prove that : cos [sin ^-1 3/5 + sin^-1 5/13] = 33/65
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prove that : cos [sin ^-1 3/5 + sin^-1 5/13] = 33/65
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[tex]\bold{To \: prove = >}\\ cos( \: {sin}^{ - 1} ( \dfrac{3}{5} ) + {sin}^{ - 1} ( \dfrac{5}{13} ) \: ) \: = \dfrac{33}{65} [/tex]
[tex]\bold{Concept \: used }= > \\ 1)sin ^{ - 1} x + {sin}^{ - 1}y = sin ^{ - 1} (x \sqrt{1 - {y}^{2} } + y \sqrt{1 - {x}^{2} } )[/tex]
[tex]2) \: {sin}^{ - 1} x = {cos}^{ - 1} \sqrt{1 - {x}^{2} } [/tex]
[tex]\bold{Solution = > } {sin}^{ - 1} \dfrac{3}{5} + {sin}^{ - 1} \dfrac{5}{13} [/tex]
[tex] = {sin}^{ - 1} ( \dfrac{3}{5} \sqrt{1 - ( \dfrac{5}{13} } )^{2} + \dfrac{5}{13} \sqrt{1 - { (\dfrac{3}{5} )}^{2} } [/tex]
[tex] = {sin}^{ - 1} ( \dfrac{3}{5} \sqrt{1 - \dfrac{25}{169} } + \dfrac{5}{13} \sqrt{1 - \dfrac{9}{25} } )[/tex]
[tex] = {sin}^{ - 1} ( \dfrac{3}{5} \sqrt{ \dfrac{169 - 25}{169} } + \dfrac{5}{13} \sqrt{ \dfrac{25 - 9}{25} } )[/tex]
[tex] = {sin}^{ - 1} ( \dfrac{3}{5} \: \dfrac{12}{13} + \dfrac{5}{13} \: \dfrac{4}{5} )[/tex]
[tex] = {sin}^{ - 1} ( \dfrac{36}{65} \: + \dfrac{20}{65} )[/tex]
[tex] = {sin}^{ - 1} \dfrac{36 + 20}{65} [/tex]
[tex] = {sin}^{ - 1} ( \dfrac{56}{65} )[/tex]
[tex] = {cos}^{ - 1} \sqrt{1 - { (\dfrac{56}{65} )}^{2} } [/tex]
[tex] = {cos}^{ - 1} \sqrt{1 - \dfrac{3136}{4225} } [/tex]
[tex] = {cos}^{ - 1} \sqrt{ \dfrac{4225 - 3136}{4225} } [/tex]
[tex] = {cos}^{ - 1} \sqrt{ \dfrac{1089}{4225} } [/tex]
[tex] {sin}^{ - 1} ( \dfrac{3}{5} ) + {sin}^{ - 1} ( \dfrac{5}{13} ) = {cos}^{ - 1} ( \dfrac{33}{65} )[/tex]
[tex] taking \: cos \: both \: sides \: we \: get[/tex]
[tex] = > cos \: ( {sin}^{ - 1} \dfrac{3}{5} + {sin}^{ - 1} \dfrac{5}{13} ) = cos \: {cos}^{ - 1} ( \dfrac{33}{65} )[/tex]
[tex] = >cos( {sin}^{ - 1} \dfrac{3}{5} + {sin}^{ - 1} \dfrac{5}{13} ) = \dfrac{33}{65} [/tex]
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