11类RD Sharma解决方案–第9章复数角和约数角的三角比–练习9.1 |套装1

证明以下身份：

问题1。√[(1-cos2x)/(1 + cos2x)] = tanx

解决方案：

Let us solve LHS,

= √[(1 – cos2x)/(1 + cos2x)]

As we know that, cos2x =1 – 2 sin²x

= 2 cos²x – 1

So,

= √[(1 – cos2x)/(1 + cos2x)] 

= √[(1 – (1 – 2sin²x))/(1 + (2cos²x – 1))]

= √(1 – 1 + 2sin²x)/(1 + 2cos²x – 1)1

= √[2 sin²x/2 cos²x]

= sinx / cosx

= tanx 

LHS = RHS

Hence proved.

问题2。sin2x/(1-cos2x)= cotx

解决方案：

Let us solve LHS,

= sin 2x/(1 – cos 2x)

As we know that,

cos 2x = 1 – 2 sin²x 

Sin 2x = 2 sin x cos x

So,

sin 2x/(1-cos 2x) = (2 sin x cos x)/(1 – (1 – 2sin²x))

= (2 sin x cos x)/(1 – 1 + 2sin²x)]

= [2 sin x cos x/2 sin²x]

= cos x/sin x

= cot x

LHS = RHS

Hence proved.

问题3. sin 2x /(1 + cos 2x)=棕褐色x

解决方案：

Let us solve LHS,

= sin 2x / (1+cos 2x) 

As we know that,

cos 2x = 1 – 2 sin²x

= 2 cos²x – 1

sin 2x = 2 sin x cos x

So,

sin 2x / (1 + cos2x) = [2 sin x cos x / (1 + (2cos²x – 1))]

= [2 sin x cos x / (1+2cos²x – 1)]

= [2 sin x cos x/2 cos²x]

= sin x/cos x

= tan x

LHS = RHS

Hence proved.

问题4。 $\sqrt{2+\sqrt{2+cos4x}}=2cosx$ ，0

解决方案：

Let us solve LHS,

$\sqrt{2+\sqrt{2+cos4x}}=\sqrt{2+\sqrt{2+2(2cos^22x-1)}}$

As we know that,

 cos 2x = 2 cos²x – 1 ⇒ cos 4x = 2 cos²2x – 1

So, 

= $\sqrt{2+\sqrt{2+4cos^22x-2}}$

= $\sqrt{2+\sqrt{4cos^22x}}$

= $\sqrt{2+2cos2x}$

= $\sqrt{2+\sqrt{2(2cos^2x-1)}}$

= $\sqrt{2+4cos^2x-2}$

= $\sqrt{4cos^2x}$

= 2 cos x

LHS = RHS

Hence proved.

问题5. [1 – cos 2x + sin 2x] / [1 + cos 2x + sin 2x] =棕褐色x

解决方案：

Let us solve LHS,

= [1 – cos 2x + sin 2x]/[1 + cos 2x + sin 2x] 

As we know that,

cos 2x = 1 – 2 sin²x

= 2 cos²x – 1

sin 2x = 2 sin x cos x

So,

= {1 – (1 – 2sin²x) + 2sinxcosx} / {1 + (2 cos²x – 1) + 2 sin x cosx}

= {1 − 1 + 2sin²x + 2sinxcosx} / {1 + 2cos²x − 1 + 2sinx cosx}

= {2 sin²x + 2sinxcosx} / {2 cos²x + 2sinxcosx}

= {2sinx (sinx + cosx)} / {2 cos x (cosx + sin x)}

= sinx/cosx

= tan x

LHS = RHS

Hence proved.

问题6 [sin x + sin 2x] / [1 + cos x + cos2x] = tanx

解决方案：

Let us solve LHS,

= [sin x + sin 2x]/[1 + cos x + cos 2x]

As we know that,

cos 2x = cos²x sin²x

sin 2x = 2 sin x cos x

So,

{sin x + sin 2x} / {1 + cos x + cos 2x} = {sin x + 2 sin x cos x} / {1 + cosx + (2cos²x − 1)}

= {sinx + 2 sinx cos x} / {1 + cosx + 2cos²x − 1}

= {sin x + 2 sin x cosx} / {cosx + 2cos²x}

= {sinx (1 + 2 cos x)} / {cosx(1 + 2cosx)}

= sinx / cosx

= tan x 

LHS = RHS

Hence proved.

问题7. cos 2x /(1+ sin 2x)= tan(π/ 4 – x)

解决方案：

Let us solve LHS,

= cos 2x / (1 + sin 2x)

As we know that,

cos 2x = cos²x – sin²x

sin 2x = 2 sin x cos x

So,

{cos 2x} / {1 + sin 2x} = {cos²x – sin²x} / {1 + 2 sin x cos x}

= {(cosx – sinx)(cosx + sinx)} / {sin²x + cos²x + 2 sin x cos x}

Since, a² – b² = (a – b)(a + b) and sin²x + cos²x = 1

So, 

= {(cosx – sinx)(cosx + sinx)} / {(sinx + cos x)²

Since, a²+ b² + 2ab = (a + b)²

So, 

 = {(cosx – sinx)(cosx + sinx)} / {(sinx + cosx)(sinx + cosx)}

= (cosx – sinx) / (sin x + cos x)

Now multiplying numerator and denominator by 1/√2, we get,

= $\frac{\frac{1}{√2}(cosx-sinx)}{\frac{1}{√2}(sinx+cosx)}$

= $\frac{(\frac{1}{√2}.cosx-\frac{1}{√2}.sinx)}{(\frac{1}{√2}.sinx+\frac{1}{√2}.cosx)}$

= $\frac{sin(\frac{π}{4})cosx-cos(\frac{π}{4})sinx}{sin(\frac{π}{4})sinx+cos(\frac{π}{4})cosx}$

Since, 1/√2 = sin π/4, so

= $\frac{sin(\frac{π}{4}-x)}{cos(\frac{π}{4}-x)}$

By using the formulas, we get

sin(A – B) = sinA cosB – sinB cosA 

cos(A – B)= cosA cosB + sinA sinB

= tan (π/4 – x)

LHS = RHS

Hence proved.

问题8. cos x /(1-sin x)= tan(π/ 4 + x / 2)

解决方案：

Let us solve LHS,

= cos x/(1 – sin x)

As we know that,

cos 2x = cos²x – sin²x

cos x = cos² x/2 – sin² x/2 

sin 2x = 2 sin x cos x

sin x = 2 sin x/2 cos x/2

So,

$\frac{cosx}{1-sinx}=\frac{cos^2(\frac{x}{2})-sin^2(\frac{x}{2})}{1-2sin(\frac{x}{2})cos(\frac{x}{2})}$

= $\frac{[cos(\frac{x}{2})-sin(\frac{x}{2})][cos(\frac{x}{2})+sin(\frac{x}{2})]}{sin^2(\frac{x}{2})+cos^2(\frac{x}{2})+2sin(\frac{x}{2})cos(\frac{x}{2})}$

By using the formulas, 

a² – b² = (a – b)(a + b) and sin²x + cos²x = 1), we get

= $\frac{[cos(\frac{x}{2})-sin(\frac{x}{2})][cos(\frac{x}{2})+sin(\frac{x}{2})]}{[sin(\frac{x}{2})+cos(\frac{x}{2})]^2}$

= $\frac{[cos(\frac{x}{2})-sin(\frac{x}{2})][cos(\frac{x}{2})+sin(\frac{x}{2})]}{[sin(\frac{x}{2})+cos(\frac{x}{2})][sin(\frac{x}{2})+cos(\frac{x}{2})]}$

= $\frac{cos(\frac{x}{2})-sin(\frac{x}{2})}{sin(\frac{x}{2})+cos(\frac{x}{2})}$

= $\frac{cos(\frac{x}{2})+sin(\frac{x}{2})}{sin(\frac{x}{2})-cos(\frac{x}{2})}$

Now multiply numerator and denominator by 1/√2, we get,

= $\frac{\frac{1}{√2}[cos(\frac{x}{2})+sin(\frac{x}{2})]}{\frac{1}{√2}[sin(\frac{x}{2})-cos(\frac{x}{2})]}$

= $\frac{(\frac{1}{√2})cos(\frac{x}{2})+(\frac{1}{√2})sin(\frac{x}{2})}{(\frac{1}{√2})sin(\frac{x}{2})-(\frac{1}{√2})cos(\frac{x}{2})}$

= $\frac{sin(\frac{π}{4})cos(\frac{x}{2})+cos(\frac{π}{4})sin(\frac{x}{2})}{sin(\frac{π}{4})sin(\frac{x}{2})-cos(\frac{π}{4})cos(\frac{x}{2})}$

= $\frac{sin(π/4-x)}{cos(π/4-x)}$

 = tan (π/4 – x)

LHS = RHS

Hence proved.

问题9. COS ^2π/ 8 + COS ^23π/ 8 + COS ^25π/ 8 + COS ^27π/ 8 = 2

解决方案：

Let us solve LHS,

= cos² π/8 + cos² 3π/8 + cos² 5π/8 + cos² 7π/8

As we know that,

cos 2x = 2cos²x – 1

cos 2x+1=2cos² x

cos²x = (Cos 2x + 1)/2

So,

= cos² π/8 + cos² 3π/8 + cos² 5π/8 + cos² 7π/8

= $\frac{1+cos(\frac{2π}{8})}{2}+\frac{1+cos(\frac{6π}{8})}{2}+\frac{1+cos(\frac{10π}{8})}{2}+\frac{1+cos(\frac{14π}{8})}{2}$

= $\frac{1+cos(\frac{2π}{8})}{2}+\frac{1+cos(π-\frac{2π}{8})}{2}+\frac{1+cos(π+\frac{2π}{8})}{2}+\frac{1+cos(2π-\frac{2π}{8})}{2}$

= $\frac{1+cos(\frac{2π}{8})}{2}+\frac{1-cos(\frac{2π}{8})}{2}+\frac{1-cos(\frac{2π}{8})}{2}+\frac{1+cos(\frac{2π}{8})}{2}$

As we know that, cos (π – A) =- cos A, cos (π+ A) = -cos A and cos (2π – A) = cos A

= 2 x {1 + cos(2π/8)/2} + 2 x {1 – cos(2π/8)/2}

= 1 + cos(2π/8) + 1 – cos(2π/8)

= 2 

LHS = RHS

Hence Proved.

问题10.罪^2π/ 8 + ²罪3π/ 8 + ²罪5π/ 8 + ²罪7π/ 8

解决方案：

Let us solve LHS,

= sin² π/8 + sin² 3π/8 + sin² 5π/8 + sin² 7π/8

As we know that, 

cos 2x = 1 – 2sin²x

2sin²x = 1 – cos 2x

sin²x = (1 – cos 2x)/2

So,

= $\frac{1-cos(\frac{2π}{8})}{2}+\frac{1-cos(\frac{6π}{8})}{2}+\frac{1-cos(\frac{10π}{8})}{2}+\frac{1-cos(\frac{14π}{8})}{2}$

= $\frac{1-cos(\frac{2π}{8})}{2}+\frac{1-cos(π-\frac{2π}{8})}{2}+\frac{1-cos(π+\frac{2π}{8})}{2}+\frac{1-cos(2π-\frac{2π}{8})}{2}$

= $\frac{1-cos(\frac{2π}{8})}{2}+\frac{1+cos(\frac{2π}{8})}{2}+\frac{1+cos(\frac{2π}{8})}{2}+\frac{1-cos(\frac{2π}{8})}{2}$

As we know, cos (π – A) = -cos A, cos (π + A) = -cos A and cos (2π – A) = cos A

= 2 x {1 – cos(2π/8)/2} + 2 x {1 + cos(2π/8)/2}

= 1 – cos(2π/8) + 1 + cos(2π/8)

= 2

LHS = RHS

Hence proved.

问题11.(cosα+ cosβ) ² +(sinα+ sinβ) ² = 4 cos ² (α–β)/ 2

解决方案：

Let us solve LHS,

= (cos α+ cos β)²+ (sin α+ sin β)²

On expanding, we get,

= cos²α + cos² β + 2 cos α cos β + sin²α+ sin²β + 2 sin α sin β

= 2+2 cos α cos β + 2 sin α sin β

= 2 (1+ cos α cos β+ sin α sin β)

= 2 (1 + cos (α – β))                        [Using, cos (A – B) = cos A cos B+ sin A sin B]

= 2 (1 + 2 cos²(α – β)/2 – 1)            [Using, cos2x = 2cos²x – 1]

= 2 (2 cos²(α – β)/2)

= 4 cos²(α – β)/2

LHS = RHS

Hence Proved.

问题12. sin ² (π/ 8 + x / 2)– sin ² (π/ 8 – x / 2)= 1 /√2sin x

解决方案：

Let us solve LHS,

= sin²(π/8 + x/2) – sin²(π/8 – x/2)

As we know that, 

sin²A – sin²B = sin (A + B) sin (A-B)

So,

sin²(π/8 + x/2) – sin²(π/8 – x/2) = sin (π/8 + x/2 + π/8 – x/2) sin (π/8 + x/2 – (π/8 – x/2))

= sin (π/8 + π/8) sin (π/8 + x/2 – π/8 + x/2)

= sin π/4 sin x

= 1/√2 sin x                       [As we knwo that, π/4 = 1/√2]

LHS = RHS

Hence proved.

问题13.1 + cos ² 2x = 2(cos ⁴ x + sin ⁴ x)

解决方案：

Let us solve LHS,

= 1 + cos²2x

As we know that, 

cos2x = cos²x sin²x

cos²x+ sin²x = 1

So,

1 + cos²2x = (cos²x+ sin²x)² + (cos²x – sin²x)² 

= (cos⁴x + sin⁴x + 2 cos²x sin²x) + (cos⁴x + sin⁴x – 2cos²x sin²x)

= cos⁴x + sin⁴x + cos⁴x + sin⁴x

= 2 cos⁴x + 2 sin⁴x

= 2 (cos⁴x + sin⁴x)

LHS = RHS

Hence proved.

问题14. cos ³ 2x + 3 cos 2x = 4(cos ⁶ x – sin ⁶ x)

解决方案：

Let us solve RHS,

= 4 (cos⁶ x – sin⁶ x)

On expanding, we get,

4 (cos⁶ x – sin⁶ x) = 4 [(cos²x)³ – (sin²x)³]

= 4 (cos²x – sin²x) (cos⁴x + sin⁴x + cos²x sin²x)

Now, using the formula, we get

a³ – b³ = (a – b) (a² + b² + ab)

= 4 cos 2x (cos⁴x + sin⁴x + cos²x sin²x + cos²x sin²x – cos²x sin²x

As we know that, 

cos 2x = cos²x – sin²x

So,

= 4 cos 2x (cos⁴x + sin⁴x + 2 cos²x sin²x – cos²x sin²x)

= 4 cos 2x [(cos²x)² + (sin²x)² + 2 cos²x sin²x – cos²x sin²x]

By using formula, 

a² + b² + 2ab = (a + b)², we get

= 4 cos 2x [(1)² – 1/4 (4 cos² x sin² x)]

= 4 cos 2x [(1)²-1/4 (2 cos x sin x)²] 

Since

sin 2x = 2sin x cos x

= 4 cos 2x [(1²) – 1/4 (sin 2x)²]

= 4 cos 2x (1 – 1/4 sin² 2x)

Since

sin² x = 1 – cos²x

= 4 cos 2x [1 – 1/4 (1 – cos² 2x)]

= 4 cos 2x [1 – 1/4 + 1/4 cos² 2x]

= 4 cos 2x [3/4 + 1/4 cos² 2x]

= 4 (3/4 cos 2x + 1/4 cos³ 2x)

= 3 cos 2x + cos³ 2x

= cos³ 2x + 3 cos 2x

LHS = RHS

Hence proved.

问题15。(正弦3A +正弦A)正弦A +(正弦3A –余弦A)余弦A = 0

解决方案：

Let us solve LHS,

= (sin 3A + sin A) sin A + (cos 3A – cos A) cos A

= (sin 3A) (sin A) + sin²A + (cos 3A) (cos A) – cos²A

= [(sin 3A) (sin A) + (cos 3A) (cos A)] + (sin²A – cos²A)

= [(sin 3A) (sin A) + (cos 3A) (cos A)] – (cos²A – sin²A)

= cos (3A – A) – cos 2A

As we know that, 

cos 2x = cos²A – sin²A

cos A cos B + sin A sin B = cos(A – B)

So,

= cos 2A – cos 2A

= 0

LHS = RHS

Hence Proved.