Prove that:
1.
Solution:
2.
Solution:
Here
= 1/2 + 4/4
= 1/2 + 1
= 3/2
= RHS
3.
Solution:
4.
Solution:
5. Find the value of:
(i) sin 75o
(ii) tan 15o
Solution:
(ii) tan 15°
It can be written as
= tan (45° – 30°)
Using formula
Prove the following:
6.
Solution:
7.
Solution:
8.
Solution:
9.
Solution:
Consider
It can be written as
= sin x cos x (tan x + cot x)
So we get
10. 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
11.
Solution:
Consider
Using the formula
12. sin2 6x – sin2 4x = sin 2x sin 10x
Solution:
13. cos2 2x – cos2 6x = sin 4x sin 8x
Solution:
We get
= [2 cos 4x cos (-2x)] [-2 sin 4x sin (-2x)]
It can be written as
= [2 cos 4x cos 2x] [–2 sin 4x (–sin 2x)]
So we get
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= RHS
14. sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x
Solution:
By further simplification
= 2 sin 4x cos (– 2x) + 2 sin 4x
It can be written as
= 2 sin 4x cos 2x + 2 sin 4x
Taking common terms
= 2 sin 4x (cos 2x + 1)
Using the formula
= 2 sin 4x (2 cos2 x – 1 + 1)
We get
= 2 sin 4x (2 cos2 x)
= 4cos2 x sin 4x
= R.H.S.
15. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution:
Consider
LHS = cot 4x (sin 5x + sin 3x)
It can be written as
Using the formula
= 2 cos 4x cos x
Hence, LHS = RHS.
16.
Solution:
Consider
Using the formula
17.
Solution:
18.
Solution:
19.
Solution:
20.
Solution:
21.
Solution:
22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
Solution:
23.
Solution:
Consider
LHS = tan 4x = tan 2(2x)
By using the formula
24. cos 4x = 1 – 8sin2 x cos2 x
Solution:
Consider
LHS = cos 4x
We can write it as
= cos 2(2x)
Using the formula cos 2A = 1 – 2 sin2 A
= 1 – 2 sin2 2x
Again by using the formula sin2A = 2sin A cos A
= 1 – 2(2 sin x cos x) 2
So we get
= 1 – 8 sin2x cos2x
= R.H.S.
25. cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
Solution:
Consider
L.H.S. = cos 6x
It can be written as
= cos 3(2x)
Using the formula cos 3A = 4 cos3 A – 3 cos A
= 4 cos3 2x – 3 cos 2x
Again by using formula cos 2x = 2 cos2 x – 1
= 4 [(2 cos2 x – 1)3 – 3 (2 cos2 x – 1)
By further simplification
= 4 [(2 cos2 x) 3 – (1)3 – 3 (2 cos2 x) 2 + 3 (2 cos2 x)] – 6cos2 x + 3
We get
= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3
By multiplication
= 32 cos6x – 4 – 48 cos4x + 24 cos2 x – 6 cos2x + 3
On further calculation
= 32 cos6x – 48 cos4x + 18 cos2x – 1
= R.H.S.