RD Sharma - Mathematics

Book: RD Sharma - Mathematics

Chapter: 7. Values of Trigonometric Functions at Sum of Difference of Angles

Subject: - Class 11th

Q. No. 15 of Exercise 7.1

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15

prove that:

sin2(n + 1)A – sin2nA = sin(2n + 1)A sinA

We know that sin2A – sin2B = sin(A +B) sin(A –B)


HereA =(n + 1)A And B = nA


LHS: sin2(n + 1)A – sin2nA = sin((n + 1)A + nA) sin((n + 1)A – nA)


= sin(nA +A + nA) sin(nA +A – nA)


= sin(2nA +A) sin(A)


= sin(2n + 1)A sinA = RHS


Hence proved.


Chapter Exercises

More Exercise Questions

6

If SinA = 1/2, cosB = , where π/2<A < π And 0 <B < π/2, find the following:

(i) tan(A +B)(ii) tan(A -B)