RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles are available here for easy reference. In this chapter, students will learn how to derive formulas that express trigonometric function values at the sum or difference of two angles in terms of their individual values. The Maths solutions are prepared in a clear and simple manner, making it easier for students to understand key concepts and formulas. These step-by-step solutions follow the latest Class 11 CBSE syllabus and can be downloaded in PDF format for offline practice.
RD Sharma Solutions Class 11 Maths -Trigonometric Ratios Of Compound Angles Question with Answers
Q1. Find sin(75°) using the sum formula.
Solution: sin(75°) = sin(45° + 30°) = sin45°cos30° + cos45°sin30° = (1/√2)(√3/2) + (1/√2)(1/2) = (√3 + 1)/(2√2)
Q2. Evaluate cos(15°).
Solution: cos(15°) = cos(45° – 30°) = (1/√2)(√3/2) + (1/√2)(1/2) = (√3 + 1)/(2√2)
Q2. Prove that sin(60° + 30°) = 1.
Solution: sin(90°) = 1
Q4. Find tan(75°) using tan(A+B) formula.
Solution: tan75° = tan(45° + 30°) = (1 + 1/√3) / (1 – 1/√3)
Q5. Simplify sin(105°).
Solution: sin105° = sin(60° + 45°) = (√6 + √2)/4
Q6. Express sin(A+B) in terms of sinA, sinB, cosA, cosB.
Solution: sin(A+B) = sinAcosB + cosAsinB
Q7. Express cos(A–B) in terms of sin and cos.
Solution: cos(A–B) = cosAcosB + sinAsinB
Q8. Evaluate cos(75°) using sum formula.
Solution: cos75° = (√3 – 1)/(2√2)
Q9. Find tan(15°).
Solution: tan15° = (1 – 1/√3)/(1 + 1/√3)
Q10. Simplify sin(120°).
Solution: sin120° = √3/2
Q11. Find cos(120°).
Solution: cos120° = -1/2
Q12. Prove sin(180° – A) = sinA.
Solution: Identity holds true
Q13. Prove cos(180° – A) = –cosA.
Solution: Identity holds true
Q14. Show tan(180° + A) = tanA.
Solution: Identity holds true
Q15. Simplify sin(225°).
Solution: sin225° = –1/√2
Q16. Find cos(330°).
Solution: cos330° = √3/2
Q17. Evaluate sin(15°) using subtraction formula.
Solution: sin15° = (√3 – 1)/(2√2)
Q18. Find tan(105°).
Solution: tan105° = (√3 + 1)/(1 – √3)
Q19. Prove sin(A+B)sin(A–B) = sin²A – sin²B.
Solution: Identity holds true
Q20. Simplify cos(300°).
Solution: cos300° = 1/2
Q21. Prove sinAcosB = ½[sin(A+B) + sin(A–B)].
Solution: Identity holds true
Q22. Evaluate tan(225°).
Solution: tan225° = 1
Q23. Find sin(390°).
Solution: sin390° = sin30° = 1/2
Q24. Prove cos(360° – A) = cosA.
Solution: Identity holds true
Q25. Simplify sin(270° – A).
Solution: sin(270° – A) = –cosA