Proofs/Verify Trigonometric Functions

FritoTaco

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Nov 14, 2015
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Hi, I don't understand how to do proofs. I know you have to look at reciprocal identities and Pythagorean identities to help solve but there's many ways to solve them and I don't understand except that I need to start on the side that looks more complicated.

5.sin(θ)csc(θ)+cos(θ)sec(θ)=1\displaystyle 5.\, \dfrac{\sin(\theta)}{\csc(\theta)}\, +\, \dfrac{\cos(\theta)}{\sec(\theta)}\, =\, 1

8.sec(β)csc(β)tan(β)+cot(β)=1\displaystyle 8.\, \dfrac{\sec(\beta)\, \csc(\beta)}{\tan(\beta)\, +\, \cot(\beta)}\, =\, 1

Can anyone please help me understand this? Thank you, I have so many problems to do so I'll just show one of them.
 
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I've moved this trig question to the "Trig" category, and have typed out the contents of the too-huge-to-display, sideways, and upside-down graphics.

...I don't understand except that I need to start on the side that looks more complicated.
So... What did you get when you started with the "more complicated" side?

5.sin(θ)csc(θ)+cos(θ)sec(θ)=1\displaystyle 5.\, \dfrac{\sin(\theta)}{\csc(\theta)}\, +\, \dfrac{\cos(\theta)}{\sec(\theta)}\, =\, 1
The left-hand side is the more complicated. What did you get when you converted everything to sines and cosines? Which identity did you then apply? Where did you get stuck?

8.sec(β)csc(β)tan(β)+cot(β)=1\displaystyle 8.\, \dfrac{\sec(\beta)\, \csc(\beta)}{\tan(\beta)\, +\, \cot(\beta)}\, =\, 1
The left-hand side is the more complicated. What did you get when you converted everything to sines and cosines? Which identity did you then apply? Where did you get stuck?

Please be complete. Thank you! ;)
 
Oops, didn't realize it got move. One questions, how did you format my questions like that? I would like to do it for another time I need to ask a question.

Also I have so far is:

5.) sinθ / 1/sinθ + cosθ / 1/cosθ = 1.

I got sinθ / 1/sinθ because in the original question, it had sinθ / cscθ and I know the reciprocal identity of cscθ is 1/sinθ.

And for the other:

8.) 1/ cosβ * 1/sinβ / sinβ/cosβ + cosβ/sinβ.

In case you don't know exactly what that looks like its the top half is 1/sinβ * 1/sinβ all over sinβ/cosβ + cosβ/sinβ = 1.
 
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Oops, didn't realize it got move. One questions, how did you format my questions like that? I would like to do it for another time I need to ask a question.
I used LaTeX. You can find various lessons on how to use LaTeX coding. In the meantime, though, please review standard web-safe formatting (here); in particular, note that grouping symbols can be crucial.

Also I have so far is:

5.) sinθ / 1/sinθ + cosθ / 1/cosθ = 1.
I will guess that the above is meant to be as follows:

. . . . .sin(θ) / (1/sin(θ)) + cos(θ) / (1/cos(θ)) = 1

...which typesets as:

. . . . .sin(θ)(1sin(θ))+cos(θ)(1cos(θ))=1\displaystyle \dfrac{\sin(\theta)}{\left(\dfrac{1}{\sin(\theta)}\right)}\, +\, \dfrac{\cos(\theta)}{\left(\dfrac{1}{\cos(\theta)}\right)}\, =\, 1

I got sinθ / 1/sinθ because in the original question, it had sinθ / cscθ and I know the reciprocal identity of cscθ is 1/sinθ.
Okay. Then you flipped and multiplied (to simplify the complex fractions), applied the obvious identity, and... then what? Where are you stuck?

And for the other:

8.) 1/ cosβ * 1/sinβ / sinβ/cosβ + cosβ/sinβ.

In case you don't know exactly what that looks like its the top half is 1/sinβ * 1/sinβ all over sinβ/cosβ + cosβ/sinβ = 1.
I think the above means this:

. . . . .(1cos(β)1sin(β))(sin(β)cos(β)+cos(β)sin(β))\displaystyle \dfrac{\left(\, \dfrac{1}{\cos(\beta)}\, *\, \dfrac{1}{\sin(\beta)}\, \right)}{\left(\, \dfrac{\sin(\beta)}{\cos(\beta)}\, +\, \dfrac{\cos(\beta)}{\sin(\beta)}\, \right)}

You multiplied on top. You converted to a common denominator underneath, combined, simplified, and then flipped and multiplied against the numerator. And... then what? Where are you stuck?

Please show all of your work and reasoning. Thank you! ;)
 
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