Triggered Trigonometry #2

Algebra Level 2

Find the smallest positive integer value of θ \theta in degrees such that 8 1 s i n 2 θ + 8 1 c o s 2 θ = 30. 81^{sin^2{\theta}} + 81^{cos^2{\theta}} = 30.


The answer is 30.

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Parag Zode by Parag Zode , 24, India

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2 solutions

Chris Sapiano
Nov 12, 2019

With the identity cos ⁡ 2 θ + sin ⁡ 2 θ = 1 \cos^2\theta + \sin^2\theta = 1

8 1 sin ⁡ 2 θ + 8 1 cos ⁡ 2 θ = 8 1 1 − cos ⁡ 2 θ + 8 1 cos ⁡ 2 θ = 81 8 1 cos ⁡ 2 θ + 8 1 cos ⁡ 2 θ 81^{\sin^2\theta} + 81^{\cos^2\theta} = 81^{1-\cos^2\theta} + 81^{\cos^2\theta} = \frac{81}{81^{\cos^2\theta}} + 81^{\cos^2\theta}

Let x = 8 1 cos ⁡ 2 θ x = 81^{\cos^2\theta}

81 x + x = 30 \frac{81}{x} + x = 30

x 2 − 30 x + 81 = 0 x^2-30x+81 = 0

( x − 3 ) ( x − 27 ) = 0 (x-3)(x-27) = 0 . Therefore 8 1 cos ⁡ 2 θ = 3 81^{\cos^2\theta} = 3 and 8 1 cos ⁡ 2 θ = 27 81^{\cos^2\theta} = 27

3 4 cos ⁡ 2 θ = 3 1 3^{4\cos^2\theta} = 3^1 and 3 4 cos ⁡ 2 θ = 3 3 3^{4\cos^2\theta} = 3^3

4 cos ⁡ 2 θ = 1 {4\cos^2\theta} = 1 then θ = 60 \theta = 60

4 cos ⁡ 2 θ = 3 {4\cos^2\theta} = 3 then θ = 30 \theta = \boxed{30}

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Akshay Sant
Mar 31, 2015

The theta Should Be 0°<x<90° Because asSine and Cosine terms Are squared x=150° also be the answer For your given data .. What you say??

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I've done by this method:-

Let t = 8 1 s i n 2 θ + 8 1 c o s 2 θ t= 81^{sin^{2}\theta} + 81^{cos^{2}\theta}

So, t + 81 t = 30 t + \dfrac{81}{t}=30

Solving this, we get a quadratic and thus we get t = 27 t=27 or t = 3 t=3 .

Now, 8 1 s i n 2 θ = 3 4 s i n 2 θ = 3 81^{sin^{2}\theta} =3^{4sin^{2}\theta} =3

Now, 4 s i n 2 θ = 1 4sin^{2}\theta=1

So, s i n θ = ± 1 2 sin\theta = \pm\dfrac{1}{2}

∴ θ = π 6 \therefore \theta = \dfrac{\pi}{6}

@Akshay Sant , I think since the range of θ \theta is between 0 ∘ 0^\circ and 9 0 ∘ 90^\circ , the angle cannot be 15 0 ∘ 150^\circ . But, the θ \theta can be 6 0 ∘ 60^\circ i.e. θ = π 3 \theta=\dfrac{\pi}{3} . Try yourself.. ⌣ ¨ \ddot\smile

Parag Zode - 6 years, 2 months ago

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I did the same thing, but I took t = 8 1 cos ⁡ 2 θ t=81^{\cos^{2}\theta} , and I came to cos ⁡ θ = 1 2 , 3 4 or − 1 2 \cos\theta=\frac{1}{2},~\frac{3}{4}~\text{or}~-\frac{1}{2} . The smallest value thus obtained is 6 0 ∘ 60^{\circ} . Could you help?

Omkar Kulkarni - 6 years, 1 month ago

Thanks! I've edited the question.

Parag Zode - 6 years, 2 months ago

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Yes... and Yes as You said x can be 60° also.. Thanks for pointing Out Good Day.

Akshay Sant - 6 years, 2 months ago

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Your Welcome! ⌣ ¨ \ddot\smile

Parag Zode - 6 years, 2 months ago

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