sin ⁡ ( x 4 ) cos ⁡ 5 ( x 4 ) + 8 sin ⁡ 5 ( x 4 ) cos ⁡ ( x 4 ) − 32 sin ⁡ 3 ( x 4 ) cos ⁡ 5 ( x 4 ) + 32 sin ⁡ 5 ( x 4 ) cos ⁡ 3 ( x 4 ) − 16 sin ⁡ 3 ( x 4 ) cos ⁡ 3 ( x 4 ) 8\sin\left(\dfrac{x}{4}\right){\cos}^5\left(\dfrac{x}{4}\right) + 8{\sin}^5\left(\dfrac{x}{4}\right)\cos\left(\dfrac{x}{4}\right) - 32{\sin}^3\left(\dfrac{x}{4}\right){\cos}^5\left(\dfrac{x}{4}\right) + 32{\sin}^5\left(\dfrac{x}{4}\right){\cos}^3\left(\dfrac{x}{4}\right) - 16{\sin}^3\left(\dfrac{x}{4}\right){\cos}^3\left(\dfrac{x}{4}\right)

Find the value of the above expression at x = 30 ∘ x = {30}^\circ . Scroll right to see the full expression if necessary. Give your answer to 3 decimal places.


The answer is 0.899.

Ashish Menon by Ashish Menon , 20, India

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2 solutions

Ashish Menon
Jul 13, 2016

8 sin ⁡ ( x 4 ) cos ⁡ 5 ( x 4 ) + 8 sin ⁡ 5 ( x 4 ) cos ⁡ ( x 4 ) − 32 sin ⁡ 3 ( x 4 ) cos ⁡ 5 ( x 4 ) + 32 sin ⁡ 5 ( x 4 ) cos ⁡ 3 ( x 4 ) − 16 sin ⁡ 3 ( x 4 ) cos ⁡ 3 ( x 4 ) = 8 sin ⁡ ( x 4 ) cos ⁡ ( x 4 ) ( cos ⁡ 4 ( x 4 ) + sin ⁡ 4 ( x 4 ) − 4 sin ⁡ 2 ( x 4 ) cos ⁡ 4 ( x 4 ) + 4 sin ⁡ 4 ( x 4 ) cos ⁡ 2 ( x 4 ) − 2 sin ⁡ 2 ( x 4 ) cos ⁡ 2 ( x 4 ) ) = 4 × 2 sin ⁡ ( x 4 ) cos ⁡ ( x 4 ) ( cos ⁡ 4 ( x 4 ) + sin ⁡ 4 ( x 4 ) − 4 sin ⁡ 2 ( x 4 ) cos ⁡ 4 ( x 4 ) + 4 sin ⁡ 4 ( x 4 ) cos ⁡ 2 ( x 4 ) − sin ⁡ 2 ( x 4 ) cos ⁡ 2 ( x 4 ) − sin ⁡ 2 ( x 4 ) cos ⁡ 2 ( x 4 ) ) = 4 sin ⁡ ( x 2 ) cos ⁡ ( x 2 ) ( ( cos ⁡ 2 ( x 4 ) − sin ⁡ 2 ( x 4 ) ) ( ( cos ⁡ 2 ( x 4 ) − sin ⁡ 2 ( x 4 ) ) − ( 4 sin ⁡ 2 ( x 4 ) cos ⁡ 2 ( x 4 ) ) ) ) = 2 × 2 sin ⁡ ( x 2 ) cos ⁡ ( x 2 ) ( cos ⁡ 2 ( x 2 ) − sin ⁡ 2 ( x 2 ) ) = 2 sin ⁡ ( x ) cos ⁡ ( x ) = sin ⁡ ( 2 x ) ∴ at x = 30°, the expression becomes sin ⁡ ( 60 ° ) = 0.866 8\sin\left(\dfrac{x}{4}\right){\cos}^5\left(\dfrac{x}{4}\right) + 8{\sin}^5\left(\dfrac{x}{4}\right)\cos\left(\dfrac{x}{4}\right) - 32{\sin}^3\left(\dfrac{x}{4}\right){\cos}^5\left(\dfrac{x}{4}\right) + 32{\sin}^5\left(\dfrac{x}{4}\right){\cos}^3\left(\dfrac{x}{4}\right) - 16{\sin}^3\left(\dfrac{x}{4}\right){\cos}^3\left(\dfrac{x}{4}\right) \\ \\ = 8\sin\left(\dfrac{x}{4}\right)\cos\left(\dfrac{x}{4}\right)\left({\cos}^4\left(\dfrac{x}{4}\right) + {\sin}^4\left(\dfrac{x}{4}\right) - 4{\sin}^2\left(\dfrac{x}{4}\right){\cos}^4\left(\dfrac{x}{4}\right) + 4{\sin}^4\left(\dfrac{x}{4}\right){\cos}^2\left(\dfrac{x}{4}\right) - 2{\sin}^2\left(\dfrac{x}{4}\right){\cos}^2\left(\dfrac{x}{4}\right)\right) \\ \\ = 4 × 2\sin\left(\dfrac{x}{4}\right)\cos\left(\dfrac{x}{4}\right)\left({\cos}^4\left(\dfrac{x}{4}\right) + {\sin}^4\left(\dfrac{x}{4}\right) - 4{\sin}^2\left(\dfrac{x}{4}\right){\cos}^4\left(\dfrac{x}{4}\right) + 4{\sin}^4\left(\dfrac{x}{4}\right){\cos}^2\left(\dfrac{x}{4}\right) - {\sin}^2\left(\dfrac{x}{4}\right){\cos}^2\left(\dfrac{x}{4}\right) - {\sin}^2\left(\dfrac{x}{4}\right){\cos}^2\left(\dfrac{x}{4}\right)\right) \\ \\ = 4\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)\left(\left({\cos}^2\left(\dfrac{x}{4}\right) - {\sin}^2\left(\dfrac{x}{4}\right)\right)\left(\left({\cos}^2\left(\dfrac{x}{4}\right) - {\sin}^2\left(\dfrac{x}{4}\right)\right) - \left(4{\sin}^2\left(\dfrac{x}{4}\right){\cos}^2\left(\dfrac{x}{4}\right)\right)\right)\right) \\ \\ = 2 × 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)\left({\cos}^2\left(\dfrac{x}{2}\right) - {\sin}^2\left(\dfrac{x}{2}\right)\right) \\ \\ = 2\sin\left(x\right)\cos\left(x\right) \\ \\ = \sin\left(2x\right) \\ \\ \therefore \text{at x = 30°, the expression becomes} \sin\left({60}^°\right) \\ \\ = \color{#3D99F6}{\boxed{0.866}}

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The second term of expression must be edited to : 8{sin(x/4)}^5 . cos(x/4) instead of 8{sin(x/4)}^5 only.

If not, the correct answer must be : 0.899 instead of 0.866

Note that the size of solution image must be decreased to adjus the screen.

thanks.

Ahmad Saad - 4 years, 10 months ago

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Thanks for mentioning. Let me think of the question for a bit and I will inform you. The answer provides a 0.3% error so I dont think it matters much. Note that I had written the complete solution but due to some bug or something, the solution is going out of the screen. Its not possible for ordinary members to make such edits, the staff should look into this matter.

Ashish Menon - 4 years, 10 months ago

The procedure of your solution is correct if you add (cos{x/4}) as a multiple to the second term [ 8 sin^5 (x/4) ] to will be 8 sin^5 (x/4)*cos(x/4)

If you review the first row of your solution , a common factor of all terms of equation is [ 8 sin(x/4) cos(x/4) ] but the remainder of second term becomes sin^4 (x/4)

this means that the orient of this term is 8 sin^5 (x/4) * cos(x/4) not 8 sin^5 only.

I wish to explain that.

Ahmad Saad - 4 years, 10 months ago

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Yes, its cos(x/4) right, I was confused when you first wrote sin (x/4) thatnks, I will change it. Sorr about the screen, I cant do anything about that.

Ashish Menon - 4 years, 10 months ago

Your 4th step is wrong. Note that cos ⁡ 2 ( x / 4 ) − sin ⁡ 2 ( x / 4 ) − 4 sin ⁡ 2 ( x / 4 ) cos ⁡ 2 ( x / 4 ) = 1 \cos^2 (x/4) - \sin^2 (x/4) - 4\sin^2 (x/4) \cos^2(x/4) =1 is not true when x = 3 0 ∘ x = 30^\circ .

Pi Han Goh - 3 years, 4 months ago

8 sin ⁡ ( x 4 ) cos ⁡ 5 ( x 4 ) + 8 sin ⁡ 5 ( x 4 ) cos ⁡ ( x 4 ) − 32 sin ⁡ 3 ( x 4 ) cos ⁡ 5 ( x 4 ) + 32 sin ⁡ 5 ( x 4 ) cos ⁡ 3 ( x 4 ) − 16 sin ⁡ 3 ( x 4 ) cos ⁡ 3 ( x 4 ) 8{\sin}\left(\dfrac{x}{4}\right){\cos}^5\left(\dfrac{x}{4}\right) + 8{\sin}^5\left(\dfrac{x}{4}\right)\cos\left(\dfrac{x}{4}\right) - 32{\sin}^3\left(\dfrac{x}{4}\right){\cos}^5\left(\dfrac{x}{4}\right) + 32{\sin}^5\left(\dfrac{x}{4}\right){\cos}^3\left(\dfrac{x}{4}\right) - 16{\sin}^3\left(\dfrac{x}{4}\right){\cos}^3\left(\dfrac{x}{4}\right)\\ L e t T = ( x 4 ) . . . . . . 2 ∗ sin ⁡ A ∗ cos ⁡ A = sin ⁡ 2 A . . . . . . . cos ⁡ 2 A − sin ⁡ 2 A = cos ⁡ 2 A s i n 4 A + c o s 4 A = ( cos ⁡ 2 A + sin ⁡ 2 A ) 2 − 2 ∗ cos ⁡ 2 A ∗ sin ⁡ 2 A = 1 − 1 2 sin ⁡ 2 A . ∴ f ( x ) = f ( T ) = 8 sin ⁡ T ∗ cos ⁡ 5 T + 8 sin ⁡ 5 T ∗ cos ⁡ T − 32 sin ⁡ 3 T ∗ cos ⁡ 5 T + 32 sin ⁡ 5 T ∗ cos ⁡ 3 T − 16 sin ⁡ 3 T ∗ cos ⁡ 3 T = 4 sin ⁡ 2 T ∗ ( sin ⁡ 4 T + cos ⁡ 4 T ) − 4 sin ⁡ 3 2 T ∗ ( cos ⁡ 2 T − sin ⁡ 2 T ) − 2 sin ⁡ 3 2 T = 4 sin ⁡ 2 T ∗ ( 1 − 1 2 sin ⁡ 2 T ) − 4 sin ⁡ 3 2 T ∗ cos ⁡ 2 T − 2 sin ⁡ 3 2 T = 4 sin ⁡ 2 T − 2 ∗ sin ⁡ 3 2 T ) − 4 sin ⁡ 3 2 T ∗ cos ⁡ 2 T − 2 sin ⁡ 3 2 T = 4 sin ⁡ 15 − 4 ∗ sin ⁡ 3 15 − 2 sin ⁡ 2 15 ∗ sin ⁡ 30 = 4 sin ⁡ 15 ∗ cos ⁡ 2 15 − ( 1 − cos ⁡ 2 15 ) = cos ⁡ 2 15 ∗ ( 4 sin ⁡ 15 + 1 ) − 1.............. O R . . . . . . . . . . 4 sin ⁡ 15 ∗ cos ⁡ 2 15 − sin ⁡ 2 15 = 0.898938 Let~T=\left(\dfrac{x}{4}\right)......2*{\sin}A*{\cos}A={\sin}2A.......{\cos}^2A-{\sin}^2A={\cos}2A\\ sin^4A+cos^4A=({\cos}^2A+{\sin}^2A)^2-2*{\cos}^2A*{\sin}^2A=1-\frac 1 2{\sin}2A.\\ \therefore ~f(x)=f(T)=8{\sin}T*{\cos}^5T +8{\sin}^5T*{\cos}T-32{\sin}^3T*{\cos}^5T+32{\sin}^5T*{\cos}^3T-16{\sin}^3T*{\cos}^3T\\ = 4{\sin}2T*( {\sin}^4T+{\cos}^4T ) -4{\sin}^3 2T*({\cos}^2T -{\sin}^2T)-2{\sin}^3 2T \\ =4{\sin}2T* (1-\frac 1 2{\sin}2T) -4{\sin}^3 2T*{\cos} 2T-2{\sin}^3 2T\\ =4{\sin}2T- 2*{\sin}^3 2T ) -4{\sin}^3 2T*{\cos} 2T-2{\sin}^3 2T \\ =4{\sin}15-4*{\sin}^3 15 -2{\sin}^2 15 *{\sin}30 \\ = 4 {\sin}15*{\cos}^2 15 - (1-{\cos}^2 15) \\ ={\cos}^2 15*(4{\sin}15 +1) -1 ..............OR..........4{\sin}15*{\cos}^2 15 ~-~{\sin}^2 15 \\ =0.898938

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