If cosθ=725 and θ is between 0 and π2, what is the value of sinθ?

pf1996

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If [FONT=MathJax_Main]cos[FONT=MathJax_Math-italic]θ[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]7[/FONT][FONT=MathJax_Main]25[/FONT] and [FONT=MathJax_Math-italic]θ[/FONT] is between [FONT=MathJax_Main]0[/FONT] and [FONT=MathJax_Math-italic]π[/FONT][FONT=MathJax_Main]2[/FONT], what is the value of [FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Math-italic]θ[/FONT]?[/FONT]

The answer must be exact!


I'm unsure how to approach this question. Do i do Cos^-1(7/25) and then plug the number into sin theta?
 
If [FONT=MathJax_Main]cos[FONT=MathJax_Math-italic]θ[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]7[/FONT][FONT=MathJax_Main]25[/FONT] and [FONT=MathJax_Math-italic]θ[/FONT] is between [FONT=MathJax_Main]0[/FONT] and [FONT=MathJax_Math-italic]π[/FONT][FONT=MathJax_Main]2[/FONT], what is the value of [FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Math-italic]θ[/FONT]?[/FONT]

The answer must be exact!


I'm unsure how to approach this question. Do i do Cos^-1(7/25) and then plug the number into sin theta?

1) This is a moderated forum. Your first few posts won't appear until they have been approved by a moderator.

2) You should have learned \(\displaystyle \sin^{2}(\theta) + \cos^{2}(\theta) = 1\ \). See where that leads you.

3) Your calculator typically will provide APPROXIMATE answers - although usually good enough for many applications. In order to find an EXACT answer, it's likely some algebra will be needed. Don't forget your algebra just because you are in trigonometry.
 
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If [FONT=MathJax_Main]cos[FONT=MathJax_Math-italic]θ[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]7[/FONT][FONT=MathJax_Main]25[/FONT] and [FONT=MathJax_Math-italic]θ[/FONT] is between [FONT=MathJax_Main]0[/FONT] and [FONT=MathJax_Math-italic]π[/FONT][FONT=MathJax_Main]2[/FONT], what is the value of [FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Math-italic]θ[/FONT]?[/FONT]

The answer must be exact!


I'm unsure how to approach this question. Do i do Cos^-1(7/25) and then plug the number into sin theta?
I assume that you meant to write [FONT=MathJax_Main]cos[FONT=MathJax_Math-italic]θ[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]7[/FONT][FONT=MathJax_Main]/25[/FONT] and [FONT=MathJax_Math-italic]not [/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Math-italic]θ[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]7[/FONT][FONT=MathJax_Main]25[/FONT].

Simply draw a right triangle, call one of the non 90 degree angles [FONT=MathJax_Math-italic]θ and label two sides so that [/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Math-italic]θ[/FONT][FONT=MathJax_Main] really equals [/FONT][FONT=MathJax_Main]7[/FONT][FONT=MathJax_Main]/25[/FONT]. Then find the third side (I think that Pythagoras would know how to do that). Then you should have enough information to find [FONT=MathJax_Main]sin[FONT=MathJax_Math-italic]θ[/FONT][/FONT] [/FONT]
 
I assume that you meant to write [FONT=MathJax_Main]cos[FONT=MathJax_Math-italic]θ[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]7[/FONT][FONT=MathJax_Main]/25[/FONT] and [FONT=MathJax_Math-italic]not [/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Math-italic]θ[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]7[/FONT][FONT=MathJax_Main]25[/FONT].

Simply draw a right triangle, call one of the non 90 degree angles [FONT=MathJax_Math-italic]θ and label two sides so that [/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Math-italic]θ[/FONT][FONT=MathJax_Main] really equals [/FONT][FONT=MathJax_Main]7[/FONT][FONT=MathJax_Main]/25[/FONT]. Then find the third side (I think that Pythagoras would know how to do that). Then you should have enough information to find [FONT=MathJax_Main]sin[FONT=MathJax_Math-italic]θ[/FONT][/FONT] [/FONT]


Thanks for your answer! So i go 24/25. I was wondering if you have any insight to whether this satisfies the condition of being between 0 and pi/2?
 
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