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प्रश्न
Find the value of the expression in terms of x, with the help of a reference triangle
sin (cos–1(1 – x))
उत्तर
sin (cos–1(1 – x)) = `sin[cos^-1 ("Adj"/"Hyp")]`
`[because cos ("Adj"/"HyP") = (1 - x)/1]`
Adj = 1 – x
Hyp = 1
Opp = `sqrt(1^2 - (1 - x)^2`
= `sqrt(1 - (1 + x^2 - 2x))`
= `sqrt(1 - 1 - x^2 + 2x)`
= `sqrt(2x - x^2`
`sin("Opp"/"Hyp") = sqrt(2x - x^2)/1`
= `sqrt(2x - x^2)`
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