DUC_123 said:Good morning, may I possibly get a clue into solving this problem : check this identity: sin[1/2 sin^−1 (x) ]=1/2 X (sqr(1+x) –sqr(1-x)). I tried to say "2y=sin^−1 (x)", Then f=sin(y), then "x=sin(2y)", x=2sin(y) X cos(y), but I can't find further. Thank you very much.
A trig identity is an equation that is true for all values of the variables involved. It is used to simplify and manipulate trigonometric expressions.
To solve a trig identity, you must use algebraic manipulations and trigonometric identities to transform the given expression into a simpler form that is easier to evaluate.
The inverse trigonometric function is the inverse of a trigonometric function. It takes a ratio of sides in a right triangle and returns the angle that would produce that ratio.
To solve for x in the given trig identity, you must first use the inverse trigonometric function to isolate the trigonometric expression. Then, use algebraic manipulations to simplify the remaining expression and solve for x.
First, use the inverse sine function to rewrite the left side of the equation as sin^−1(x/2). Then, use the double angle formula for sine to simplify the expression to sin^−1(x/2) = 1/2(sin^−1(x) + sin^−1(x)). Next, use the identity sin^−1(x) = x to simplify the expression to 1/2(x + x) = x. Finally, use the identity sqr(1-x) = 1-x to rewrite the right side of the equation as x(sqr(1+x) – 1+x). The final step is to distribute the x and simplify the expression to x = 1/2(sqr(1+x) – sqr(1-x)).