Question

2sin15°cos75° =?????????????????? ​

Answer 1

As we know that,

• 2sinA•cosB = sin(A + B) + sin(A – B)

If A = 15° and B = 75° , then

= 2sin15°cos75° = sin(15° - 75°) + sin(15° + 75°)

= 2sin15°cos75° = sin(-60°) + sin90°

= 2sin15°cos75° = -sin60° + sin90°

= 2sin15°cos75° = -√3/2 + 1

= 2sin15°cos75° = 1 - √3/2

= 2sin15°cos75° = (2 - √3)/2

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Answer 2

$2 \sin15 \cos(90 - 15) \\ \\ = > 2 \sin15 \sin15 \\ \\ = > 2 { \sin }^{2}15 = 2 {( \sin(45 - 30)) }^{2} \\ \\ = > 2 {( \sin45 \cos30 - \cos45 \sin30) }^{2} \\ \\ = > 2 {( \frac{1}{ \sqrt{2}} \times \frac{ \sqrt{3} }{2} - \frac{1}{ \sqrt{2} } \times \frac{1}{2} )}^{2} \\ \\ = > 2 {( \frac{ \sqrt{3} }{2 \sqrt{2} } - \frac{1}{2 \sqrt{2} }) }^{2} \\ \\ = > 2 {( \frac{ \sqrt{3} - 1}{2 \sqrt{2} }) }^{2} \\ \\ = > 2 \times \frac{ {( \sqrt{3} - 1)}^{2} }{8} \\ \\ = > \frac{3 - 2 \sqrt{3} + 1 }{4} \\ \\ = > \frac{4 - 2 \sqrt{3} }{4} = 1 - \frac{ \sqrt{3} }{2}$