Question

find value of sin thetha

if 8sin thetha = 4+cos thetha find sin thetha. (with steps)

Answer 1

Hey!!!!

Good Evening

Welcome to my 1200 th Brainly answer

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let thetha = A (for easy typing)

Then we have

=> 8sinA = 4 + cosA

=> 8sinA - 4 = cosA -------(1)

Now we know

=> sin²A + cos²A = 1

=> cos²A = 1 - sin²A

$= > \cos(a) = \sqrt{1 - { \sin(a) }^{2} }$

Substituting both the cosA

=> 8sinA - 4 = √1 - sin²A

Squaring both sides

=> (8sinA - 4)² = 1 - sin²A

=> 64sin²A + 16 - 64sinA = 1 - sin²A

=> 65sin²A - 64sinA + 15 = 0

let sinA = x

=> 65x² - 64x + 15 = 0 (easy ?)

=> 65x² - 39x - 25x + 15 = 0 (splitting method)

=> 13x(5x - 3) - 5(5x - 3) = 0

=> (13x - 5)(5x - 3) = 0

Thus 13x - 5 = 0
=> x = 5/13

=> sinA = 5/13 <<<<<<< Answer

Or

=> 5x - 3 = 0
=> x = 3/5

=> sinA = 3/5 <<<<<<<<<Answer

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Hope this helps ✌️

Good Night :-)

Answer 2

Hi,

Here is your answer !
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8sinθ = 4 + cosθ

8sinθ - 4 = cosθ

Squaring both sides, We get

(8sinθ - 4)² = cos²θ

64sin²θ - 64sinθ +16 = 1 - sin²θ

64sin²θ + sin²θ - 64sinθ + 16 - 1 = 0

65sin²θ - 64sinθ + 15 = 0

65sin²θ - 39sinθ - 25sinθ + 15 = 0

13sinθ (5sinθ - 3) - 5 (5sinθ - 3) = 0

(13sinθ - 5) (5sinθ - 3) = 0

=> 13sinθ - 5 = 0 or 5sinθ - 3 = 0

=> sinθ = 5/13 or sinθ = 3/5

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Formula used :

★ (a + b)² = a² + 2ab + b²

★ sin²A + cos²A = 1

=> cos²A = 1 - sin²A