Trigonometric identities problem

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Given sinθ = 0.6, the problem involves calculating tanθ using trigonometric identities. One participant successfully applied the identity sin^2(θ) + cos^2(θ) = 1 to find one value, while discussing the limitations of distributing cosine over addition. It was clarified that cos(90 + θ) equals -sin(θ), not sin(θ), and that functions cannot be distributed over addition unless they are linear. Additionally, a graphical approach using the unit circle was suggested as an alternative method to find the other side of the triangle. The discussion emphasized the importance of correctly applying trigonometric identities and understanding their properties.
angelcase
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Given sinθ = 0.6, calculate tanθ without using the inverse sine function, but instead by using one or more trigonometric identities. You will find two possible values.

I found one of the values using sin^2 (theta) + cos^2 (theta) = 1

I tried using cos (90 + theta)= sin theta to find the second one, but couldn't remember if you were able to distribute the cos...since addition is communitive or whatever that property is called...and get cos 90 + cos theta= sin theta
 
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I'd use these:

tan(x) = sin(x)/cos(x)
Sin = Opposite / Hypotinuse
Cos =Adjacent / Hypotinuse
Opposite^2 + Adjacent^2 = Hypotinuse^2
.6 = 6/10
 


angelcase said:
I tried using cos (90 + theta)= sin theta to find the second one, but couldn't remember if you were able to distribute the cos...since addition is communitive or whatever that property is called...and get cos 90 + cos theta= sin theta
I should talk about this too. You can't distrubte any given function over addition. And cos(90 +x) isn’t sin(x), it’s –sin(x). But in general cos(u + v) = cos(u)cos(v) – sin(u)sin(v)
 


Thank you JonF...the cos(90 + theta)= sin theta was in my text as a trig equation to use...I didn't make it up...I know that the derivative of cos is -sin and the derivative of sin is cos, and tan is sec^2...I get all the derivative stuff..Just seem to have an issue with the basics, which to me is pretty pathetic..On my part.
 


JonF said:
You can't distrubte any given function over addition.
You can if the function in question happens to be linear. The cosine function is apparently not linear so you can not distribute.
 


To the OP, you should find two possible values from sin2θ + cos2θ = 1. This is because there is both a positive and negative square root.
 


Moderator's note: thread moved from General Math to Homework & Coursework Questions area.[/color]
 


Forget identities! Draw a unit circle with two right triangles in it! The angle at the origin will be θ and the hypotenuse (radius) will be 1. If sinθ = 0.6, where will the 0.6 go? And, given a hypotenuse of 1 and one side, could you find the other side, considering the Pythagorean theorem?
 


Unit said:
Forget identities! Draw a unit circle with two right triangles in it! The angle at the origin will be θ and the hypotenuse (radius) will be 1. If sinθ = 0.6, where will the 0.6 go? And, given a hypotenuse of 1 and one side, could you find the other side, considering the Pythagorean theorem?
The OP has already solved half of the problem using the identity
sin^2 (theta) + cos^2 (theta) = 1​
She could either use your graphical method, or she could consider the two solutions for cos(θ) in that equation by taking a negative square root.
 
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