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```
find the value of sin48.cos78=? a.√5+1/8 b.1+√5/8 c.1-√5/8 d.√5-1/8 the question is correct please give solution of this question..please
find the value of sin48.cos78=?a.√5+1/8b.1+√5/8c.1-√5/8d.√5-1/8the question is correct please give solution of this question..please

```
2 years ago

kkbisht
90 Points
```							Correct answer is option no (d) Solution: sin48. cos78=sin48.cos(90-12) Now using only  Cos(A-B)  Trigonometric-formula we have sin48( cos90. cos12+ sin90.sin12)= sin48.(0.cos12 + 1.sin12)=sin48.sin12Now multiply and divide by 2 the last expression= 1/2(2 sin48 sin12)= 1/2(cos(48-12) + cos(48+12))=1/2(cos36-cos60)=1/2{(root5 +1)/4-1/2) = 1/2{(root5+1)-2)/4= 1/8(root5 -1)  as  sinB= Cos(A-B) - Cos(A+B))by : kkbisht
```
2 years ago
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### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions