2015 Countdown Problem #26: Trig-gy Trigonometry

Geometry Level 4

Consider the following two functions:

f ( x ) = 1860 sin ⁡ ( 2 π x 2015 ) f(x)=1860\sin(\frac{2\pi x}{2015}) g ( x ) = 775 cos ⁡ ( 2 π x 2015 ) g(x)=775\cos(\frac{2\pi x}{2015})

When the graph y = f ( x ) + g ( x ) y=f(x)+g(x) is plotted against the x x and y y axes, the resulting trigonometric graph has an amplitude of α \alpha , a period of β \beta and a y-intercept of γ \gamma .

Also, the negative x x -intercept of the graph nearest to the y y -axis can be expressed as σ π tan ⁡ − 1 ϵ \frac{\sigma}{\pi} \tan^{-1}{\epsilon} where σ \sigma is an integer.

Determine the value of − β σ α γ ϵ 2 -\frac{\beta\sigma}{\alpha\gamma{\epsilon}^2}

This problem is part of the set 2015 Countdown Problems .


The answer is 65.

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Wee Xian Bin by Wee Xian Bin , 25, Singapore

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1 solution

Wee Xian Bin
Dec 28, 2014

f ( x ) + g ( x ) = 1860 sin ⁡ ( 2 π x 2015 ) + 775 cos ⁡ ( 2 π x 2015 ) f(x)+g(x) = 1860\sin(\frac{2\pi x}{2015})+775\cos(\frac{2\pi x}{2015})

= 1860 2 + 775 2 sin ⁡ ( 2 π x 2015 + tan ⁡ − 1 ( 775 1860 ) ) = \sqrt{{1860}^2+{775}^2}\sin(\frac{2\pi x}{2015} + \tan^{-1} {(\frac{775}{1860}))}

= 2015 sin ⁡ ( 2 π x 2015 + tan ⁡ − 1 ( 5 12 ) ) =2015\sin(\frac{2\pi x}{2015} + \tan^{-1} {(\frac{5}{12}))}

Hence α = 2015 \alpha=2015 , β = 2015 \beta=2015 , γ = 775 \gamma=775

To find the required x x -intercept we need to solve

sin ⁡ ( 2 π x 2015 + tan ⁡ − 1 ( 5 12 ) ) = 0 \sin(\frac{2\pi x}{2015} + \tan^{-1} {(\frac{5}{12}))}=0

By considering special angles you would arrive at

x = − 2015 tan ⁡ − 1 ( 5 12 ) 2 π x=-\frac{2015 \tan^{-1}{(\frac{5}{12})}} {2\pi}

Let u = 1 2 tan ⁡ − 1 5 12 u=\frac{1}{2} \tan^{-1} \frac{5}{12}

tan ⁡ − 1 5 12 = 2 u \tan^{-1} \frac{5}{12}=2u

5 12 = tan ⁡ 2 u = 2 tan ⁡ u 1 − tan ⁡ 2 u \frac{5}{12}=\tan{2u}=\frac{2\tan u}{1-\tan^{2} u}

By considering the quadratic equation you would arrive at

x x -intercept= ( − 2015 π ) tan ⁡ − 1 ( 1 5 ) ⇒ σ = − 2015 , ϵ = 1 5 (-\frac{2015}{\pi}) \tan^{-1}(\frac{1}{5}) \Rightarrow \sigma=-2015, \epsilon=\frac{1}{5}

− β σ α γ ϵ 2 = − − ( 2015 ) 2 ( 2015 ) ( 775 ) ( 1 5 ) 2 = 65 -\frac{\beta \sigma}{\alpha \gamma {\epsilon}^2}=-\frac{-(2015)^2}{(2015)(775)(\frac{1}{5})^2}=65

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I think you should prove atleast σ \sigma and ϵ \epsilon .

Pranjal Jain - 6 years, 5 months ago

We could also have σ = 2015 \sigma=2015 and ϵ = − 1 5 \epsilon=-\frac{1}{5} , best to specify their respective signs in the question since you could also give − 65 -65 as a valid answer as well.

Jared Low - 6 years, 5 months ago

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