Question

#QualityQuestion
@Trigonometry

$\bf{Prove ~that~:}$
sin(105°) = ( sqrt(6) + sqrt(2) ) / 4​

Answer 1

SOLUTION--;

$sin \:(90°+θ)=cos(θ) \:$

$sin(105°)=cos(15°)=y</p><p></p><p>$

$cos(2θ)=2cos²(θ)−1$

$⟹2y² - 1 = \frac{3}{ \sqrt{2} }$

$2y²= \frac{3}{ \sqrt{2} } + 1$

$y= \frac{ \sqrt{2 + \sqrt{3} } }{2} </p><p>$

$= \frac{ \sqrt{3 + 1} }{2 \times \sqrt[]{2} }$

$= \frac{ \sqrt{6 + \sqrt{2} } }{4}$

Hence proved✓

Answer 2

☆To Prove

$\sin(105°) = \frac{ \sqrt{6} + \sqrt{2} }{4}$

☆Solution

$\sin(105°) = \sin(180° - 105°) = \sin(75°) = \cos(15°)$

Now,

$\frac{ \sqrt{3} }{2} = \cos(30°) = \cos(2.15°) = {\cos}^{2}(15°) - {\sin}^{2}(15°) = 2{\cos}^{2} - 1.$

⇒ ${\sin}^{2}(105°) = \frac{\frac{\sqrt{3}}{2} + 1}{2} = \frac{\sqrt{3} + 2}{4} = \frac{4\sqrt{3} + 8}{16} \frac{2 + 6 + 2 \sqrt{12} }{16} = \frac{ ({ \sqrt{6} + \sqrt{2} })^{2} }{16} .... (1)$

Now, because $\sin(105°) > 0$ , let us take the positive root of (1)

$\sin(105°) = \sqrt{ \frac{ ({ \sqrt{6} + \sqrt{2}) }^{2} }{16} }$

= $\sf\green{\frac{\sqrt{6} + \sqrt{2}}{4}}$

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