VALUES OF TRIGONOMETRIC RATIOS FOR DIFFERENT ANGLES

Example 1 :

Find the value of cos 105°.

Solution :

cos 105°  =  cos (60° + 45°)

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

cos (60 + 45)  =  cos 60° cos 45° - sin 60° sin 45°  ----(1)

sin 45°  =  1/√2

sin 60  =  √3/2

cos 45°  =  1/√2

cos 60°  =  1/2

By applying the above values in the first equation, we get

cos (60 + 45)  =  (1/2) (1/√2) - (√3/2)(1/√2)

=  (1/2√2) - (√3/2√2)

=  (1 - √3)/2√2

So, the value of cos 105° is (1 - √3)/2√2.

Example 2 :

Find the value of sin 105°.

Solution :

sin 105°  =  sin (60 + 45)

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

sin (60 + 45)  =  sin 60° cos 45° + cos 60° sin 45°  ----(1)

sin 60°  =  √3/2

sin 45°  =  1/√2

cos 60°  =  1/2

cos 45°  =  1/√2

By applying the above values in the first equation, we get

sin (60 + 45)  =  (√3/2)(1/√2) + (1/√2)(1/2)

=  (√3/2√2) + (1/2√2)

=  (√3 + 1)/2√2

So, the value of sin 105° is (√3 + 1)/2√2.

Example 3 :

Find the value of tan 7π/12.

Solution :

 tan 7π/12  =  tan 105°

tan 105°  =  tan (60° + 45°)

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

  =  (tan 60 + tan 45)/(1 - tan 60 tan 45)

  =  (√3 + 1) / (1 - √3(1))

  =  (1 + √3) / (1 - √3)

Multiply the above fraction by its conjugate.

  =  [(1 + √3) / (1 - √3)] ⋅ [ (1 + √3) /  (1 + √3)]

  =  (1 + √3)2 / (1 - √3) (1 + √3)

  =  (1 + 2√3 + 3) / (1 - 3)

  =  (4 + 2√3) / (-2)

  =  -2 (2 + √3)/ 2

  =  -(2 + √3) 

So, the value of tan 7π/12 is -(2 + √3).

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