Finding Trig Identities: How Do I Solve Trig Equations with Given Values?

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SUMMARY

The discussion focuses on solving trigonometric equations given the value of tan(theta) = -4/5 and the condition that sin(theta) > 0. The key steps involve using the identity cos^2(t) + sin^2(t) = 1 to derive sin(theta) and cos(theta). The user initially struggles with substituting sin(t) into the identity but later realizes the connection to the Pythagorean theorem. The geometric definitions of trigonometric functions are also emphasized for clarity.

PREREQUISITES
  • Understanding of trigonometric identities, specifically cos^2(t) + sin^2(t) = 1
  • Knowledge of tangent function and its relationship with sine and cosine
  • Familiarity with the Pythagorean theorem as it applies to trigonometry
  • Ability to visualize trigonometric functions in a right triangle context
NEXT STEPS
  • Study the derivation and application of trigonometric identities
  • Learn how to apply the Pythagorean theorem in trigonometric contexts
  • Explore the geometric definitions of sine, cosine, and tangent
  • Practice solving trigonometric equations with various given values
USEFUL FOR

Students studying trigonometry, educators teaching trigonometric identities, and anyone looking to improve their problem-solving skills in trigonometric equations.

Mike_Winegar
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Homework Statement


Problem 6. If you know that tan(theta) = -4/5 and sin(theta) > 0, find:
(a) sin(theta)
(b) cos(theta)
(c) tan(theta + pi)

Homework Equations


cos^2(t)+sin^2(t)=1
tan(t)=sin(t)/cos(t)

The Attempt at a Solution


My teacher went over this today, but likes to skip over steps that I just don't understand. What I've done so far...

-4/5=sin(t)/cos(t)

sin(t)=(-4cos(t)/5)
solved for sin(t)

((-4cos(t)/5)/cos(t))
plugged sin(t) equation back into original equation

ends up being
(-4cos^2(t)/5)=-4/5

We then have the identity that cos^2 + sin^2 = 1

Here's where I get lost. My teacher jumped it directly to:
(-4/5 cos(t))^2 +cos^2(t)=1

I assume you would plug in our original sin(t)=(-4cos(t)/5) into the above identity, but I have no clue as to how she did that without having an additional 4/5th where she substituted the cos side of the equation in for the sin portion of the identity.

Any help would be appreciated.

Lollol

I think I just figured it out actually. Was my teacher using the Pythagorean theorem and not a trig identity?
So:
(-4/5 cos)^2 + cos^2 =r^2
(-4/5 cos)^2 + cos^2 =1?
 
Last edited:
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To do these, it will help if you recall the geometric definitions of the trig functions. tan is opposite over adjacent. So, think of an angle in a triangle, with opposite=-4 and adjacent=5. Find the hypotenuse, then use SOH-CAH-TOA.
 

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