Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
65 views
in Trigonometry by (114k points)
closed by
Find sin 15° .
1. \(\frac{\sqrt3-1}{2\sqrt2}\)
2. \(​​​​\frac{\sqrt3-2}{\sqrt2}\)
3. \(\frac{\sqrt3}{2\sqrt2}\)
4. \(\frac{3-\sqrt2}{\sqrt2}\)

1 Answer

0 votes
by (115k points)
selected by
 
Best answer
Correct Answer - Option 1 : \(\frac{\sqrt3-1}{2\sqrt2}\)

Concept:

sin (A – B) = sinA cosB – cosA sinB

Calculation:

sin 15° = sin (45 – 30)

⇒ sin (45 – 30) = sin 45 cos 30 – cos45 sin30

⇒ sin (45 – 30) = (1/√2) × (√3/2) – (1/√2) × (1/2)

⇒ sin (45 – 30) = \(\frac{\sqrt3}{2\sqrt2}-\frac{1}{2\sqrt2}\)

⇒ sin 15° = \(\frac{\sqrt3-1}{2\sqrt2}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...