EVALUATING TRIGONOMETRIC FUNCTIONS OF SPECIAL ANGLES

Example 1 :

Evaluate :

sin45° + cos45°

Solution :

sin45° = 1/√2

cos45° = 1/√2

sin45° + cos45° = 1/√2 + 1/√2

= (1 + 1)/√2

 = 2/√2

= (√2 ⋅ √2) / √2

= √2

Example 2 :

Evaluate :

sin60° tan30°

Solution :

sin60° = √3/2

tan30° = 1/√3

sin60°cos30° = (√3/2) ⋅ (1/√3)

= 1/2

Example 3 :

Evaluate :

tan45°/(tan30° + tan60°)

Solution :

tan45° = 1

tan30° = 1/2

tan60° = √3

tan45°/(tan30° + tan60°) = 1/[(1/√3) + √3]

  = 1/[(1 + 3)/√3]

= √3/4

Example 4 :

Evaluate :

tan²60° - 2tan²45° - cot²30° + 2sin²30

Solution :

tan²60° = (tan60°)²  = (√3)² = 3

tan²45° = (tan45°)²  = (1)² = 1

cot²30° = (cot30°)²  = (√3)² = 3

sin²30° = (sin30°)²  = (1/2)² = 1/4

 = 3 - 2 (1) - 3 + 2(1/4)

= -2 + 1/2

= (-4 + 1)/2 =

-3/2

Example 5 :

Evaluate :

4(sin⁴30° + cos⁴60°) - 3(cos²45° - sin²90°)

Solution :

sin⁴30° = (sin30°)⁴  = (1/2)⁴ = 1/16

cos⁴60° = (cos60°)⁴  = (1/2)⁴ = 1/16

cos²45° = (cos45°)² = (1/√2)² = 1/2

sin²90° = (sin90°)² = (1)² = 1

= 4 [(1/16) + (1/16)] - 3[(1/2) - 1]

= 4(2/16)  -  3 (-1/2)

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

= (1 + 3)/2

= 4/2

= 2

Example 6 :

Evaluate :

6cos²90° + 3sin²90° + 4tan²45°

Solution :

cos²90° = (cos90°)² = (0)² = 0

sin²90° = (sin90°)² = (1)² = 1

tan²45° = (tan45°)² = (1)² = 1

6cos²90° + 3sin²90° + 4tan²45° = 6(0) + 3(1) + 4(1)

= 0 + 3 + 4

= 7

Example 7 :

Evaluate :

4cot²45  - sec²60 + sin²60 + cos²60

Solution :

cot²45° = (cot 45°)² = (1)² = 1

sec²60 = (sec 60°)² = (2)² = 4

sin²60 = (sin 60°)² = (√3/2)² = 3/4

cos²60 = (cos 60°)² = (1/2)² = 1/4

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

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

= 4/4

= 1

Example 8 :

Evaluate :

sin 30°cos60° + cos30°sin60°

Solution :

sin30° = 1/2

cos60° = 1/2

cos30° = √3/2

sin60° = √3/2

sin30°cos60° + cos30°sin60° :

= (1/2)(1/2) + (√3/2)(√3/2)

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

= (1 + 3)/4

= 4/4

= 1

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