Question

Guys please answer this question​

Answer 1

Answer :

1st Option : (2 - √3)/2

Note:

★ sin(A + B) = sinA•cosB + cosA•sinB

★ sin(A – B) = sinA•cosB – cosA•sinB

★ cos(A + B) = cosA•cosB – sinA•sinB

★ cos(A – B) = cosA•cosB + sinA•sinB

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

★ sin(A + B) – sin(A – B) = 2cosA•sinB

★ cos(A + B) + cos(A – B) = 2cosA•cosB

★ cos(A + B) – cos(A – B) = - 2sinA•sinB

Solution :

• Solution : To find : 2sin15°cos75° = ?

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

Hence ,

2sin15°cos75° = (2 - √3)/2

Answer 2

$\underline\blue{\bold{Given \: Question :- }}$

$\bf \:Evaluate : 2 sin 15° cos 75°$

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$\huge \orange{AηsωeR} ✍$

${ \boxed {\bf{Formula \: used :- }}}$

$\bf \:sin(x - y) = sinx \:cosy - siny \: cosx$

$\bf \:cos(x + y) = cosx \: cosy - sinx \: siny$

$\bf \: {2sin}^{2} x = 1 - cos2x$

$\bf \: 2 \: sinx \: cosy \: = sin(x + y) +\: sin(x - y)$

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$\bf \:\large \red{Method :- 1} ✍$

$\bf \:  ⟼ 2 sin 15° cos 75°$

$\bf \:  ⟼ 2 sin 15° cos (90° - 15°)$

$\bf \:  ⟼ 2 sin 15° \times sin 15°$

$\bf \:  ⟼ 2 {sin}^{2} 15°$

$\bf \:  ⟼ 1 - cos30°$

$\bf \:  ⟼ 1 - \dfrac{ \sqrt{3} }{2}$

$\bf \:  ⟼ \dfrac{2 - \sqrt{3} }{2}$

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$\bf \:\large \red{Method :- 2} ✍$

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We first find the value of sin 15° and then value of cos 75°.

$\begin{gathered}\bf\blue{Now,}\end{gathered}$

$\bf \:Consider \: sin15°$

$\bf \:  ⟼ sin(45° - 30°)$

$\bf \: ⟼sin45° \: cos30° - cos45° \: sin30°$

$\bf \:  ⟼ \dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} - \dfrac{1}{ \sqrt{2} } \times \dfrac{1}{2}$

$\bf \:  ⟼ \dfrac{ \sqrt{3} - 1}{2 \sqrt{2} }$

$\begin{gathered}\bf\green{Now,}\end{gathered}$

$\bf \:  ⟼ Consider \: cos75°$

$\bf \:  ⟼ cos(45° + 30°)$

$\bf \: ⟼cos45° \: cos30° - sin45° \: sin30°$

$\bf \:  ⟼ \dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} - \dfrac{1}{ \sqrt{2} } \times \dfrac{1}{2}$

$\bf \:  ⟼ \dfrac{ \sqrt{3} - 1}{2 \sqrt{2} }$

$\begin{gathered}\bf\red{Now,}\end{gathered}$

$\bf \:Consider \: 2 sin 15° cos 75°$

$\bf \:  ⟼ 2 \times \dfrac{ \sqrt{3} - 1 }{2 \sqrt{2} } \times \dfrac{ \sqrt{3} - 1}{2 \sqrt{2} }$

$\bf \:  ⟼ \dfrac{ {( \sqrt{3} - 1)}^{2} }{4}$

$\bf \:  ⟼ \dfrac{3 + 1 - 2 \sqrt{3} }{4}$

$\bf \:  ⟼ \dfrac{4 - 2 \sqrt{3} }{4}$

$\bf \:  ⟼ \dfrac{2(2 - \sqrt{3} )}{4}$

$\bf \:  ⟼ \dfrac{2 - \sqrt{3} }{2}$

$\large{\boxed{\boxed{\bf{Option \: (a) \: is \: correct}}}}$

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$\bf \:\large \red{Method : 3} ✍$

$\bf \:Consider \: 2 sin 15° cos 75°$

$\bf \:  ⟼ sin(15° + 75°) + sin(15° - 75°)$

$\bf \: ⟼sin90° + sin( - 60°)$

$\bf \:  ⟼ 1 - sin60°$

$\bf \:  ⟼ 1 - \dfrac{ \sqrt{3} }{2}$

$\bf \:  ⟼ \dfrac{2 - \sqrt{3} }{2}$

$\large{\boxed{\boxed{\bf{Option \: (a) \: is \: correct}}}}$

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$\large \blue{\bf \: Explore \: more } ✍$

Trigonometry Formulas

• sin(−θ) = −sin θ
• cos(−θ) = cos θ
• tan(−θ) = −tan θ
• cosec(−θ) = −cosecθ
• sec(−θ) = sec θ
• cot(−θ) = −cot θ

Product to Sum Formulas

• sin x sin y = 1/2 [cos(x–y) − cos(x+y)]
• cos x cos y = 1/2[cos(x–y) + cos(x+y)]
• sin x cos y = 1/2[sin(x+y) + sin(x−y)]
• cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

• sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]
• sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]
• cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]
• cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

• cos (A + B) = cos A cos B – sin A sin B
• cos (A – B) = cos A cos B + sin A sin B
• sin (A+B) = sin A cos B + cos A sin B
• sin (A -B) = sin A cos B – cos A sin B
• tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]
• tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]
• cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]
• cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]
• cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A
• sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

• sin2A = 2sinA cosA = [2tan A /(1+tan²A)]
• cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]
• tan 2A = (2 tan A)/(1-tan²A)