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2

证明以下身份：

问题16。cos ² (π/ 4 – x)– sin ² (π/ 4 – x)= sin 2x

解决方案：

Let us solve LHS,

= cos² (π/4 – x) – sin²(π/4 – x)

As we know that,

cos² A – sin² A = cos 2A

So,

= cos² (π/4 – x) sin² (π/4 – x)

= cos 2 (π/4 – x)

= cos (π/2 – 2x)

= sin 2x                       [As we know that, cos (π/2 – A) = sin A]

LHS = RHS

Hence Proved.

问题17. cos 4x = 1 – 8 cos ² x + 8 cos ⁴ x

解决方案：

Let us solve LHS,

= cos 4x

As we know that,

cos 2x = 2 cos²x – 1

So,

cos 4x = 2 cos² 2x – 1

= 2(2 cos² 2x – 1)² – 1 

= 2[(2 cos² 2x)² + 1² – 2 × 2 cos²x] – 1

= 2(4 cos⁴ 2x + 1 – 4 cos²x) – 1

= 8 cos⁴ 2x + 2 – 8 cos² x – 1

= 8 cos⁴ 2x + 1 – 8 cos²x

LHS = RHS

Hence Proved.

问题18. sin 4x = 4 sin x cos ³ x – 4 cos x sin ³ x

解决方案：

Let us solve LHS,

= sin 4x

As we know that,

sin 2x = 2 sin x cos x

cos 2x = cos²x – sin²x

So,

sin 4x = 2 sin 2x cos 2x

= 2 (2 sin x cos x) (cos² x – sin² x)

= 4 sin x cos x (cos² x – sin² x)

= 4 sin x cos³ x – 4 sin³ x cos x

LHS = RHS

Hence proved.

问题19. 3(sin x – cos x) ⁴ + 6(sin x + cos x) ² + 4(sin ⁶ x + cos ⁶ x)= 13

解决方案：

Let us solve LHS,

= 3(sin x – cos x)⁴ + 6 (sin x + cos x)² + 4 (sin⁶ x + cos⁶ x) 

As we know that,

(a + b)² = a² + b² + 2ab

(a – b)² = a² + b² – 2ab

a³ + b³ = (a+b)(a² + b² – ab)

So,

= 3{(sinx – cosx)²}² + 6 {(sinx)² + (cosx)² + 2 sinx cosx} + 4 {(sin²x)³ + (cos²x)³}

= 3{(sinx)² + (cosx)² – 2 sinx cosx}² + 6(sin²x + cos²x + 2 sinx cosx) + 4{(sin²x + cos²x)(sin⁴x + cos⁴x – sin²x cos²x)}

= 3(1 – 2 sinx cosx)² + 6(1 + 2 sinx cosx) + 4{(1)(sin⁴x + cos⁴x – sini²x cos²x)}

Since, 

sin²x + cos²x = 1

So,

= 3{1² + (2 sinx cosx)² – 4 sinx cosx} + 6(1 + 2 sinx cosx) + 4{(sin²x)² + (cos²x)² + 2 sin²x cos²x – 3 sin²x cos²x}

= 3{1 + 4 sin²x cos²x – 4 sinx cosx} + 6(1 + 2 sinx cosx) + 4{(sin²x + cos²x)² – 3 sin²x cos²x}

= 3 + 12 sin²x cos²x – 12 sinx cosx + 6 + 12 sinx cosx + 4{(1)² – 3 sin²x cos²x}

= 9 + 12 sin²x cos²x + 4(1 – 3 sin²x cos²x)

= 9 + 12 sin²x cos²x + 4 – 12 sin²x cos²x

= 13

LHS = RHS

Hence proved.

问题20. 2(sin ⁶ x + cos ⁶ x)– 3(sin ⁴ x + cos ⁴ x)+1 = 0

解决方案：

Let us solve LHS,

= 2(sin⁶ x + cos⁶ x) – 3(sin⁴ x + cos⁴ x) + 1 

As we know that,

(a + b)² = a² + b² + 2ab

a³ + b³ = (a + b) (a² + b² – ab) 

So,

 = 2(sin⁶x + cos⁶x) – 3(sin⁴x + cos⁴x) + 1 

= 2{(sin²x)³ + (cos²x)³} – 3{(sin²x)² + (cos²x)²} + 1

= 2((sin²x + cos²x)(sin⁴x + cos⁴x – sin²x cos²x) – 3{(sin²x)² + (cos²x)² + 2 sin²x cos²x – 2sin²x cos²x} + 1

= 2{(1)(sin⁴x + cos⁴x + 2 sin²x cos²x – 3 sin²x cos²x) – 3((sin²x + cos²x)² – 2sin²x cos²x) + 1

Since

sin²x + cos²x = 1

So, 

= 2{(sin²x + cos²x)² – 3 sin²x cos²x} – 3{(1)² – 2 sin²x cos²x} + 1

= 2{(1)² – 3 sin²x cos²x} – 3(1 – 2 sin²x cos²x) + 1

= 2(1 – 3 sin²x cos²x) – 3 + 6 sin²x cos²x + 1

= 2 – 6 sin²x cos²x – 2 + 6 sin²x cos²x

= 0 

LHS = RHS

Hence Proved.

问题21. cos ⁶ x – sin ⁶ x = cos 2x(1 – 1/4 sin ² 2x)

解决方案：

Let us solve LHS,

= cos⁶ x – sin⁶ x

As we know that,

(a + b)² = a² + b² + 2ab

a³ – b³ = (a – b) (a² + b² + ab)

So,

cos⁶ x – sin⁶ x = (cos² x)³ – (sin² x)³

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

As we know that,

cos 2x = cos²x – sin²x

So,

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

= cos 2x [(cos²x)² + (sin²x)² – 1/4 × 4 cos²x sin²x]

As we know that,

sin²x + cos²x = 1

So,

= cos2x [(1)² – 1/4 × (2 cosx sinx)²]

As we know that,

sin2x = 2 sinx cosx

So,

= cos 2x [1 – 1/4 × (sin 2x)²]

= cos 2x [1 – 1/4 × sin²2x]

LHS = RHS

Hence proved.

问题22.棕褐色(π/ 4 + x)+棕褐色(π/ 4 – x)= 2秒2x

解决方案：

Let us solve LHS,

= tan (π/4 + x) + tan (π/4 – x)

As we know that,

tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

tan (A – B) = (tan A – tan B)/(1 + tan A tan B) 

So, 

= $\frac{tan(\frac{π}{4})+tanx}{1-tan(\frac{π}{4})tanx}+\frac{tan(\frac{π}{4})-tanx}{1+tan(\frac{π}{4})tanx}$

Since, tan π/4 = 1

So,

= $\frac{1+tanx}{1-tanx}+\frac{1-tanx}{1+tanx}$

= $\frac{(1+tanx)^2+(1-tanx)^2}{(1-tanx)(1+tanx)}$

By using formulas, we get

(a – b)(a + b) = a² – b²

(a + b)² = a² + b² + 2ab &  (a – b)² = a²+ b² – 2ab

So,

= $\frac{1^2+tan^2x+2tanx+1^2+tan^2x-2tanx}{1^2-tan^2x}$

= $\frac{1+tan^2x+1+tan^2x}{1-tan^2x}$

= $\frac{2(1+tan^2x)}{1-tan^2x}$

As we know that,

tan x = sin x/cos x

So,

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

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

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

By using the formulas, we get

cos²x+ sin²x = 1 & cos 2x = cos²x – sin²x

So,

= $\frac{2(\frac{1}{cos^2x})}{\frac{cos2x}{cos^2x}}$

= 2/cos2x

= 2 sec2x

LHS = RHS

Hence Proved.

问题23. cot ² x –棕褐色² x = 4cot2x cosec2x

解决方案：

Let us solve LHS,

= cot²x  – tan²x

= cos²x/sin²x – sin²x/cos²x

= [(cos²x)² – (sin²x)²] / sin²xcos²x

= [(cos²x + sin²x)(cos²x – sin²x)] / sin²xcos²x

= (1 × cos2x) / sin²xcos²x

= 4cos2x / 4sin²xcos²x

= 4(cos2x) / (sin2x)²

= 4(cos2x) / (sin2x) × 1 / (sin2x)

= 4 cot2x cosex2x 

LHS = RHS

Hence Proved

问题24. cos4x –cos4α= 8(cosx –cosα)(cosx +cosα)(cosx –sinα)(cosx +sinα)

解决方案：

Let us solve RHS,

= 8(cosx – cosα)(cosx + cosα)(cosx – sinα)(cosx + sinα)

= 8(cos²x – cos²α)(cos²x – sin²α)

= 8(cos⁴x – cos²x × sin²α – cos²α × cos²x + cos²α × sin²α)

= 8{cos⁴x – cos²x(sin²α + cos²α) + cos²α × sin²α}

= 8{cos⁴x – cos²x + cos²α × (1 – cos²α)}

= 8{cos⁴x – cos²x + cos²α – cos⁴α)}

= 8{cos²x(cos²x – 1) + cos²α × (1 – cos²α)}

= 8{1/2 cos²x (2cos²x – 1 – 1) – 1/2 cos²α (2cos²α – 1 -1)}

= 8{1/2 cos²x (cos2x – 1) – 1/2cos²α (cos²α – 1)}

= 8[1/4 {2cos²x (cos2x – 1) – 2cos²x (cos2α – 1)}]

= 8[1/4 {(1 + cos2x)(cos2x – 1) – (1 + cos2α)(cos2α – 1)}]

= 8[1/4 { cos²2x – 1 – cos²2α + 1}]

= 8[1/8 {2cos²2x – 2cos²2α}]

= [{(1 + cos4x) – (1 + cos4α)}]

= [1 + cos4x – 1 – cos4α]

= cos4x – cos4α 

LHS = RHS

Hence proved

问题25. sin3x + sin2x – sinx = 4 sinx cos(x / 2)cos(3x / 2)

解决方案：

Let us solve LHS,

= sin3x + sin2x – sinx

= sin3x + 2sin(2x – x)/2 cos(2x + x)/2

= sin3x + 2sin(x/2) cos(3x/2)

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

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

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

= 2cos(3x/2)[2sinx cos(x/2)]

= 4 sinx cos(x/2) cos(3x/2)

LHS = RHS

Hence proved.

问题26。 $tan82\frac{1}{2}°$ =(√3+√2)(√2+1)=√2+√3+√4+√6

解决方案：

Let us solve LHS,

tan(82.5)° = tan(90 – 7.5)° = cot(7.5)° = 1/ tan(7.5)°

We have,

tan(x/2) = sinx/(1 + cosx)

Now on putting x = 15°, we get

tan(15/2) = sin15°/(1 + cos15°)

= sin(45-30)°/{1 + cos(45-30)°}

 = (sin45°cos30° – sin30°cos45°) / (1 + cos45° sin30°)

= $\frac{(\frac{1}{√2})(\frac{√3}{2})-(\frac{1}{2})(\frac{1}{√2})}{1+(\frac{1}{√2})(\frac{√3}{2})+(\frac{1}{√2})(\frac{1}{2})}$

= $\frac{\frac{√3}{2√2}-\frac{1}{2√2}}{1+\frac{√3}{2√3}+\frac{1}{2√2}}$

= $\frac{√3-1}{2√2+√3+1}$

Now,

tan(82.5)° = 1/tan(7.5)°

 = (2√2 + √3 + 1)/(√3 – 1)

= (2√2 + √3 + 1)/(√3 – 1) × (√3 + 1)/(√3 + 1)

= [√3 + 1(2√2 + √3 + 1)] / [(√3)² – 1²]

= (2√6 + 3 + √3 + 2√2 + √3 + 1) / (3 – 1)

= (2√6 + 2√3 + 2√2 + 4) / (2)

= √6 + √3 + √2 + 2

= √2 + √3 + √4 + √6        …..(i)

= √6 + √3 + 2 + √2

= √3(√2 + 1) + √2(√2 + 1)

= (√3 + √2)(√2 + 1)    …..(ii)

From equ (i) and (ii), we get 

tan(82.5)° = (√3 + √2)(√2 + 1) = √2 + √3 + √4 + √6

LHS = RHS

Hence proved

问题27。 $cot22\frac{1}{2}°$ =√2+1

解决方案：

As we know that, π/8 = $22\frac{1}{2}°$ = 45°

Let A = $22\frac{1}{2}°$

By using the identity cot2A = (cot²A – 1)/2cotA, we get

cot45° = {cot²($22\frac{1}{2}$)° – 1} / 2cot($22\frac{1}{2}°$)°

⇒ 1 = {cot²($22\frac{1}{2}$)° – 1} / 2cot($22\frac{1}{2}$)°

⇒ 2cot($22\frac{1}{2}$)° – cot²($22\frac{1}{2}$)° + 1 = 0

⇒ cot²($22\frac{1}{2}$)° – 2cot($22\frac{1}{2}$)° – 1 = 0

⇒ { cot²($22\frac{1}{2}$) – 2cot($22\frac{1}{2}$)° + 1} – 2 = 0

⇒ { cot($22\frac{1}{2}$)° – 1}² = 2

⇒ cot($22\frac{1}{2}$)° – 1 = √2

⇒ cot($22\frac{1}{2}$)° = √2 + 1

LHS = RHS

Hence proved

问题28(i)。如果cosx =(-3/5)并且x位于第三象限，则找到cos(x / 2)，sin(x / 2)，sin2x的值。

解决方案：

Given that, 

cosx = (-3/5)

⇒  cosx = cos²(x/2) – sin²(x/2)

⇒  -3/5 = 2cos²(x/2) – 1

⇒ 1 – 3/5 = 2cos²(x/2)

⇒ 2/5 = 2cos²(x/2)

⇒ 1/5 = cos²(x/2)

⇒ cos(x/2) = ± √(1/5)

Also, given that x lies in 3rd quardant, so x/2 lies in 2nd quadrant.

cos(x/2) = – √(1/5)

 Again,

cosx = cos²(x/2) – sin²(x/2)

⇒ -3/5 = (- √(1/5))² – sin²(x/2)

⇒ – 3/5 = 1/5 – sin²(x/2)

⇒ -1/5 -3/5 = -sin²(x/2)

⇒ 4/5 = sin²(x/2)

⇒ sin(x/2) = ± 2/√5

It is given x lies in 3rd quardant, so x/2 lies in 2nd quadrant.

sin(x/2) = 2/√5

Now,

sinx = √(1 – cos²x)

= √(1 – (-3/5))²

= √(1 – 9/25)

=  ± 4/5

It is given x llies in 3rd quardant, so sinx is negative.

sinx = – 4/5

sin2x = 2 sinx cosx

= 2 (-4/5) (-3/5)

= 24/25

Hence, the value of cos(x/2) = – √(1/5), sin(x/2) = 2/√5, and sin2x = 24/25.

问题28(ii)。如果cosx =(-3/5)并且x位于第三象限，则找到sin2x和sin(x / 2)的值。

解决方案：

Given that, 

 cosx = (-3/5)

 sinx = $\sqrt{1-cos^2x}=\sqrt{1-(-3/5)^2}$

⇒ sinx = ± 4/5

Here, x lies in the second quadant

So, sinx = 4/5

As we know that, 

sin2x = 2 sinx cosx

sin2x = 2 × 4/5 × (-3/5) = (-24/25)

Now,

cosx = 1 – 2 sin²(x/2)

⇒ 2sin2(x/2) = 1 – (-3/5) = 8/5

⇒ sinx2(x/2) = 4/5

⇒ sin(x/2) = ± 2/√5

Since x lies in the second quadrant,

x/2 lies in the first quadrant

So, sin(x/2) = 2/√5

Hence, the value of sin2x = (-24/25) and sin(x/2) = 2/√5

问题29.如果sinx =√5/ 3并且x位于第二象限，则求出cos(x / 2)，sin(x / 2)和tan(x / 2)的值。

解决方案：

Given that, sinx = √5/3

As we know that sinx = P/H 

So, P = √5, H = 3 and B = 2

Now, cosx = B/H = -2/3

So, 

cos(x/2) = √{(1 + cosx)/2} = √{(1 – 2/3)/2} = 1/√6

sin(x/2) = √{(1 – cosx)/2} = √{(1 + 2/3)/2} = √(5/6)

tan(x/2) = sin(x/2)/cos(x/2) = {√(5/6)} / (1/√6) = √5