Finding the Exact Value of sin[2arcsin(3/5)] with Inverse Trig Functions

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SUMMARY

The exact value of the expression sin[2arcsin(3/5)] can be determined using the double angle formula sin(2u) = 2sin(u)cos(u). By letting u = arcsin(3/5), we find sin(u) = 3/5. To find cos(u), a right triangle is constructed, yielding cos(u) = 4/5. Substituting these values into the double angle formula results in sin(2u) = 2 * (3/5) * (4/5) = 24/25.

PREREQUISITES
  • Understanding of inverse trigonometric functions, specifically arcsin.
  • Familiarity with the double angle formulas in trigonometry.
  • Basic knowledge of right triangle properties and Pythagorean theorem.
  • Ability to manipulate trigonometric identities.
NEXT STEPS
  • Study the derivation and applications of the double angle formulas in trigonometry.
  • Explore the properties and graphs of inverse trigonometric functions.
  • Practice solving problems involving arcsin and related trigonometric identities.
  • Learn how to construct right triangles based on given trigonometric ratios.
USEFUL FOR

Students studying trigonometry, educators teaching inverse trigonometric functions, and anyone looking to enhance their problem-solving skills in trigonometric equations.

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Homework Statement


Find the exact value of the expression:


Homework Equations


sin[2arcsin(3/5)]


The Attempt at a Solution


I know you're supposed to use sin2x=2sinxcosx somehow but not sure how to start.
 
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page13 said:

Homework Statement


Find the exact value of the expression:


Homework Equations


sin[2arcsin(3/5)]


The Attempt at a Solution


I know you're supposed to use sin2x=2sinxcosx somehow but not sure how to start.

Start by letting u = arcsin(3/5). Then your expression is sin(2u) = ?

It will be helpful to draw a right triangle where u is one of the acute angles. Label the sides and hypotenuse so that sin(u) = 3/5.
 
Ah, OK! 24/25 correct?
 

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