11类RD Sharma解–第7章复合角的三角比–习题7.1 |套装1

问题1.如果sin A = 4/5且cos B = 5/13，其中0

(i)罪(A + B)

(ii)cos(A + B)

(iii)罪(A – B)

(iv)cos(A – B)

解决方案：

Given that

sin A = 4/5 and cos B = 5/13

As we know, cos A = (1 – sin²A) and sin B = (1 – cos²B), where 0

Now we find the value of cosA and sinB

cos A = √(1 – sin²A)

= √(1 – (4/5)²) 

= √(1 – (16/25))

= √((25 – 16)/25)

= √(9/25)

= 3/5

sin B = √(1 – cos²B)

= √(1 – (5/13)²)

= √(1 – (25/169))

= √(169 – 25)/169)

= √(144/169)

= 12/13

(i) sin (A + B)

sin (A + B) = sinA cosB + cosA sinB

= 4/5 × 5/13 + 3/5 x 12/13

= 20/65 + 36/65 

= (20 + 36)/65

= 56/65

(ii) cos (A + B)

cos (A + B) = cosA cosB – sinA sinB

= 3/5 × 5/13 –  4/5 x 12/13

= 15/65 – 48/65

= -33/65

(iii) sin (A – B)

sin (A – B) = sinA cosB – cosA sinB

= 4/5 × 5/13 – 3/5 x 12/13

= 20/65 – 36/65

= – 16/65

(iv) cos (A – B)

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

= 3/5 x 5/13 + 4/5 x 12/13

= 15/65 + 48/65

= 63/65

问题2。(a)如果sinA = 12/13和sin B = 4/5，其中π/ 2

(i)罪(A + B)

(ii)cos(A + B)

解决方案：

We have,

sinA = 12/13 and sinB = 4/5, where π/2 < A < and 0 < B < π/2

As we know, cosA = – √(1 – sin²A) and cosB = √(1 – sin²B)

Now we find the value of cosA and cosB

cosA = – √(1 – sin²A)

= – √(1 – (12/13)²)

= – √(1 – 144/169)

= – √((169 – 144)/169)

= – √(25/169)

= – 5/13

cosB = √(1 – sin²B)

= √(1 – (4/5)²)

= √(1 – 16/25)

= √((25 – 16)/25)

= √(9/25)

= 3/5

(i) sin (A + B)

Since, sin (A + B) = sinA cosB + cosA sinB

= 12/13 x 3/5 + (-5/13) x 4/5

= 36/65 – 20/65

= 16/65

(ii) cos (A + B)

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

= – 5/13 x 3/5 – 12/13 x 4/5

= – 15/65 – 48/65

= – 63/65

(b)如果sinA = 3/5，cosB = 12/13，其中A和B都位于第二象限，则求出sin(A + B)的值。

解决方案：

We have,

sinA = 3/5, cosB = -12/13, where A and B, both lie in second quadrant.

As we know cosA = – √(1- sin²A) and sinB = √(1 – cos²B)

Now we find the value of cosA and sinB

cos A = – √(1 – sin²A)

= -√(1 – (3/5)²)

= -√(1 – 9/25)

= – √((25 – 9)/25)

= – √(16/25)

= – 4/5

sinB = √(1 – cos²B)

= √(1 – (-12/13)2)

= √(1 – 144/169)

= √((169 – 144)/169)

= √(25/169)

= 5/13

We need to find the value of sin (A + B)

Since, sin (A + B) = sinA cosB + cosA sinB

= 3/5 × (-12/13) + (-4/5) x 5/13

= -36/65 – 20/65

= -56/65

问题3.如果cosA = -24/25且cosB = 3/5，其中π

(i)罪(A + B)

(ii)cos(A + B)

解决方案：

We have,

cosA = -24/25 and cosB = 3/5, where π < A < 3π/2 and 3π/2 < B < 2π

As we know that A is present in third quadrant, B is 

present in fourth quadrant, so the sine function is Negative.

By using the formulas, sinA = √(1 – cos²A) and sinB = -√(1 – cos²B)

We find the value of sinA and sinB

sinA = – √(1 – cos²A)

= – √(1 – (-24/25)²)

= – √(1 – 576/625)

= – √((625 – 576)/625)

= – √(49/625)

= – 7/25

sinB = – √(1 – cos²B)

= – √(1 – (3/5)²)

= – √(1 – 9/25) 

= – √((25 – 9)/25)

= – √(16/25)

= – 4/5

(i) sin (A + B)

Since, sin (A + B) = sinA cosB + cosA sinB

= -7/25 x 3/5 + (-24/25) x (-4/5)

= -21/125 + 96/125

= 75/125

= 3/5

(ii) cos (A + B)

Since, cos (A + B) = cosA cosB – sinA sinB

= (-24/25) x 3/5 – (-7/25) × (-4/5)

= – 72/125 – 28/125

=- 100/125

= – 4/5

问题4.如果tanA = 3/4，cosB = 9/41，其中π

解决方案：

We have,

tanA = 3/4 and cosB = 9/41, where π < A < 3π/2 and 0 < B < π/2

As we know that, A is present in third quadrant, B is present in first quadrant

So, tan and sin functions are positive.

Now by using the formula,

sinB = √(1 – cos²B)

We find the value of sin B.

sinB = √(1 – cos²B)

= √(1 – (9/41)²)

= √(1 – 81/1681)

= √((1681 – 81)/1681)

= √(1600/1681)

= 40/41

As we know that, tanB = sinB/cosB, so

= (40/41)/(9/41)

= 40/9

Since, tan(A + B) = (tanA + tanB)/(1 – tanA tanB), so

= (3/4 + 40/9)/(1 – 3/4 x 40/9)

= (187/36)/(1 – 120/36)

= (187/36)/((36 – 120)/36)

= (187/36)/(- 84/36)

= -187/84

Hence, tan(A + B) = -187/84

问题5.如果sinA = 1/2，cosB = 12/13，其中π/ 2

解决方案：

We have,

sinA = 1/2, cosB = 12/13, where π/2 < A < π and 3π/2 < B < 2π

As we know that, A is present in second quadrant and B is present in fourth quadrant.

So, the sine function is positive, cosine and tan functions are negative in second quadrant

and the sine and tan functions are negative, cosine function is positive in the fourth quadrant

Now by using the following formulas,

cosA = -√(1 – sin²A) and sinB = -√(1 – cos²B)

We find the value of cosA and sinB

cosA = – √(1 – sin²A)

= – √(1 – (1/2)²)

= – √(1 – 1/4)

= – √((4 – 1)/4)

= – √(3/4)

= – √3/2

sinB = – √(1 – cos²B)

= – √(1 – (12/13)²)

= – √(1 – 144/169)

= – √((169 – 144)/169)

= – √(25/169)

= – 5/13

As we know, tanA = sinA/cosA and tanB = sinB / cosB

tanA = (1/2)/(-√3/2) = -1/√3 and

tanB = (-5/13)/(12/13) = -5/12

Since, tan (A – B) = (tanA – tanB) / (1 + tanA tanB), so

= ((-1/√3) – (-5/12)) / (1 + (-1/√3) x (-5/12))

= ((-12 + 5√3)/12√3) / (1 + 5/12√3)

= ((-12 + 5√3)/12√3) / ((12√3 + 5)/12√3)

= (5√3 – 12)/(5 + 12√3)

Hence, tan (A – B) = (5√3 – 12)/(5 + 12√3)

问题6.如果sinA = 1/2，cosB =√3/ 2，其中π/ 2

(i)棕褐色(A + B)

(ii)棕褐色(A – B)

解决方案：

We have,

SinA = 1/2 and cosB = √3/2, where π/2 < A < π and 0 < B < π/2

As we know that, A is in second quadrant, B is in first quadrant.

So, all functions are positive in first quadrant and sine function is positive, 

cosine and tan functions are negative in the second quadrant. 

So, by using the following formulas,

cosA = – √(1 – sin²A) and sinB = √(1 – cos²B)

We find the value of cosA and sinB

cosA = – √(1 – sin²A)

= – √(1 – (1/2)²)

= – √(1 – 1/4)

= – √((4 – 1)/4)

= – √(3/4)

= – √3/2

sinB = √(1 – cos²B)

= √(1 – (√3/2)²)

= √(1 – 3/4)

= √((4 – 3)/4)

= √(1/4)

= 1/2

As we know that, tanA = sinA / cosA and tanB = sinB / cosB

So, tanA = (1/2)/(-√3/2) = -1/√3 and

tanB = (1/2)/(√3/2) = 1/√3

(i) Since, tan(A + B) = (tanA + tanB)/(1 – tanA tanB), so

= (-1/√3 + 1/√3)/(1 – (-1/√3) × 1/√3)

= 0/(1 + 1/3)

= 0

Hence, tan(A + B) = 0

(ii) Since, tan(A – B) = (tanA – tanB)/(1 + tanA tanB), so

= ((-1/√3) – (1/√3))/(1 + (-1/√3) x (1/√3))

= ((-2/√3)/(1 – 1/3)

= ((-2/√3)/(3 – 1)/3)

= ((-2/√3)/2/3)

= -√3

Hence, tan(A – B) = -√3

问题7：评估以下内容：

(i)sin 78°cos18⁰-cos 78°sin18⁰

(ii)cos 47°cos13⁰– sin47⁰sin13⁰

(iii)sin 36°cos9⁰+ cos 36°sin9⁰

(iv)cos 80°cos20⁰+ sin 80°sin20⁰

解决方案：

(i) sin 78° cos 18° – cos 78° sin 18°

Since, sinAcosB – cosAsinB = sin(A – B)

So

sin 78° cos 18° – cos 78° sin 18° = sin(78 – 18)°

= sin 60°

= √3/2

(ii) cos 47° cos 13° – sin 47° sin 13°

Since, cosA cosB – sinA sinB = cos(A + B)

So, cos 47° cos 13° – sin 47° sin 13° = cos (47 + 13)°

= cos 60°

= 1/2

(iii) sin 36° cos 9° + cos 36° sin 9°

Since, sin A cos B + cos A sin B = sin (A + B) 

So, sin 36° cos 9° + cos 36° sin 9° = sin (36 + 9)°

= sin 45°

= 1/√2

(iv) cos 80° cos 20° + sin 80° sin 20⁰

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

So, cos 80° cos 20° + sin 80° sin 20° = cos (80 – 20)°

= cos 60°

= 1/2

问题8.如果cosA = -12/13并且cotB = 24/7，其中A位于第二象限中，而B位于第三象限中，请找到以下值：

(i)罪(A + B)

(ii)cos(A + B)

(iii)棕褐色(A + B)

解决方案：

We have,

cosA = -12/13 and cotB= 24/7

It is given that, A lies in second quadrant, B in the third quadrant.

So, sine function is positive in second quadrant and both sine and cosine 

functions are negative in third quadrant.

So, by using the following formulas,

sinA = √(1 – cos²A), sinB = 1/√(1 + cot²B) and cosB = -√(1 – sin²B),

We find the value of sinA and sinB

sinA = √(1 – cos²A)

= √(1 -(-12/13)²)

= √(1 – 144/169) 

= √((169 – 144)/169)

= √(25/169)

= 5/13

sinB = -1/√(1 + cot²B)

= -1/√(1 + (24/7)²)

= -1/√(1 + 576/49)

= -1/√((49 + 576)/49)

= -1/√(625/49)

= -1/(25/7)

= -7/25

cosB = -√(1 – sin²B)

= -√(1 -(-7/25)²)

= -√(1 – (49/625))

= -√((625 – 49)/625)

= -√(576/625)

= -24/25

(i) sin(A + B)

Since, sin(A + B) = sinA cosB + cosA sinB

So, 

= 5/13 x (-24/25) + (-12/13) x (-7/25)

= -120/325 + 84/325

= -36/325

(ii) cos(A + B)

Since, cos(A + B) = cosA cosB – sinA sinB

So, 

= -12/13 x (-24/25) – (5/13) x (-7/25)

= 288/325 + 35/325

= 323/325

(iii) tan(A + B)

Since, tan(A + B) = sin(A + B) / cos(A + B)

So, 

= (-36/325)/(323/325)

=-36/323

问题9.证明：cos7π/ 12 + cosπ/ 12 = sin5π/ 12 – sinπ/ 12

解决方案：

As we know that, 

7π/12 = 105°, π/12 = 15°, 5π/12 = 75°

Now, L.H.S = cos 105° + cos 15°

= cos (90° + 15°) + sin (90° – 75°)

= -sin 15° + sin 75°

= sin 75° – sin 15°

= RHS

So, LHS = RHS

Hence proved.

问题10。证明：(tanA + tanB)/(tanA – tanB)= sin(A + B)/ sin(A – B)

解决方案：

Let solve, LHS = (tanA + tanB) / (tanA – tanB)

= $\frac{\frac{sinA}{cosA}+\frac{sinB}{cosB}}{\frac{sinA}{cosA}-\frac{sinB}{cosB}}$

= $\frac{\frac{sinAcosB+cosAsinB}{cosAcosB}}{\frac{sinAcosB-cosAsinB}{cosAcosB}}$

As we know that, 

sin(A ± B) = sinA cosB ± cosA sinB 

So, 

= {sin(A + B)} / {sin(A – B)}

= RHS

So, LHS = RHS

Hence proved.

问题11：证明：

(i)(cos 11°+ sin 11°)/(cos 11°– sin 11°)= tan 56°

(ii)(cos9⁰+ sin 9″)/(cos 9″-sin 9°)= tan 54°

(iii)(cos 8°– sin 8°)/(cos 8°+ sin 8°)=棕褐色37⁰

解决方案：

(i) (cos 11° + sin 11°) / (cos 11° – sin 11°) = tan 56°

Let solve, LHS = (cos 11° + sin 11°)/(cos 11° – sin 11°)

Now divide the numerator and denominator by cos 11° we get,

(cos 11° + sin 11°)/(cos 11° – sin 11°) = (1 + tan 11°)/(1 – tan 11°)

= (1+ tan 11°)/(1 – 1 x tan 11°)

= (tan 45° + tan 11°)/(1 – tan 45° x tan 11°)

As we know that, 

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

So,

(tan 45° + tan 11°)/(1 – tan 45° x tan 11°) = tan (45° + 11°)

= tan 56°

= RHS

So, LHS = RHS

Hence proved.

(ii) (cos 9° + sin 9°)/(cos 9° – sin 9°) = tan 54°

Let solve, LHS = (cos 9° + sin 9°)/(cos 9° – sin 9°)

Now divide the numerator and denominator by cos 9° we get, 

(cos 9° +  sin 9°)/(cos 9° – sin 9°) = (1 + tan 9°)/(1 – tan 9°)

= (1 + tan 9°) / (1 – 1 x tan 9°)

= (tan 45° + tan 9°)/(1 – tan 45° x tan 9°)

As we know that

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

So, 

(tan 45° + tan 9°)/(1 – tan 45° x tan 9°) = tan (45° + 9°)

= tan 54°

= RHS

So, LHS = RHS

Hence proved.

(iii) (cos 8° – sin 8°)/(cos 8° + sin 8°) = tan 37⁰

Let solve, LHS = (cos 8° – sin 8°) / (cos 8° + sin 8°)

Now divide the numerator and denominator by cos 8° we get,

(cos 8° – sin 8°) / (cos 8° + sin 8°) = (1 – tan 8°)/(1 + tan 8°) 

= (1 – tan 8°)/(1 + 1 x tan 8°)

= (tan 45° – tan 8°) / (1 + tan 45° x tan 8″)

As we know that

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

So,

(tan 45° – tan 8°)/(1 + tan 45° x tan 8°) = tan (45° – 8°)

= tan 37°

 = RHS

LHS = RHS

Hence proved.

问题12：证明：

(i)sin(π/ 3 – x)cos(π/ 6 + x)+ cos(π/ 3 – x)sin(π/ 6 + x)= 1

(ii)sin(4π/ 9 + 7)cos(π/ 9 + 7)– cos(4π/ 9 + 7)sin(π/ 9 + 7)=√3/ 2

(iii)sin(3π/ 8 – 5)cos(π/ 8 + 5)+ cos(3π/ 8 – 5)sin(π/ 8 + 5)= 1

解决方案：

(i) sin (π/3 – x) cos(π/6 + x) + cos (π/3 – x) sin(π/6 + x) = 1

Let solve, LHS = sin (π/3 – x) cos(π/6 + x) + cos (π/3 – x) sin(π/6 + x)

As we know that

sin(A + B) = sinA cosB + cosA sinB

sin (π/3 – x) cos(π/6 + x) + cos (π/3 – x) sin(π/6 + x) = sin (π/3 – x + π/6 + x)

= sin ((2π + π)/6)

= sin (π/2) 

= sin 90°

= 1

= RHS

LHS = RHS

Hence proved.

(ii) sin (4π/9 + 7) cos(π/9 + 7) – cos (4π/9 + 7) sin (π/9 + 7) = √3/2

Let solve, LHS = sin (4π/9 + 7) cos(π/9 + 7) – cos (4π/9 + 7) sin (π/9 + 7)

As we know that

sin(A – B) = sinA cosB – cosA sinB

sin (4π/9 + 7) cos(π/9 + 7) – cos (4π/9 + 7) sin (π/9 + 7) = sin (4π/9 + 7 – π/9 – 7)

= sin (3π/9)

= sin (π/3)

= sin 60° 

= √3/2 

= RHS

LHS = RHS

Hence proved.

(iii) sin ( 3π/8 – 5) cos (π/8 + 5) + cos (3π/8 – 5) sin(π/8 + 5) = 1

Let solve, LHS = sin ( 3π/8 – 5) cos (π/8 + 5) + cos (3π/8 – 5) sin(π/8 + 5) 

As we know that

sin(A + B) = sinA cosB + cosA sinB

sin ( 3π/8 – 5) cos (π/8 + 5) + cos (3π/8 – 5) sin(π/8 + 5) = sin (3π/8 – 5 + π/8 + 5)

= sin ((3π + π)/8)

= sin (4π/8)

= sin (π/2)

= sin 90°

= 1

= RHS

LHS = RHS

Hence proved.

问题13.证明：(tan 69°+ tan 66°)/(1 – tan 69°tan 66°)= -1

解决方案：

Let solve, LHS = (tan 69°+tan 66°)/(1-tan 69° tan 66°)

As we know that

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

= (tan 69° + tan 66°)/(1 – tan 69° tan 66°)

= tan (69 +66)°

= tan 135⁰

= – tan 45º

= -1

= RHS

LHS = RHS

Hence proved.

问题14.(i)如果tanA = 5/6且tanB = 1/11，则证明(A + B)=π/ 4

(ii)如果tanA = m /(m-1)且tanB = 1 /(2m – 1)，则证明(A – B)=π/ 4

解决方案：

(i) We have,

tanA = 5/6 and tanB = 1/11

As we know that

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

= [(5/6) + (1/11)] /[1 – (5/6) x (1/11)]

= (55 + 6)/(66 – 5)

= 61/61

= 1

= tan 45° or tan π/4

So, tan(A + B) = tan π/4

(A + B) = π/4

Hence proved.

(ii) We have,

tanA = m/(m – 1) and tanB = 1/(2m – 1)

As we know that

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

= $\frac{\frac{m}{m-1}-\frac{1}{2m-1}}{1+\frac{m}{m-1}\frac{1}{2m-1}}$

= (2m² – m – m + 1)/(2m² – m – 2m + 1 + m)

= (2m² – 2m + 1)/(2m² – 2m + 1)

= 1

= tan 45° or tan π/4

So, tan(A – B) = tan π/4

(A – B) = π/4

Hence proved.

问题15：证明：

(i)cos ^2π / 4 – sin ^2π / 12 =√3/ 4

(ii)sin ² (n + 1)A – sin ² nA = sin(2n +1)A sin A

解决方案：

(i) cos² π/4 – sin² π/12 = √3/4

Let solve, LHS = cos² π/4 sin² π/12

As we know that

cos² A – sin² B = cos (A + B) cos (A – B)

So, 

cos² π/4 – sin² π/12 = cos (π/4 + π/12) cos (π/4 – π/12)

= cos 4π/12 cos 2π/12

= cos π/3 cos π/6

= 1/2 x √3/2

= √3/4 = RHS

LHS = RHS 

Hence proved.

(ii) sin² (n + 1)A – sin²nA = sin (2n + 1)A sinA

Let solve, LHS = sin² (n+1)A – sin²nA

As we know that

sin²A – sin²B = sin(A + B) sin(A – B)

Where, A = (n + 1)A and B = nA

So,

sin²(n + 1)A – sin²nA = sin((n + 1)A + nA) sin((n + 1)A – nA)

= sin(nA + A + nA) sin (nA + A – nA)

= sin(2nA + A) sinA

= sin(2n + 1)A sinA 

= RHS

LHS = RHS

Hence proved.

问题16：证明：

(i){sin(A + B)+ sin(A – B)} / {cos(A + B)+ cos(A – B)} = tanA

解决方案：

Prove: {sin(A + B) + sin(A – B)}/{cos(A + B) + cos(A – B)} = tanA

Proof:

Let solve, LHS = {sin(A + B) + sin(A – B)}/{cos(A + B) + cos(A – B)}

As we know that, sin (A ± B) = sinA cosB ± cosA sinB and cos(A ± B) = cosA cosB ± sinA sinB

So,

= $\frac{sin(A+B)+sin(A-B)}{cos(A+B)+cos(A-B)}$

= $\frac{sinAcosB+cosAsinB+sinAcosB-cosAsinB}{cosAcosB-sinAsinB+cosAcosB+sinAsinB}$   

= (2cosA cosB)(2cosA cosB)

= tan A

= RHS

LHS = RHS

Hence proved.

(ii){sin(A – B)/cosA.cosB} + {sin(B – C)/cosB.cosC} + {sin(C – A)/cosC.cosA} = 0

解决方案：

Prove: {sin(A – B)/cosA.cosB} + {sin(B – C)/cosB.cosC} + {sin(C – A)/cosC.cosA} = 0

Proof:

Let solve, LHS = {sin(A – B)/cosA.cosB} + {sin(B – C)/cosB.cosC} + {sin(C – A)/cosC.cosA}

As we know that

sin(A – B) = sinA cosB – cosA sinB

So, 

= {sin(A – B)/cosA.cosB} + {sin(B – C)/cosB.cosC} + {sin(C – A)/cosC.cosA}

= (sinA cosB – cosA sinB)/(cosA cosB) + (sinB cosC – cosB sinC)/(cosB cosC) + 

   (sinC cosA – cosC sinA)/(cosC cosA)

= (sinA cosB)/(cosA sinB) – (cosA sinB)/(cosB cosC) + (sinB cosC)/(cosB cosC) – 

  (cosB sinC)/(cosB cosC) + (sinC cosA)/(cosC cosA)

= tanA – tanB + tanB – tanC + tanC – tanA

= 0

= RHS

LHS = RHS

Hence proved.

(iii){sin(A – B)/sinA.sinB} + {sin(B – C)/sinB.sinC} + {sin(C – A)/sinC.sinA} = 0

解决方案：

Prove: {sin(A – B)/sinA.sinB} + {sin(B – C)/sinB.sinC} + {sin(C – A)/sinC.sinA} = 0

Proof:

Let solve, LHS = {sin(A – B)/sinA.sinB} + {sin(B – C)/sinB.sinC} + {sin(C – A)/sinC.sinA}

As we know that

sin(A – B) = sinA cosB – cosA sinB

= {sin(A – B)/sinA.sinB} + {sin(B – C)/sinB.sinC} + {sin(C – A)/sinC.sinA}

= (sinA cosB – cosA sinB)/(sinA sinB) + (sinB cosC – cosB sinC)/(sinB sinC) + 

    (sinC cosA- cosC sinA)/(sinC sinA)

= (sinA.cosB)/(sinA.sinB) – (cosA.sinB)/(sinA.sinB) + (sinB.cosC)/(sinB.sinC) – 

   (cosB.sinC)/(sinB.sinC) +(sinC.cosA)/(sinC.sinA) – (cosC.sinA)/(sinC.sinA)

= cotB – cotA + cotC – cotB + cotA – cot C

= 0

= RHS

LHS = RHS

Hence proved.

(iv)sin ² B = sin ² A + sin ² (A – B)– 2sinA cosB sin(A – B)

解决方案：

Prove: sin²B = sin²A + sin²(A – B) – 2sinA cosB sin(A – B)

Proof:

Let’s solve RHS = sin²A + sin²(A – B) – 2 sinA cosB sin(A – B)

= sin²A+ sin(A -B)[sin(A – B) – 2 sinA cosB]

As we know that

sin(A – B) = sinA cosB –  cosA sinB

So,

= sin²A + sin(A -B)[sinA cosB – cosA sinB – 2 sinA cosB]

= sin²A + sin(A-B)[-sinA cosB – cosA sinB] 

= sin²A – sin(A -B)[sinA cosB + cosA sinB]

As we know that

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

So,

= sin²A – sin(A – B) sin(A + B).

= sin²A – sin²A + sin²B

= sin²B

= LHS

LHS = RHS

Hence proved.

(v)cos ² A + cos ² B – 2 cosA cosB cos(A + B)= sin ² (A + B)

解决方案：

Prove: cos²A + cos²B – 2 cosA cosB cos(A + B) = sin²(A + B)

Proof:

Let solve LHS = cos²A + cos²B – 2 cosA cosB cos(A + B)

= cos²A + 1 – sin²B – 2 cosA cosB cos(A + B)

= 1 + cos²A – sin²B – 2 cosA cosB cos(A + B)

As we know that cos²A – sin²B = cos(A + B) cos(A – B)

So,

= 1 + cos(A + B) cos(A – B) – 2 cosA cosB cos(A + B)

= 1 + cos(A + B)[cos(A – B) – 2 cosA cosB]

Also, 

cos(A – B) = cosA cosB + sinA sinB

So,

= 1 + cos(A +B)[cosA cosB + sinA sinB – 2 cosA cosB]

= 1 + cos(A +B)[-cosA cosB + sinA sinB]

= 1 cos(A +B)[cosA cosB – sinA sinB]

Also,

cos(A + B) = cosA cosB – sinA sinB

So,

1 – cos²(A + B) = sin²(A + B) = RHS

LHS = RHS

Hence proved.

(vi)tan(A + B)/ cot(A – B)=(tan ² A – tan ² B)/(1 – tan ² A tan ² B)

解决方案：

Prove: tan(A + B)/cot(A – B) = (tan²A – tan²B)/(1 – tan²A tan²B)

Proof:

Let solve LHS = tan(A + B)/cot(A – B)

As we knwo that

tan (A ± B) = (tanA ± tanB) / (1 ± tanA tanB)

So, 

 $\frac{tan(A+B)}{1/tan(A-B)}=\frac{\frac{tanA+tanB}{1-tanAtanB}}{1/\frac{tanA-tanB}{1+tanAtanB}}$

= $\frac{tanA+tanB}{1-tanAtanB}\frac{tanA-tanB}{1+tanAtanB}$

We know that, (x+y) (x – y) = x²– y²

So,

$\frac{tanA+tanB}{1-tanAtanB}\frac{tanA-tanB}{1+tanAtanB}=\frac{tan^2A-tan^2B}{1-tan^2Atan^2B}$ 

= RHS

LHS = RHS 

Hence proved.