Question 1:   Prove that:  2 cos(π /13)cos(9π /13)+ cos(3π /13)+ cos(5π /13)=0

Answer 1: L.H.S

=2 cos(π /13)cos(9π /13)+ cos(3π /13)+ cos(5π /13)=0

=2 cos(π /13)cos(9π /13)+ 2cos(4π /13)+ cos(-π /13)

=2 cos(π /13)cos(9π /13)+2 cos(4π /13)+ cos(π /13)

=2 cos(π /13)[cos(9π /13)+ cos(4π /13)]

=2 cos(π /13)[2 cos(π /2)cos(5π /26)]

=2 cos(π /13) x 2 x 0 x cos(5π /26)

=0=R.H.S

Question 2: Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

Answer 2: L.H.S. = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

= sin 3x sin x + sin² x + cos 3x cos x – cos² x

= cos 3x cos x + sin 3x sin x=(cos² x – sin² x)

=cos(3x-x)-cos 2x               [cos(A-B)=cos A cos B + sin A sin B]

=cos 2x – cos 2x

=0

=R.H.S

Question 3:   Prove that:  (cos x + cos y)² + (sin x – sin y)² = 4 cos² (x + y/2)

 Answer 3: L.H.S = (cos x + cos y)² + (sin x – sin y)²

= cos² x + cos² y +2 cos x cos y + sin² x + sin²y – 2 sin x sin y

= (cos² y + cos² x)+(cos² x + cos² y)+2 (cos x cos y - sin x sin y)

= 1 + 1 + 2 cos (x + y)                    [cos (A+B)=(cos A cos B-sin A sin B)]

= 2 + 2 cos (x + y)

= 2 [1 + cos (x + y)]

= 2 [1 + 2 cos² (x + y/2)-1]           [cos 2A=2 cos² A -1]

= 4 cos2 (x + y/2)

= R.H.S

Question 4:  Prove that:  (cos x – cos y)² + (sin x – sin y)² = 4 sin² (x-y/2)

Answer 4:  L.H.S. =(cos x – cos y)² + (sin x – sin y)²

= cos² x + cos² y – 2 cos x cos y + sin² x + sin² y – 2 sin x sin y

= (cos² x + sin² x) – (cos² y + sin² y) – 2[cos x cos y + sin x sin y]

=  1+ 1 – 2[ cos(x-y)]                                 [cos(A-B)=cos A cos B + sin A sin B]

=2[1-cos(x-y)]

=2[1-{1-2 sin²(x-y/2)}]                  [cos 2A=1-2 sin² A]

=4 sin2 (x-y/2)=R.H.S

Question 5:  Prove that: sin x + sin 3x + sin 5X + sin 7x=4 cos x cos 2x sin 4x

Answer 5: sin A + sin B=2 sin(A + B/2). cos(A – B/2).

It is known that

L.H.S= sin x + sin 3x + sin 5x + sin 7x

=(sin x + sin 5x) + (sin 7x)

=2 sin(x+5x/2).cos(x-5x/2)+ 2 sin(3x + 7x/2)cos(3x -7x/2)

= 2 sin 3x cos(-2x)+2 sin 5x cos(-2x)

=2 sin 3x cos 2x + 2 sin 5x cos 2x

=2 cos 2x [sin 3x + sin 5x]

=2 cos 2x [2 sin(3x + 5x/2). cos(3x – 5x/2)]

=2 cos 2x[2 sin 4x.cos(-x)]

= 4 cos 2x sin4x cos x=R.H.S

Question 6:  Prove that:

Answer 6: It is known that

sin A + sin B=2 sin(A + B/2).cos(A-B/2), cos A + cos B=2 cos(A + B/2).cos(A-B/2)                    

L.H.S=

 

= tan 6x

=R.H.S

Question 7:   Prove that:  sin 3x + sin 2x – sin x=4 sin x cos(x/2)cis(3x/2)

Answer 7: L.H.S. =sin 3x + sin 2x – sin x

=sin 3x + (sin 2x – sin x)

=sin 3x+[2 cos(2x+x/2)sin(2x-x/2)]       [sin A-sin B=2 cos(A+B/2)sin(A-B/2)]

=sin 3x +[2  cos(3x/2)sin(x/2)]

=sin 3x + 2 cos(3x/2)sin(x/2)]

=2 sin(3x/2).cos(3x/2)+2 cos(3x/2)sin(x/2)   [sin 2A=2 sin A. cos B]

=2 cos(3x/2).[ sin(3x/2)+ sin(x/2)]

=2 cos(3x/2).2 sin x cos(x/2)

=4 sin x cos(x/2) cos(3x/2)=R.H.S.

Question 8: Find sin x/2, cos x/2, and tan x/2, if tan x=-4/3, x in quadrant II

Answer 8: Here, x is in quadrant II.

i.e., π/2 < x < π

⇒ π/4 < x/2 < π/2

Therefore, sin(x/2), cos(x/2) and tan(x/2)

all lies in first quadrant.

It is given that tan x=-4/3.

sec² x=1 + tan²x=a+(-4/3)²=1+16/9=25/9

Therefore cos²x=9/25

⇒ cos x=±3/5

As x is in 2^nd quadrant , cos x is negative.

cos x = -3/5

Now, cos x=2 cos² (x/2-1)

⇒-3/5=2 cos² (x/2-1)

⇒2 cos² (x/2=2/5)

      [Because cos(x/2) is positive]

Therefore

sin²x/2 + cos²x/2=1

⇒sin²(x/2)=1-1/5=4/5

                                [Because   sin(x/2) is positive]

Thus, the respective values of sin(x/2), cos(x/2) and tan(x/2) are

Question 9: Find , sin(x/2), cos(x/2) and tan(x/2) for cos x=-1/3, x in quadrant III

Answer 9: Here, x is in quadrant III.

i.e., π < x < 3π/2

⇒ π /2 < x/2 < 3π/4

Therefore, cos(x/2)  and tan(x/2)    are negative, where as is positive. sin(x/2)

It is given that cos x=-1/3.

cos x=1-2sin²x/2

⇒sin²x/2=(1-cos x/2)

                           [Because   sin(x/2) is positive]

Now

cos x=2 cos²(x/2)-1

               [Because  cos(x/2) is negative]

Thus, the respective values of sin(x/2), cos(x/2) and tan(x/2) are

Question 10: Find sin(x/2), cos(x/2) and tan(x/2), for x=1/4, x in quadrant II

Answer 10: Here, x is in quadrant II.

i.e., π/2 < x < π

⇒ π/2 < x/2 < π/2

Therefore, sin(x/2), cos(x/2), tan(x/2) are all positive.

It is given that sin x=1/4.

cos²x=1-sin²x=1-(1/4)²=1-1/16=15/16

     (cos x is negative in quadrant II]

                               [Because sin(x/2) is positive]

                                  [Because sin(x/2) is positive]

Thus, the respective values of sin(x/2), cos(x/2), tan(x/2)

are