anonymous
  • anonymous
how do i verify cot (x- pi/2)=-tanx i dont know how to do this so pls be patient with me!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jdoe0001
  • jdoe0001
hmmm \(\bf cot\left( x\cfrac{\pi }{2} \right) = -tan(x)?\)
anonymous
  • anonymous
cot (x- pi/2)=-tanx
anonymous
  • anonymous
sorry about that!

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jdoe0001
  • jdoe0001
k
jdoe0001
  • jdoe0001
hmmm what's the \(\bf cos\left(\frac{\pi }{2} \right)\quad and \quad sin\left(\frac{\pi }{2} \right)\)
anonymous
  • anonymous
what?
jdoe0001
  • jdoe0001
ehehh, what are, check your Unit Circle, the \(\bf cos\left(\frac{\pi }{2} \right)\quad and \quad sin\left(\frac{\pi }{2} \right)?\)
anonymous
  • anonymous
-1/2 and 0 ?
jdoe0001
  • jdoe0001
-1/2 and 0? check your unit circle closer
anonymous
  • anonymous
(0,1) ? i dont know i just started
jdoe0001
  • jdoe0001
hmmm, well, \(\bf cos\left(\frac{\pi }{2} \right)=0\quad and \quad sin\left(\frac{\pi }{2} \right)=1\) so we know that much
anonymous
  • anonymous
how is that relevant?
jdoe0001
  • jdoe0001
\(\bf cot\left( x-\frac{\pi }{2} \right)\implies \cfrac{cos\left( x-\frac{\pi }{2} \right)}{sin\left( x-\frac{\pi }{2} \right)} \\ \quad \\ \cfrac{cos(x)cos\left(\frac{\pi }{2} \right)+sin(x)sin\left(\frac{\pi }{2} \right)}{sin(x)cos\left(\frac{\pi }{2} \right)-cos(x)sin\left(\frac{\pi }{2} \right)} \\ \quad \\ \cfrac{cos(x)\cdot {\color{brown}{ 0}}+sin(x)\cdot {\color{brown}{ 1}}}{sin(x)\cdot {\color{brown}{ 0}}-cos(x)\cdot {\color{brown}{ 1}}}\implies \cfrac{sin(x)}{-cos(x)}\implies ?\)
anonymous
  • anonymous
wait wait wait how did you get that huge second fraction
jdoe0001
  • jdoe0001
recall \(\bf \textit{Sum and Difference Identities} \\ \quad \\ sin({\color{brown}{ \alpha}} - {\color{blue}{ \beta}})=sin({\color{brown}{ \alpha}})cos({\color{blue}{ \beta}})- cos({\color{brown}{ \alpha}})sin({\color{blue}{ \beta}}) \\ \quad \\ cos({\color{brown}{ \alpha}} - {\color{blue}{ \beta}})= cos({\color{brown}{ \alpha}})cos({\color{blue}{ \beta}}) + sin({\color{brown}{ \alpha}})sin({\color{blue}{ \beta}})\)
anonymous
  • anonymous
a=x and b=pi/2 ???
jdoe0001
  • jdoe0001
yeap
anonymous
  • anonymous
so sin/-cos=-tan because tan=sin/cos right
jdoe0001
  • jdoe0001
\(\bf \textit{Sum and Difference Identities} \\ \quad \\ sin({\color{brown}{ x}} - {\color{blue}{ \frac{\pi }{2}}})=sin({\color{brown}{ x}})cos({\color{blue}{ \frac{\pi }{2}}})- cos({\color{brown}{ x}})sin({\color{blue}{ \frac{\pi }{2}}}) \\ \quad \\ cos({\color{brown}{ x}} - {\color{blue}{ \frac{\pi }{2}}})= cos({\color{brown}{ x}})cos({\color{blue}{ \beta}}) + sin({\color{brown}{ x}})sin({\color{blue}{ \frac{\pi }{2}}})\)
jdoe0001
  • jdoe0001
yeap
jdoe0001
  • jdoe0001
\(\bf \cfrac{cos(x)\cdot {\color{brown}{ 0}}+sin(x)\cdot {\color{brown}{ 1}}}{sin(x)\cdot {\color{brown}{ 0}}-cos(x)\cdot {\color{brown}{ 1}}}\implies \cfrac{sin(x)}{-cos(x)}\implies \cfrac{sin(x)}{-1\cdot cos(x)} \\ \quad \\ \cfrac{1}{-1}\cdot \cfrac{sin(x)}{cos(x)}\implies - \cfrac{sin(x)}{cos(x)}\implies -tan(x)\)
anonymous
  • anonymous
thank you sooo much
jdoe0001
  • jdoe0001
yw

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