I Need A Trigonometric Table

Geometry Level 1

sin ⁡ π 2 , sin ⁡ π 3 , sin ⁡ π 4 , sin ⁡ π 5 \large \sin \dfrac\pi2, \quad \sin \dfrac\pi3, \quad \sin \dfrac\pi4, \quad \sin \dfrac\pi5

The above shows all irrational numbers except for one. Which one?

sin ⁡ π 2 \sin\frac\pi2 sin ⁡ π 3 \sin\frac\pi3 sin ⁡ π 4 \sin\frac\pi4 sin ⁡ π 5 \sin\frac\pi5

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Chung Kevin by Chung Kevin , 45, Australia

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1 solution

sin ⁡ ( π 2 ) = 1 \sin(\dfrac{\pi}{2})=1 which is rational.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

The answer is correct, but you have to prove that the other numbers are irrational numbers. I know that sin ⁡ π 3 = 3 2 and sin ⁡ π 4 = 2 2 \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \text{ and }\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} are irrational numbers, but why? and could you also prove that sin ⁡ π 5 \sin \frac{\pi}{5} is an irrational number , please?. It can be easy for you (even for me too), but I think it's not easy or clear for everybody...

Guillermo Templado - 5 years, 2 months ago

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Actually it is given in the question that the other numbers are irrational but anyway here is a brief outline to calculate sin (π/5):

Take x=π/10 radians.5x=π/2

Do some manipulation to get sin(2x)=cos(3x)

Use the double and triple angle formula for L.H.S and R.H.S respectively and divide both sides by cos(x). Bring all terms to one side of the equation and convert the cos ⁡ 2 ( x ) \cos^2(x) to 1 − sin ⁡ 2 ( x ) 1-\sin^2(x) .

You get the quadratic equation in sin(x): 4 sin ⁡ 2 ( x ) + 2 sin ⁡ ( x ) − 1 = 0 4\sin^2(x)+2\sin(x)-1=0

Solving you get sin(x)= − 1 ± 5 4 \dfrac{-1 \pm \sqrt5}{4}

sin(x) must be positive as it lies on the first quadrant.

Hence sin(x)=sin(π/10)= − 1 + 5 4 \dfrac{-1+\sqrt5}{4}

Now find cos(π/5) using the formula cos ⁡ ( 2 x ) = 1 − 2 sin ⁡ 2 ( x ) \cos(2x)=1-2\sin^2(x) .

You get cos(π/5)= 1 + 5 4 \dfrac{1+\sqrt5}{4}

Now, find sin (π/5)= 1 − cos ⁡ 2 ( π / 5 ) \sqrt{1-\cos^2(π/5)}

you get sin(π/5)= 10 − 2 5 4 \dfrac{\sqrt{10-2\sqrt5}}{4} which is irrational

A Former Brilliant Member - 5 years, 2 months ago

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Yup, I use a lot times your suggestion... when a question says which option of the next statements is the right? and you find one which is right, then the other options have to be false. It is a resource used for me and for other people, and it generally works, but it doesn`t mean we are right.... Anyway, althought there is a small typo , cos ⁡ 2 x \cos 2x isn't 1 − sin ⁡ 2 x 1 - \sin^2 x ,is it? it's ok, don't worry, thank you very much. I really appreciate your effort

Guillermo Templado - 5 years, 2 months ago

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@Guillermo Templado – Yes it was a typo...

My pleasure.

A Former Brilliant Member - 5 years, 2 months ago

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