Question

sin15/cose15=2-√3 to prove that
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Answer 1

QUESTION:

prove that

$\frac{\sin15°}{\cos15°} = 2 - \sqrt{3} \:$

USED FORMULAS:

sin(A - B) = sinAcosB - cosBsinA

cos(A - B) = cosAcosB - sinAsinB

(a - b)² = a² - 2ab + b²

(a + b)(a - b) = a² - b²

ANSWER:

in the question taking LHS( left hand side)

we know that

sin15° = sin(45° - 30°)

it is in the form of sin(A - B)

using the first formula

sin15° = sin45°cos30° - cos45°sin30°

sin45° = 1/√2 = cos45°

cos30° = √3/2

sin30° = 1/2

$\sin(15°) = \frac{1}{ \sqrt{2} } \times \frac{ \sqrt{3} }{2} - \frac{1}{ \sqrt{2} } \times \frac{1}{2}$

$\sin(15°) = \frac{ \sqrt{ 3} }{2 \sqrt{2} } - \frac{1}{2 \sqrt{2} } \\ \\ \sin(15°) = \frac{ \sqrt{3} - 1 }{2 \sqrt{2} }$

cos15° = cos(45° - 30°)

using the second formula

cos15° = cos45°cos30° + sin45°sin30°

cos45° = 1/√2

cos30° = √3/2

sin45° = 1/√2

sin30° = 1/2

$\cos(15°) = \frac{1}{ \sqrt{2} } \times \frac{ \sqrt{3} }{2} + \frac{1}{ \sqrt{2} } \times \frac{1}{2} \\ \\ \cos(15°) = \frac{ \sqrt{3} }{2 \sqrt{2} } + \frac{1}{2 \sqrt{2} } \\ \\ \cos(15°) = \frac{ \sqrt{3} + 1 }{2 \sqrt{2} }$

to find:

$\frac{ \sin(15°) }{ \cos(15°) }$

substituting the values

$\frac{ \sin(15°) }{ \cos(15°) } = \frac{ \frac{ \sqrt{3} - 1 }{2 \sqrt{2} } }{ \frac{ \sqrt{3 } + 1 }{2 \sqrt{2} } } \\ \\ \\ \frac{ \sin(15°) }{ \cos(15°) } = \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 }$

rationalize the denominator

$\frac{ \sin(15°) }{ \cos(15°) } = \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } \times \frac{ \sqrt{3} - 1 }{ \sqrt{3 } - 1}$

using third formula & fourth formula

$\frac{ \sin(15°) }{ \cos(15°) } = \frac{ {( \sqrt{3 } - 1) }^{2} }{ {( \sqrt{3}) - }^{2} {(1)}^{2} }$

$\frac{ \sin(15°) }{ \cos(15°) } = \frac{4 - 2 \sqrt{3} }{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{2(2 - \sqrt{3} )}{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2 - \sqrt{3}$

LHS = RHS

hence proved