Math, asked by umiko28, 8 months ago

${sin}^{ - 1}( - x ) = { - sin}^{ - 1}x \\ \\ prove$

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Answered by Anonymous

Step-by-step explanation:

$\sf\ {sin}^{ - 1}( - x) = { - sin}^{ - 1}x \\ \\ \sf\ solution \\ \\ \sf\ let \\ \tt\ \implies: {sin}^{ - 1}( - x) = y \\ \\\sf\ \implies: - x = sin \bf\ y \\ \\ \sf\ \implies:x = - sin \: y \\ \\\sf\ \implies:x = sin( - y) \pink{ \: \: \: \: \: \: \: \: \because \: - sin \theta = sin( - \theta)} \\ \\\sf\ \implies: {sin}^{ - 1}x = - y \\ \\ \sf\ \implies: {sin}^{ - 1} x - {sin}^{ - 1}( - x) \\ \\ \boxed{\sf\ \implies: { - sin}^{ - 1}x = {sin}^{ - 1}( - x) }$

Answered by Anonymous

$<body bgcolor = "red">\sf\{sin}^{ - 1}( - x) = { - sin}^{ - 1}x } \\ \\ \sf\ \: solution \\ \\ \sf\ let \\ \tt\ \: \implies: {sin}^{ - 1}( - x) = y\\ \\\sf\ \: \implies: - x = sin \bf\ y \\ \\ \sf\ \: \implies:x = - sin \: y \\ \\\sf\ \: \implies:x = sin( - y) \pink{ \: \: \: \: \: \: \: \: \because \: - sin \theta = sin( - \theta)} \\ \\\sf\ \implies: {sin}^{ - 1}x = - y \\ \\ \sf\ \implies: {sin}^{ - 1} x - {sin}^{ - 1}( - x) \\ \\ \boxed{\sf\ \implies: { - sin}^{ - 1}x = {sin}^{ - 1}( - x) }$

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Step-by-step explanation:

${sin}^{ - 1}( - x ) = { - sin}^{ - 1}x \\ \\ prove$