证明:
问题1. 2cos(π/ 13)cos(9π/ 13)+ cos(3π/ 13)+ cos(5π/ 13)= 0
解决方案:
Let’s take L.H.S = 2cos(π/13)cos(9π/13) + cos(3π/13) + cos(5π/13)
Rearranging terms, we get
= 2cos(π/13)cos(9π/13)+ cos(5π/13) + cos(3π/13)
By using factorization formula,
cos A + cos B = 2 cos((A + B) / 2) cos((A – B) / 2)
We get
= 2cos(π/13)cos(9π/13)+ 2cos((5π + 3π)/(2 x13))cos((5π – 3π)/(2×13))
= 2cos(π/13)cos(9π/13)+ 2cos((8π)/(2×13))cos((2π)/(2×13))
= 2cos(π/13)cos(9π/13)+ 2cos(4π/13)cos(π/13)
= 2cos(π/13)(cos(9π/13) + cos(4π/13))
= 2cos(π/13)(2cos((9π + 4π)/(2 x13))cos((9π – 4π)/(2×13)))
= 2cos(π/13)(2cos(π/2)cos(5π /(2×13)))
As we know that,
cos (π /2) = 0
= 2cos(π/13)(0 x cos(5π /(2×13)))
= 0
L.H.S = R.H.S
Hence Proved.
问题2.(sin 3x + sin x)sin x +(cos 3x – cos x)cos x = 0
解决方案:
Let’s take L.H.S = (sin 3x + sin x)sin x + (cos 3x – cos x)cos x
By using factorization formula,
sin A + sin B = 2 sin((A + B) / 2) cos((A – B) / 2)
and
cos A – cos B = -2 sin((A + B) / 2) sin((A – B) / 2)
We get
= (2 sin((3x + x) / 2) cos((3x – x) / 2))sin x + (-2 sin((3x + x) / 2) sin((3x – x) / 2))cos x
= (2 sin((4x) / 2) cos((2x) / 2))sin x + (-2 sin((4x) / 2) sin((2x) / 2))cos x
= 2 sin2x cos x sin x – 2 sin 2x sin x cos x
= 0
L.H.S = R.H.S
Hence Proved.
问题3.(cos x + cos y) 2 +(sin x – sin y) 2 = 4 cos 2 ((x + y)/ 2)
解决方案:
Let’s take L.H.S = (cos x + cos y)2 + (sin x – sin y)2
By using factorization formula,
cos A + cos B = 2 cos((A + B) / 2) cos((A – B) / 2)
and
sin A – sin B = 2 cos((A + B) / 2) sin((A – B) / 2)
We get
= (2 cos((x + y) / 2) cos((x – y) / 2)))2 + (2 cos((x + y) / 2) sin((x – y) / 2)))2
= 4cos2((x + y) / 2) cos2((x – y) / 2)) + 4cos2((x + y) / 2) sin2((x – y) / 2))
= 4cos2((x + y) / 2) (cos2((x – y) / 2) + sin2((x – y) / 2))
As we know that,
sin2 X + cos2 X = 1
Therefore,
= 4cos2((x + y) / 2) (1)
= 4cos2((x + y) / 2)
L.H.S = R.H.S
Hence Proved.
问题4.(cos x – cos y) 2 +(sin x – sin y) 2 = 4 sin 2 ((x – y)/ 2)
解决方案:
Let’s take L.H.S = (cos x – cos y)2 + (sin x – sin y)2
By using factorization formula,
cos A – cos B = 2 sin((A + B) / 2) sin((A – B) / 2)
and
sin A – sin B = 2 cos((A + B) / 2) sin((A – B) / 2)
We get
= (-2 sin((x + y) / 2) sin((x – y) / 2)))2 + (2 cos((x + y) / 2) sin((x – y) / 2)))2
= 4sin2((x + y) / 2) sin2((x – y) / 2)) + 4cos2((x + y) / 2) sin2((x – y) / 2))
= 4sin2((x – y) / 2) (sin2((x + y) / 2) + cos2((x + y) / 2))
As we know that,
sin2 X + cos2 X = 1
Therefore,
= 4sin2((x – y) / 2) (1)
= 4sin2((x – y) / 2)
L.H.S = R.H.S
Hence Proved.
问题5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x
解决方案:
Let’s take L.H.S = sin x + sin 3x + sin 5x + sin 7x
= sin 7x + sin x + sin 5x + sin 3x
By using factorization formula,
sin A + sin B = 2 sin((A + B) / 2) cos((A – B) / 2)
We get
L.H.S = 2 sin((7x + x) / 2) cos((7x – x) / 2) + 2 sin((5x + 3x) / 2) cos((5x – 3x) / 2)
= 2 sin((8x) / 2) cos((6x) / 2) + 2 sin((8x) / 2) cos((2x) / 2)
= 2 sin 4x (cos 3x + cos x)
Again by using factorization formula,
cos A + cos B = 2 cos((A + B)/2) cos((A – B)/2)
We get
= 2 sin 4x(2 cos((3x + x)/2) cos((3x – x)/2))
= 4 sin 4x cos (4x/2) cos (2x/2)
= 4 cos x cos 2x sin 4x
L.H.S = R.H.S
Hence Proved.
问题6。
解决方案:
Let’s take L.H.S =
By using factorization formula.
and
We get
=
=
=
=
=
=
= sin 6x/cos 6x (∵ sinθ/cosθ = tanθ)
= tan 6x
L.H.S = R.H.S
Hence Proved.
问题7. sin 3x + sin 2x – sin x = 4 sin x cos(x / 2)cos(3x / 2)
解决方案:
Lets take L.H.S = sin 3x + sin 2x – sin x
By using factorization formula,
sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2)
we get
= sin 3x + (2 cos((2x + x)/2) sin((2x – x)/2))
= sin 3x + 2 cos(3x/2) sin(x/2)
= sin 2(3x/2) + 2 cos(3x/2) sin(x/2)
We know that,
sin 2a = 2 sin a cos a
Therefore,
= 2 sin(3x/2 cos (3x/2) + 2 cos(3x/2) sin(x/2)
= 2 cos(3x/2) (sin(3x/2) + sin(x/2))
Again by using factorization formula,
sin A + sin B = 2 sin ((A + B)/2) cos ((A – B)/2)
= 2 cos(3x/2) (2 sin ((3x/2 + x/2)/2) cos ((3x/2 – x/2)/2))
= 2 cos(3x/2) (2 sin ((4x/2)/2) cos ((2x/2)/2)
= 4 sin x cos (x/2) cos (3x/2)
L.H.S = R.H.S
Hence Proved.
在下面的每一个中找到sin(x / 2),cos(x / 2),tan(x / 2)
问题8. tan x = -4 / 3,x在象限II中
解决方案:
Given: x is in 2nd Quadrant
i.e
90° < x < 180°
On dividing by 2 throughout we get,
90°/2 < x/2 < 180°/2
45° < x/2 < 90°
Thus x/2 lies in 1st Quadrant.
Since Sine, cosine and tangent of any angle is positive in1st Quadrant.
Therefore, sin x/2, cos x/2, tan x/2 are positive.
Given:
tan x = – 4/3 ..(1)
By using double angle formula,
tan 2x = 2 tan(x) / (1 – tan2(x))
i.e tan (2x/2) = 2 tan(x/2) / (1 – tan2(x/2))
i.e tan (x) = 2 tan(x/2) / (1 – tan2(x/2)) ..(2)
From eq(1) and (2) we have
– 4/3 = 2 tan(x/2) / (1 – tan2(x/2))
-4 + 4tan2(x/2) = 6 tan x/2
2tan2(x/2) – 3 tan(x/2) – 2 = 0
2tan2(x/2) – 4tan(x.2) + tan(x/2) – 2 = 0
2tan(x/2)(tan(x/2) – 2) + 1(tan(x/2) – 2) = 0
(2tan(x/2) + 1) (tan(x/2) – 2) = 0
2tanx(x/2) + 1 = 0 or tan(x/2) – 2 = 0
tan x/2 = -1/2 or tan x/2 = 2
But tan x/2 is Positive,
Therefore
tan (x/2) = 2 ..(3)
By identity,
1 + tan2a = sec2a
Therefore,
1 + tan2x/2 = sec2x/2
sec2x/2 = 1 + 22
sec2x/2 = 1 + 4
sec x/2 = ± (5)1/2
Since sec x/2 lies in 1st Quadrant.
Therefore, sec x/2 = (5)1/2
We know that
cos a = 1 / sec a
Therefore,
cos x/2 = 1 / sec (x/2)
cos x/2 = 1/√5 = √5/5
Also we have,
sin2a + cos2a = 1
sin2x/2 = (1 – cos2x/2)
sin2x/2 = (1 – 1/5)
sin x/2 = ±(4/5)1/2
Since sin x/2 is in 1st Quadrant,
Therefore, sin x/2 = 2/√5 = 2√5/5
Hence, the values are,
sin x/2 = 2√5/5, cos x/2 = √5/5, tan x/2 = 2
问题9. cos x = -1 / 3,x在象限III中
解决方案:
Given: x is in 3rd Quadrant.
i.e 180° < x < 270°
i.e 180°/2 < x/2 < 270°.2
i.e 90° < x/2 < 135°
Thus, x/2 lies in 2nd Quadrant.
Therefore, sin x/2 is positive while cos x/2, tan x/2 are negative.
Here,
cos x = -1/3 ..(1)
By using double angle formula
cos 2x = 2cos2x – 1
cos 2(x/2) = 2cos2(x/2) – 1
cos x = 2cos2(x/2) – 1
2cos2(x/2) = 1 + cos x
2cos2(x/2) = 1 – 1/3 (From eq(1))
2cos2(x/2) = (3 – 1)/3
cos2(x/2) = 1/3
cos x/2 = ±(1/3)1/2
Since cos x/2 is negative
Therefore
cos x/2 = -1/√3 = -√3/3
We know that,
sin2a + cos2a = 1
Therefore,
sin2(x/2) + cos2(x/2) = 1
sin2(x/2) = 1 – cos2(x/2)
sin2(x/2) = 1 – (-1/√3)2
sin2(x/2) = (3 -1)/3
sin x/2 = ±(2/3)1/2
But sin x/2 is positive.
Therefore, sin x/2 = (2/3)1/2 = √6/3
Since tan a = sin a / cos a
Therefore,
tan x/2 = sin x/2 / cos x/2
tan x/2 = (2/3)1/2/ -(1/3)1/2
tan x/2 = -√2
Therefore, the values are,
sin x/2 = √6/3, cos x/2 = -√3/3, tan x/2 = -√2
问题10. sin x = 1/3,x在象限II中
解决方案:
Given: x is in 2nd Quadrant
i.e
90° < x < 180°
On dividing by 2 throughout we get,
90°/2 < x/2 < 180°/2
45° < x/2 < 90°
Thus x/2 lies in 1st Quadrant.
Since Sine, cosine and tangent of any angle is positive in1st Quadrant.
Therefore, sin x/2, cos x/2, tan x/2 are positive.
We know that,
cos2x + sin2x = 1
cos2x = 1 – sin2x
cos2x = 1 – (1/4)2 (Given sin x = 1/4)
cos2x = 1 – 1/16
cos2x = (16 – 1)/16
cos2x = 15/16
cos x = ± (√15/4)
Since x is in 2nd Quadrant,
Therefore, cos x = – (√15/4)
By using double angle formula,
cos 2x = 1 – 2sin2x
cos 2(x/2) = 1 – 2sin2(x/2)
cos x = 1 – 2sin2(x/2)
sin2(x/2) = (1 – cos x)/2
sin2(x/2) = (1 – (-√15/4)/2
sin2(x/2) = (4 + √15)/8
sin x/2 = ±((4 + √15)/8)1/2
But sin x/2 is Positive in 1st Quadrant.
Therefore, sin x/2 = ((4 + √15)/8)1/2
Also,
sin2x/2 + cos2x/2 = 1
cos2x/2 = 1 – sin2x/2
cos2x/2 = 1 – (4 + √15)/8
cos2x/2 = (8 – (4 + √15))/8
cos2x/2 = (8 – 4 – √15)/8
cos2x/2 = (8 – 4 – √15)/8
cos x/2 = ±((4 – √15)/8)1/2
But cos x/2 is negative in 1st Quadrant.
Therefore, cos x/2 = ((4 – √15)/8)1/2
Since,
tan x/2 = sin x/2 / cos x/2
Therefore,
tan x/2 = ((4 + √15)/8)1/2/-((4 – √15)/8)1/2
tan x/2 = ((4 + √15) / (4 – √15))1/2
tan x/2 = (((4 + √15)(4 + √15)) / ((4 – √15)(4 + √15)))1/2
tan x/2 = ((4 + √15)2/(42 – 15))1/2
tan x/2 = (4 + √15)
Therefore, the values are,
sin x/2 = ((4 + √15)/8)1/2, cos x/2 = ((4 – √15)/8)1/2, tan x/2 = (4 + √15)