Simplifying sin(3pi/2-x): An Exploration

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SUMMARY

The discussion focuses on simplifying the expression sin(3π/2 - x) using trigonometric identities. Participants clarify that the goal is to express it in a simpler form, such as A*sin(x) + B*cos(x), rather than sin(θ + x). Key identities mentioned include sin(π + x) = -sin(x) and sin(π/2 - x) = cos(x). The consensus is that while simplification can be subjective, the intended transformation aligns with established trigonometric principles.

PREREQUISITES
  • Understanding of basic trigonometric identities
  • Familiarity with angle subtraction formulas
  • Knowledge of sine and cosine functions
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the derivation and applications of sine and cosine addition formulas
  • Learn how to apply trigonometric identities in simplifying expressions
  • Explore advanced trigonometric transformations and their proofs
  • Practice problems involving angle subtraction and addition in trigonometry
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Students of mathematics, educators teaching trigonometry, and anyone looking to deepen their understanding of trigonometric simplifications.

rachael
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What does "simplify" means in this case:
Simplify each of the following:
1. sin(3pi/2-x)
 
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rachael said:
What does "simplify" means in this case:
Simplify each of the following:
1. sin(3pi/2-x)
Do you mean something like this:
\sin \left( \frac{3 \pi}{2 - x} \right)?
If you mean that, then I don't think it can be simplified any more. :wink:
 
nope i mean sin [(3pi/2)-x]
 
rachael said:
nope i mean sin [(3pi/2)-x]
Ahh, I see. I'll give you some hints. Do you know something like:
\sin (\pi + x) = - \sin x and
\sin \left( \frac{\pi}{2} - x \right) = \cos x?
 
In other words, "simplify" here means "write in a simpler form" which would be something like Asin(x)+ Bcos(x) rather than sin(\theta+x). In my opinion you would certainly be justified in arguing whether or not that is simpler but it is clearly what is intended.
 
In general:
sin(A \pm B)=sinA \, cosB \pm sinB \, cosA
cos(A \pm B)=cosA \, cosB \mp sinA \, sinB

-Dan
 
thank you...
 

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