类11 NCERT解决方案-第3章函数-练习3.2

问题。在练习1至5中找到其他五个三角函数的值。

1. cos x = –1 / 2，x位于第三象限。

解决方案：

Since cos x = (-1/2) 

We have sec x = 1/cos x = -2

Now sin² x + cos² x = 1

i.e., sin² x = 1 – cos² x 

or sin² x = 1 – (1/4) = (3/4) 

Hence sin x = ±(√3/2) 

Since x lies in third quadrant, sin x is negative. Therefore

sin x = (–√3/2) 

Which also gives

cosec x = 1/sin x = (–2/√3) 

Further, we have

tan x = sin x/cos x

         = (–√3/2)/(-1/2)

         = √3 

and cot x = 1/tanx = (1/√3) 

2. sin x = 3/5，x位于第二象限。

解决方案：

Since sin x = (3/5) 

we have cosec x = 1/sin x = (5/3) 

Now sin² x + cos² x = 1

i.e., cos² x = 1 – sin² x

or cos²  x = 1 – (3/5) 

              = 1 – (9/25) 

              = (16/25) 

Hence cos x = ±(4/5) 

Since x lies in second quadrant, cos  x is negative. 

Therefore

cos x = (–4/5) 

which also gives

sec x = 1/cos x= (-5/4) 

Further, we have

tan x =  sin x/cos x

         =  (3/5)/(-4/5) 

         =  (-3/4) 

and cot x = 1/tan x = (-4/3)

3. cot x = 3/4，x位于第三象限。

解决方案：

Since cot x = (3/4) 

we have tan x = 1 / cot x = (4/3) 

Now sec² x = 1 + tan² x

                  = 1 + (16/9)

                  = (25/9) 

Hence sec x = ±(5/3) 

Since x lies in third quadrant, sec x  will be negative. Therefore

sec x = (-5/3) 

which also gives

cos x = (-3/5) 

Further, we have

sin x = tan x  * cos x

        = (4/3) x (-3/5)

        = (-4/5) 

and cosec x = 1/sin x

                   = (-5/4) 

4.秒x = 13/5，x位于第四象限。

解决方案：

Since sec x = (13/5)

we have cos x = 1/secx = (5/13) 

Now sin² x + cos² x = 1

i.e., sin² x = 1 – cos² x

or sin² x = 1 – (5/13)²

              = 1 – (25/169)

              = 144/169

Hence sin x = ±(12/13)

Since x lies in forth quadrant, sin x is negative. 

Therefore

sin x = (–12/13)

which also gives

cosec x = 1/sin x = (-13/12)

Further, we have

tan x =  sin x/cos x

         =  (-12/13) / (5/13)

         =  (-12/5)

and cot x = 1/tan x = (-5/12) 

5. tan x = –5 / 12，x位于第二个象限中。

解决方案：

Since tan x = (-5/12)

we have cot x = 1/tan x = (-12/5)

Now sec² x = 1 + tan² x

                  = 1 + (25/144)

                  = 169/144

Hence sec x = ±(13/12)

Since x lies in second quadrant, sec x will be negative. Therefore

sec x = (-13/12)

which also gives

cos x = 1/sec x = (-12/13)

Further, we have

sin x = tan x  * cos x

        = (-5/12) x (-12/13) 

        = (5/13)

and cosec x = 1/sin x = (13/5) 

问题。在练习6至10中找到三角函数的值。

6.正弦(765°)

解决方案：

We known that the values of sin x repeats after an interval of 2π or 360^∘. 

So, sin(765°)

= sin(720° + 45°) { taking nearest multiple of 360 }

= sin(2 × 360° + 45°) 

= sin(45°) 

= 1/√2

Hence, sin(765°) = 1/√2

7. cosec(–1410°)

解决方案：

We known that the values of cosec  x  repeats after an interval of 2π or 360^∘. 

So, cosec(-1410°) 

= – cosec(1410°) 

= – cosec(1440° – 30°)    { taking nearest multiple of 360 }

= – cosec(4 × 360° – 30°) 

= cosec(30°) 

= 2

Hence, cosec(–1410°) = 2.

8.棕褐色(19π/ 3)

解决方案：

We known that the values of tan x  repeats after an interval of π or 180^∘.

So, tan(19π/3) 

 = tan(18π/3 + π/3) { breaking into nearset integer } 

 = tan(6π + π/3)

 = tan(π/3) 

 = tan(60°) 

 = √3

Hence, tan(19π/3) = √3. 

9. sin(–11π / 3)

解决方案：

We known that the values of sin x  repeats after an interval of 2π or 360^∘.

So, sin(-11π/3)

= -sin(11π/3) 

= -sin(12π/3 – π/3) { breaking nearest multiple of 2π divisible by 3}

= -sin(4π – π/3)

= -sin(-π/3) 

= -[-sin(π/3)] 

= sin(π/3) 

= sin(60°)

= √3/2

Hence, sin(-11π/3) = √3/2.

10.轻便小床(–15π / 4)

解决方案：

We known that the values of cot x repeats after an interval of π or 180°. 

So, cot(-15π/4)

= -cot(15π/4)

= -cot(16π/4 – π/4) { breaking into nearset multiple of π  divisible by 4 }

= -cot(4π – π/4)

= -cot(-π/4)

= -[-cot(π/4)] 

= cot(45°)

= 1

Hence, cot(-15π/4) = 1.