How do you verify this identity?
\[\tan \Theta * \sin \Theta + \cos \Theta = \sec \Theta?\]
What did you try?
Did you look at it in terms of sign and cosine? Perhaps try to get it into a fractional form and do the addition?
yes all you have to do is change tangent into sin/cos then multiply cos by cos/cos to get a common denominator moral of the story you will get sin^2+cos^2 over cos which wil give you sec... if that makes sense
You can start with the basic Pythagorean identity... \[\large \sin^2(\theta) + \cos^2(\theta) = 1\] Divide everything by \(\cos (\theta)\) \[\large \frac{\sin^2(\theta)}{\cos(\theta)}+\frac{\cos^2(\theta)}{\cos(\theta)}=\frac1{\cos(\theta)}\] Simplify... \[\large \color{green}{\frac{\sin^2(\theta)}{\cos(\theta)}=\frac{\sin(\theta)\sin(\theta)}{\cos(\theta)}=\frac{\sin(\theta)}{\cos(\theta)}\times\sin(\theta)}=\color{blue}{\tan(\theta)\sin(\theta)}\] Also... \[\large\color{green}{\frac{\cos^2(\theta)}{\cos(\theta)}}=\color{blue}{\cos(\theta)}\] And of course \[\large \color{green}{\frac1{\cos(\theta}}=\color{blue}{\sec(\theta)}\] And therefore, it follows that \[\huge \tan(\theta)\sin(\theta)+\cos(\theta)=\sec(\theta)\]
Use: \[\tan (\theta) = \frac{\sin(\theta)}{\cos(\theta)}\]\[\frac{\sin(\theta) \cdot \sin(\theta)}{\cos(\theta)} + \cos(\theta)\] Now go ahead..
@terenzreignz You wrote a small math book there. Sweet!
I have that unhealthy tendency... to be thorough, @e.mccormick :D
hey it is usually better to be that way especially when it comes to proving something
Thanks! @e.mccormick @jonsysibye @terenzreignz @waterineyes I wasn't exactly sure how to solve it, i only had the sec = 1/cos part, and tan = sin/cos
@terenzreignz I have done that too. Turned one problem into a PDF with LaTeX so I could ue PGF to roperly graph the points to show what negative inverse slope was. You are welcome toadytica305. have fun!
@e.mccormick Just saw you're a CS major.. Me too! :D
Fun fun. machines do exactly what we tell them... even if we tell them to do the wrong thing... hehe. Code is fun!
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