Question

prove that cos 75= \sqrt{3} - 1 \div 2 \sqrt{2}
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Answer 1

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

$\sf \: Prove \: that \: cos75\degree = \dfrac{ \sqrt{3} - 1}{2 \sqrt{2} }$

$\large\underline{\sf{Solution-}}$

We know that,

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

$\bf \: Now, \: Put \: x = 45\degree \: and \: y = 30\degree, \: we \: get$

$\rm :\longmapsto\:cos(45\degree + 30\degree) = cos45\degree \: cos30\degree - sin45\degree \: sin30\degree$

$\rm :\longmapsto\:cos75\degree = \dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} - \dfrac{1}{ \sqrt{2} } \times \dfrac{1}{2}$

$\rm :\longmapsto\:cos75\degree = \dfrac{ \sqrt{3} }{2 \sqrt{2} } - \dfrac{1}{2 \sqrt{2} }$

$\rm :\longmapsto\:cos75\degree = \dfrac{ \sqrt{3} - 1}{2 \sqrt{2} }$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

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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)