Exercise 3.3
Q5. Find the value of
(i) sin 75° (ii) tan 15°
Solution:
(i) sin 75° = sin (45° + 30°) [∵ sin(x + y) = sin x cos y + cos x sin y ]
= sin 45° cos 30° + cos 45° sin 30°
(ii) tan 15°
Solution:
(ii) tan 15° = tan (45° - 30°)
Q10. sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x = cos x
Solution:
LHS = sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x
Or cos(n + 2)x cos(n + 1)x + sin(n + 2)x sin(n + 1)x
Let be the A = (n + 2)x, B = (n + 1)x
Now we have,
LHS = cos A cos B + sin A sin B
= cos ( A - B) [∵ sin(x + y) = sin x cos y + cos x sin y ]
= cos [(n + 2)x - (n + 1)x]
= cos [nx + 2x - (nx + x) ]
= cos [nx + 2x - nx - x ]
= cos x
LHS = RHS
Q12. sin2 6x - sin2 4x = sin 2x sin 10x
Solution:
Sin2 A - sin2B = sin(A + B) sin(A - B)
LHS = sin2 6x - sin2 4x
= sin (6x + 4x) sin(6x - 4x)
= sin 10x sin 2x
= sin 2x sin 10x
LHS = RHS
Q13. cos2 2x - cos2 6x = sin 4x sin 8x
Solution:
LHS = cos2 2x - cos2 6x
= (1 - sin2 2x) - (1 - sin2 6x)
= (1 - sin2 2x - 1 + sin2 6x)
= - sin2 2x + sin2 6x
= sin2 6x - sin2 2x
= sin (6x + 2x) sin(6x -2x)
= sin 10x sin 4x
Q24. cos 4x = 1 – 8 sin2 x cos2 x
Solution:
LHS = cos 2(2x)
= cos 2A [ Let be A = 2x ]
= 1 - 2 sin2 A
= 1 - 2 sin2 2x [Putting A = 2x ]
= 1 - 2 [sin 2x]2
= 1 - 2 [2 sin x cos x ]2
= 1 - 2 [4 sin2 x cos2 x]
= 1 - 8 sin2 x cos2 x
LHS = RHS
Q25. cos 6x = 32 cos6 x – 48cos4 x + 18 cos2 x – 1
Solution:
LHS = cos 6x = cos 3(2x)
= cos 3A [Let be A = 2x]
= 4 cos3 A - 3 cos A
= 4 cos3 2x - 3 cos 2x [Putting A = 2x]
= 4 [cos 2x]3 - 3 [cos 2x]
= 4 [2cos2 x – 1]3 - 3 [2cos2 x – 1]
= 4 [(2cos2 x)3 - 13 - 3(2cos2 x)2 (1) + 3 (2cos2 x)(1)2 ] - 3 [2cos2 x – 1]
= 4 [8 cos6 x - 1 - 12 cos2 x + 6 cos2 x] - 6 cos2 + 3
= 32 cos6 x - 4 - 48 cos2 x + 24 cos2 x - 6 cos2 + 3
= 32 cos6 x - 48 cos2 x + 18 cos2 x - 1
LHS = RHS
