FIND THE VALUES OF THE OTHER FIVE TRIGONOMETRIC FUNCTIONS OF THETA

In six trigonometric ratios sin, cos, tan, csc, sec and cot, if the value of one of the ratios is given, we can find the values of the other five functions. 

The following steps will be useful in the above process.  

Step 1 :

The given given trigonometric ratio has to compared with one of the formulas given below.

sin θ = Opposite side/hypotenuse side

cos θ = Adjacent side/hypotenuse side

tan θ = Opposite side/Adjacent side

cosec θ = Hypotenuse side/Opposite side

sec θ = Hypotenuse side/Adjacent side

cot θ = Adjacent side/Opposite side 

Step 2 :

Now, find the missing side from the given known sides.

Step 3 :

By using the above formula, find the values of the other trigonometric ratios.

Examples

Example 1 :

If sin A  =  9/15, find the other trigonometric ratios

Solution :

sin θ  =  Opposite side/hypotenuse side

sin A  =  9/15

Opposite side  =  9, Hypotenuse side  =  15

(Hypotenuse side)2  =  (Opposite side)2 + (Adjacent side)2

Adjacent side  =  √152 - 92

  =  √225 - 81

  =  √144

Adjacent side  =  12

cos A  =  Adjacent side/hypotenuse side  =  12/15

tan A  =  Opposite side/Adjacent side  =  9/12

cosec A  =  Hypotenuse side/Opposite side  =  15/9

sec A  =  Hypotenuse side/Adjacent side  =  15/12

cot A  =  Adjacent side/Opposite side  =  12/9

Example 2 :

If cos A  =  15/17, find the other trigonometric ratios

Solution :

cos θ  =  Adjacent side/hypotenuse side

cos A  =  15/17

Adjacent side  =  15, Hypotenuse side  =  17

(Hypotenuse side)2  =  (Opposite side)2 + (Adjacent side)2

Opposite side  =  √172 - 152

  =  √289 - 225

  =  √64

Opposite side  =  8

sin A  =  Opposite side/hypotenuse side  =  8/17

tan A  =  Opposite side/Adjacent side  =  8/15

cosec A  =  Hypotenuse side/Opposite side  =  17/8

sec A  =  Hypotenuse side/Adjacent side  =  17/15

cot A  =  Adjacent side/Opposite side  =  15/8

Example 3 :

If sec θ  =  17/8, find the other trigonometric ratios

Solution :

sec θ  =  Hypotenuse side/Adjacent side

sec θ  =  17/8

Hypotenuse side  =  17, Adjacent side  =  8

(Hypotenuse side)2  =  (Opposite side)2 + (Adjacent side)2

Opposite side  =  √172 - 82

  =  √289 - 64

  =  √225

Opposite side  =  15

sin θ  =  Opposite side/hypotenuse side  =  15/17

cos θ  =  Adjacent side/hypotenuse side  =  8/17

tan θ  =  Opposite side/Adjacent side  =  15/8

cosec θ  =  Hypotenuse side/Opposite side  =  17/15

cot θ  =  Adjacent side/Opposite side  =  8/15

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Digital SAT Math Problems and Solutions (Part - 213)

    Jul 13, 25 09:51 AM

    digitalsatmath292.png
    Digital SAT Math Problems and Solutions (Part - 213)

    Read More

  2. Digital SAT Math Problems and Solutions (Part - 212)

    Jul 13, 25 09:32 AM

    digitalsatmath290.png
    Digital SAT Math Problems and Solutions (Part - 212)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 211)

    Jul 11, 25 08:34 AM

    digitalsatmath289.png
    Digital SAT Math Problems and Solutions (Part - 211)

    Read More