Trigonometry

Geometry Level 3

If sin ⁡ 2 θ + 3 cos ⁡ θ − 2 = 0 \sin^2 \theta + 3\cos\theta - 2 = 0 , find the value of cos ⁡ 3 θ + sec ⁡ 3 θ \cos^3 \theta + \sec^3 \theta .


The answer is 18.

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Aira Thalca by Aira Thalca , 24, Indonesia

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2 solutions

Md Zuhair
Mar 12, 2017

sin ⁡ 2 θ + 3 cos ⁡ θ − 2 = 0 \sin^2 \theta + 3 \cos \theta -2 = 0 So Now 1 − cos ⁡ 2 θ + 3 cos ⁡ θ − 2 = 0 1 - \cos ^2 \theta + 3 \cos \theta -2 = 0

So cos ⁡ 2 θ − 3 cos ⁡ θ + 1 = 0 \cos^2 \theta - 3 \cos \theta +1 = 0

So cos ⁡ θ + sec ⁡ θ = 3 \cos \theta + \sec \theta = 3

So now , ( cos ⁡ θ + sec ⁡ θ ) ( cos ⁡ 2 θ + sec ⁡ 2 θ − 1 ) (\cos \theta + \sec \theta )(\cos^2 \theta + \sec^2 \theta - 1)

now cos ⁡ θ + sec ⁡ θ = 3 \cos \theta + \sec \theta = 3 , Now squaring cos ⁡ 2 θ + sec ⁡ 2 θ = 7 \cos^2 \theta + \sec^2 \theta = 7

putting values We 3.6 3 . 6 = 18 \boxed{18}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

nice solution! did almost the same, instead used ,

cos ⁡ 3 θ + sec ⁡ 3 θ = ( cos ⁡ θ + sec ⁡ θ ) 3 − 3 cos ⁡ θ × sec ⁡ θ ( cos ⁡ θ + sec ⁡ θ ) cos ⁡ 3 θ + sec ⁡ 3 θ = 3 3 − 3 × 3 = 27 − 9 = 18 \begin{aligned} \cos^3\theta+\sec^3\theta&=(\cos\theta+\sec\theta)^3-3\cos\theta\times\sec\theta(\cos\theta+\sec\theta)\\ \cos^3\theta+\sec^3\theta &=3^3-3\times3 =27-9=18\end{aligned}

  • :)

Anirudh Sreekumar - 4 years, 3 months ago

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Thanks Anirudh

Md Zuhair - 4 years, 3 months ago

Better simpler. +1)

Niranjan Khanderia - 4 years, 3 months ago

Nicee, please follow me okay(:

Aira Thalca - 4 years, 3 months ago
Chew-Seong Cheong
Mar 13, 2017

sin ⁡ 2 θ + 2 cos ⁡ θ − 2 = 0 1 − cos ⁡ 2 θ + 3 cos ⁡ θ − 2 = 0 cos ⁡ 2 θ + 1 = 3 cos ⁡ θ Dividing both sides by cos ⁡ θ cos ⁡ θ + sec ⁡ θ = 3 Cubing both sides cos ⁡ 3 θ + 3 cos ⁡ θ + 3 sec ⁡ θ + sec ⁡ 3 θ = 27 cos ⁡ 3 θ + 3 ( 3 ) + sec ⁡ 3 θ = 27 cos ⁡ 3 θ + sec ⁡ 3 θ = 27 − 9 = 18 \begin{aligned} \sin^2 \theta + 2\cos \theta - 2 & = 0 \\ 1 - \cos^2 \theta + 3\cos \theta -2 & =0 \\ \cos^2 \theta + 1 & = 3 \cos \theta & \small \color{#3D99F6} \text{Dividing both sides by }\cos \theta \\ \cos \theta + \sec \theta & = 3 & \small \color{#3D99F6} \text{Cubing both sides} \\ \cos^3 \theta + {\color{#3D99F6}3\cos \theta + 3\sec \theta} + \sec^3 \theta & = 27 \\ \cos^3 \theta + {\color{#3D99F6}3(3)} + \sec^3 \theta & = 27 \\ \cos^3 \theta + \sec^3 \theta & = 27 - 9 \\ & = \boxed{18} \end{aligned}

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Niceee, please follow me okay

Aira Thalca - 4 years, 2 months ago

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