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katovenus [111]
3 years ago
7

Verify the identity cot(x-pi/2)=-tan x

Mathematics
1 answer:
vampirchik [111]3 years ago
4 0
To verify the identity we need the following identities:

i) \displaystyle{ \cot(x)= \frac{\cos x}{\sin x}

ii) \displaystyle{ \sin (x-y)=\ sinx\cdot\ cosy -\ siny\cdot\ cosx

iii) \displaystyle{ \cos (x-y)=\ cosx\cdot \ cosy +\ sinx\cdot\ siny.

Also, we have know that \displaystyle{ \sin \frac{ \pi }{2}=1 and \displaystyle{ \cos \frac{ \pi }{2}=0.


Thus, \displaystyle{ \cot(x-\frac{\pi}{2})= \frac{\cos (x-\frac{\pi}{2})}{\sin (x-\frac{\pi}{2})}

By (ii) and (iii) we have:

\displaystyle{ \frac{\cos (x-\frac{\pi}{2})}{\sin (x-\frac{\pi}{2})}= \frac{\ cosx\cdot \ cos\frac{\pi}{2} +\ sinx\cdot\ sin\frac{\pi}{2}}{\ sinx\cdot\ cos\frac{\pi}{2} -\ sin\frac{\pi}{2}\cdot\ cosx} = \frac{\ sinx}{-\cos x}

by simplifying \displaystyle{ \sin \frac{ \pi }{2}=1 and \displaystyle{ \cos \frac{ \pi }{2}=0.

Now, \displaystyle{ \frac{\ sinx}{-\cos x} is clearly -tanx.
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= 60 × \frac{\pi }{180} ( divide ( cancel) 60 and 180 by 60 )

= \frac{\pi }{3}

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I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).
Aleksandr-060686 [28]

Answer:

\frac{7\pi}{24} and \frac{31\pi}{24}

Step-by-step explanation:

\sqrt{3} \tan(x-\frac{\pi}{8})-1=0

Let's first isolate the trig function.

Add 1 one on both sides:

\sqrt{3} \tan(x-\frac{\pi}{8})=1

Divide both sides by \sqrt{3}:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

Now recall \tan(u)=\frac{\sin(u)}{\cos(u)}.

\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}

or

\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}

The first ratio I have can be found using \frac{\pi}{6} in the first rotation of the unit circle.

The second ratio I have can be found using \frac{7\pi}{6} you can see this is on the same line as the \frac{\pi}{6} so you could write \frac{7\pi}{6} as \frac{\pi}{6}+\pi.

So this means the following:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

is true when x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

where n is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

Add \frac{\pi}{8} on both sides:

x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi

Find common denominator between the first two terms on the right.

That is 24.

x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi

x=\frac{7\pi}{24}+n \pi (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval [0,2\pi).

So if \sqrt{3} \tan(x-\frac{\pi}{8})-1=0 and we let u=x-\frac{\pi}{8}, then solving for x gives us:

u+\frac{\pi}{8}=x ( I just added \frac{\pi}{8} on both sides.)

So recall 0\le x.

Then 0 \le u+\frac{\pi}{8}.

Subtract \frac{\pi}{8} on both sides:

-\frac{\pi}{8}\le u

Simplify:

-\frac{\pi}{8}\le u

-\frac{\pi}{8}\le u

So we want to find solutions to:

\tan(u)=\frac{1}{\sqrt{3}} with the condition:

-\frac{\pi}{8}\le u

That's just at \frac{\pi}{6} and \frac{7\pi}{6}

So now adding \frac{\pi}{8} to both gives us the solutions to:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}} in the interval:

0\le x.

The solutions we are looking for are:

\frac{\pi}{6}+\frac{\pi}{8} and \frac{7\pi}{6}+\frac{\pi}{8}

Let's simplifying:

(\frac{1}{6}+\frac{1}{8})\pi and (\frac{7}{6}+\frac{1}{8})\pi

\frac{7}{24}\pi and \frac{31}{24}\pi

\frac{7\pi}{24} and \frac{31\pi}{24}

5 0
3 years ago
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