Question

cos 15°-cos 75° =
a) whole root 3/2 b)root 3/2 c) 1/2 d) 1/root 2
​

Answer 1

Answer:

hey here is your solution

pls mark it as brainliest

Step-by-step explanation:

$To \: find = value \: of \: cos \: 15 - cos \: 45 \\ \\ so \: cos \: 15 \: can \: be \: written \: as \: cos(45 - 30) \\ and \: similarly \: cos \: 75 \: as \: cos \: (45 + 30) \\ \\ so \: here \: cos \: (45 - 30) \: and \: cos \: (45 + 30) \: are \: in \: \\ form \: \: cos \: (A - B) \: and \: cos \: (A + B) \: respectively \\ \\ so \: we \: know \: that \\ cos \: (A - B) = cos \: A.cos \: B \: + \: sin \: A.sin \: B \\ \\ cos \: (A + B) = cos \: A.cos \: B \: - \: sin \: A.sin \: B$

$hence \: then \: accordingly \\ cos \: 15 - cos \: 75 \\ = cos \: 45.cos \: 30 \: + \: sin \: 30.sin \: 45 \: - \: (cos \: 30.cos \: 45 \: - \: sin \: 30.sin \: 45) \\ = cos \: 30.cos \: 45 \: + \: sin \: 30.sin \: 45 \: - \: cos \: 30.cos \: 45 + sin \: 30.sin \: 45 \\ = sin \: 30.sin \: 45 + sin \: 30.sin \: 45 \\ = 2.sin \: 30.sin \: 45 \\ = 2 \times \frac{1}{2} \times \frac{1}{ \sqrt{2} } \\ since \: sin \: 30 = \frac{1}{2} \\ \\ sin \: 45 = \frac{1}{ \sqrt{2} } \\ \\ thus \: then \\ = 1 \times \frac{1}{ \sqrt{2} } \\ \\ = \frac{1}{ \sqrt{2} } \\ \\ hence \: cos \: 15 \: - \: cos \: 75 = \frac{1}{ \sqrt{2} }$