SINE COSINE PROBLEMS

Problem 1 : 

Prove :

sin 105° + cos 105°  =  cos 45°

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)

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

Hence the value of sin 105°  =  (√3 + 1)/2√2.

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° -----(2) 

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

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

=  (1/2√2) - (√3/2√2)

=  (1 - √3)/2√2

Hence the value of cos 105°  =  (1 - √3)/2√2.

sin 105° +  cos 105°  =  (√3 + 1)/2√2  +  (1 - √3)/2√2

  =  (√3 + 1) + (1 - √3)/2√2

  =  2/2√2

  =  1/√2  =  cos 45°  ----> R.H.S

Problem 2 :

Prove : 

sin 75° − sin 15°  =  cos 105° + cos 15°

Solution :

L.H.S

sin 75°  =  sin (45° + 30°)

By finding expansion using compound angle formula, we get

  =  sin 45° cos 30° + cos 45° sin 30°

  =  (1/√2)(√3/2) + (1/√2)(1/2)

sin 75°  =  (√3 + 1)/2√2  ----(1)

sin 15°  =  sin (45° - 30°)

  =  sin 45° cos 30° - cos 45° sin 30°

  =  (1/√2)(√3/2) - (1/√2)(1/2)

sin 15°  =  (√3 - 1)/2√2  ----(2)

(1) - (2)

sin 75° - sin 15°  =  1/√2  ----> L.H.S

R.H.S

cos 105°  =  (1 - √3)/2√2   -----(3)

cos 15°  =  cos (45° - 30°)

  =  cos 45° cos 30° + sin 45° sin 30°

  =  (1/√2)(√3/2) + (1/√2)(1/2)

cos 15°  =  (√3 + 1)/2√2   -----(4)

(3) + (4)

 cos 105° + cos 15°  =  1/√2  ---> R.H.S

Hence proved.

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