| sin [ tan-1(4⁄3) + cos-1 (-12⁄13) ] |
| Find the value without using the tables or a calculator. |
Strategy
| - The rule "Must have" of Mr.Zhang ® |
| Identity; sin(α + β) = sinαcosβ + cosαsinβ |
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| Solution |
| Use the identity and inverse properties of Trigonometry; |
  |
| Name | Notation | Definition | Domain of x | Range |
| arctangent | y=arctan x | x=tan y | all real numbers | -π/2 < y < π/2 |
| arccosine | y=arccos x | x=cos y | -1 ≤ x ≤ 1 | 0 ≤ y ≤ π |
tan-1(4⁄3) = arctan(4⁄3)
= α
α = arctan(4⁄3)
∴ tanα = 4⁄3, sinα = 4⁄5, cosα = 3⁄5
cos-1(-12⁄13) = arccos(-12⁄13) = β
β = arccos(-12⁄13)
∴ cosβ = -12⁄13, sinβ = 5⁄13
α = arctan(4⁄3)
∴ tanα = 4⁄3, sinα = 4⁄5, cosα = 3⁄5
cos-1(-12⁄13) = arccos(-12⁄13) = β
β = arccos(-12⁄13)
∴ cosβ = -12⁄13, sinβ = 5⁄13
| sin [ tan-1(4⁄3) + cos-1 (-12⁄13) ] | = | sin(α + β) |
| = | sinα cosβ + cosα sinβ | |
| = | (4⁄5)(-12⁄13) + (3⁄5) (5⁄13) | |
| Ans. | = | -33⁄65 |
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