10类RD Sharma解决方案–第5章三角比–练习5.2 |套装1

评估以下各项(1-13)

问题1.正弦45°正弦30°+余弦45°余弦30°

解决方案：

Given: sin 45° sin 30° + cos 45° cos 30°         -(1)

Putting the values of sin 45° = cos 45°= 1/√2, sin 30° = 1/2, cos 30° = √3/2 in eq(1)

= (1/√2)(1/2) + (1/√2})(√3/2)           

= (1/2√2) + (√3/2√2)

= (1 + √3)/2√2

问题2. sin60°cos30°+ cos60°sin30°

解决方案：

Given: sin60°cos30° + cos60°sin30°          -(1) 

Putting the values of sin 60° = cos 30° = √3/2, sin 30° = cos 60° = 1/2 in eq(1)

= (√3/2)(√3/2) + (1/2)(1/2)

= 3/4 + 1/4

= 1

问题3. cos60°cos45°− sin60°sin45°

解决方案：

Given: cos60°cos45° − sin60°sin45°         -(1) 

Putting the values of sin 45° = cos 45° = 1/√2, sin 60° = √3/2, cos 60° = 1/2 in eq(1)

= (1/2)(1/√2) – (√3/2)(1/√2)  

= (1/2√2) – (√3/2√2) 

= (1 -√3)/2√2

问题4.正弦² 30°+正弦² 45°+正弦² 60°+正弦² 90°

解决方案：

Given: sin²30° + sin²45° + sin²60° + sin²90°         -(1) 

Putting the values of sin 45° = 1/√2, sin 30° = 1/2, sin 60° = √3/2, sin 90° = 1 in eq(1)

= (1/2)² + (1/√2)² + (√3/2)² + 1² 

= 1/4 + 1/2 + 3/4 + 1

= 1/4 + 2/4 + 3/4 + 4/4

= (1 + 2 + 3 + 4)/4

= 10/4

= 5/2 

问题5. cos ² 30°+ cos ² 45°+ cos ² 60°+ cos ² 90°

解决方案：

Given: cos²30° + cos²45° + cos²60° + cos²90°          -(1) 

Putting the values of cos 45° = 1/√2, cos 60° = 1/2, cos 30° = √3/2, cos 90° = 0 in eq(1)

= (√3/2)² +(1/√2)² + (1/2)² + 0²

= 3/4 + 1/2 + 1/4

= 3/4 + 2/4 + 1/4

= (1 + 2 + 3)/4

= 6/4

= 3/2

问题6. tan ² 30°+ tan ² 60°+ tan ² 45°

解决方案：

Given: tan²30° + tan²60° + tan²45°           -(1) 

Putting the values of tan 45° = 1, tan 30° = 1/√3, tan 60° = √3 in eq(1)

= (1/√3)² + (√3)² + (1)²

= 1/3 + 3 + 1

= (1 + 12)/3

= 13/3 

问题7. 2sin ² 30°– 3cos ² 45°+ tan ² 60°

解决方案：

Given: 2sin²30° – 3cos²45° + tan²60°          -(1) 

Putting the values of tan 60° = √3, cos 45° = 1/√2, sin 30° = 1/2 in eq(1)

= 2(1/2)² -3 (1/√2)² + (√3)²

= 2/4 – 3/2 + 3

= 1/2 – 3/2 + (3×2)/2

= 1/2 – 3/2 + 6/2

= 4/2

= 2

问题8.正弦² 30°cos ² 45°+ 4tan ² 30°+(1/2)sin ² 90°– 2cos ² 30°+(1/24)cos ² 0°

解决方案：

Given: sin²30°cos²45° + 4tan²30° + (1/2)sin²90° – 2cos²30° + (1/24)cos²0°         

= (1/2)²(1/√2)² + 4(1/√3)² + (1/2)(1)² – 2(0)² + (1/24)(1)²  

= 1/8 + 4/3 + 1/2 + 1/24

= 3/24 + 32/24 + 12 + 24 + 1/24

= 48/24

= 2

问题9. 4(SIN ⁴ 60°+ COS ⁴ 30°) – ^3-(2黄褐色60° -褐色² 45°)+ 5cos ² 45°

解决方案：

Given: 4(sin⁴60° + cos⁴30°) − 3(tan²60° − tan²45°) + 5cos²45°

= 4(√3/2)⁴ + (√3/2)⁴) – 3((√3)² – (1)²) + 5(1/√2)² 

= 4(9/16 + 9/16) – 3(3 – 1) + 5/2

= 4(18/16) – 3(2) + 5/2

= 9/2 – 12/2 + 5/2

= (9 – 12 + 5)/2

= 2/2

= 1

问题10.(cosec ² 45°sec ² 30°)(sin ² 30°+ 4cot ² 45°– sec ² 60°)

解决方案：

Given: (cosec²45°sec²30°)(sin²30° + 4cot²45° – sec²60°)

= ((√2)²(2/√3)²((1/2)² + 4(1)² – (2)²)

= (8/3) × (1/4) + 4 – 4

= (8/3) × (1/4)

= 2/3 

问题11. cosec ³ 30°cos60°tan ³ 45°sin ² 90°sec ² 45°cot30°

解决方案：

Given: cosec³30°cos60°tan³45°sin²90°sec²45°cot30°  

= (2)³(1/2)(1)³(1)²(√2)²(√3)  

= (8)(1/2)(2)(√3)

= 8√3

问题12.婴儿床² 30°– 2cos ² 60°–(3/4)sec ² 45°– 4sec ² 30°

解决方案：

Given: cot²30° – 2cos²60° – (3/4)sec²45° – 4sec²30°  

= (√3)² – 2(1/2)² – (3/4)(√2)² – 4(2/√3)²

= 3 – 2/4 – 6/4 – 16/3

= 3 – 1/2 – 3/2 – 16/3

= (18 – 3 – 9 – 32)/6

= -26/6

= -13/3

问题13.(cos0°+ sin45°+ sin30°)(sin90°– cos45°+ cos60°)

解决方案：

Given: (cos0° + sin45° + sin30°)(sin90° – cos45° + cos60°)

= (1 + 1/√2 + 1/2)(1 – 1/√2 + 1/2) 

= (3/2 + 1/√2)(3/2 – 1/√2)

Using identity (a + b)(a – b) = a² – b²

= (3/2)² – (1/√2)² 

= 9/4 – 1/2

= (9 – 2)/4

= 7/4