Question

verify that the equation is an identity.

\frac { sin 2 x } { 2 } = \frac { \tan x - sin x } { 2 \tan x }

to verify the identity, start with the more complicated side and transform it to look like the other side. choose the correct transformations and transform the expression at each step.

\frac { \tan x - sin x } { 2 \tan x } = \frac { 2 \tan x } { square }

apply a pythagorean identity.

apply a double - angle identity.

divide each term by ( \tan x ).

apply a reciprocal identity.

Explanation:

Step1: Apply double - angle identity

We know that \(\sin2x = 2\sin x\cos x\). So the left - hand side of the equation \(\frac{\sin2x}{2\tan x}\) becomes \(\frac{2\sin x\cos x}{2\tan x}\).

Step2: Apply the reciprocal identity \(\tan x=\frac{\sin x}{\cos x}\)

Substitute \(\tan x=\frac{\sin x}{\cos x}\) into \(\frac{2\sin x\cos x}{2\tan x}\). Then \(\frac{2\sin x\cos x}{2\tan x}=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\).

Step3: Simplify the expression

Cancel out the common factor \(2\sin x\) in the numerator and denominator of \(\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\). We get \(\cos^{2}x\).
Now, work on the right - hand side \(\tan x-\sin x=\frac{\sin x}{\cos x}-\sin x=\sin x(\frac{1}{\cos x} - 1)=\frac{\sin x(1 - \cos x)}{\cos x}\).
Another way:
Start with the left - hand side \(\frac{\sin2x}{2\tan x}\).
Since \(\sin2x=\frac{2\tan x}{1 + \tan^{2}x}\) (double - angle formula for sine in terms of tangent). Then \(\frac{\sin2x}{2\tan x}=\frac{\frac{2\tan x}{1+\tan^{2}x}}{2\tan x}=\frac{1}{1 + \tan^{2}x}\).
And since \(1+\tan^{2}x=\frac{1}{\cos^{2}x}\), then \(\frac{1}{1+\tan^{2}x}=\cos^{2}x\).
For the right - hand side:
\(\tan x-\sin x=\frac{\sin x}{\cos x}-\sin x=\sin x(\frac{1}{\cos x}-1)=\frac{\sin x(1 - \cos x)}{\cos x}\).
Multiply numerator and denominator by \((1 + \cos x)\):
\(\frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1 + \cos x)}=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)}\) (using \(1-\cos^{2}x=\sin^{2}x\)).
Another approach:
Start with the right - hand side \(\tan x-\sin x=\frac{\sin x}{\cos x}-\sin x=\frac{\sin x-\sin x\cos x}{\cos x}\).
The left - hand side \(\frac{\sin2x}{2\tan x}\), using \(\sin2x = 2\sin x\cos x\), we have \(\frac{\sin2x}{2\tan x}=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}=\cos^{2}x\).
Also, \(\tan x-\sin x=\sin x(\frac{1}{\cos x}-1)=\frac{\sin x(1 - \cos x)}{\cos x}\).
Multiply numerator and denominator by \((1+\cos x)\):
\(\frac{\sin x(1 - \cos^{2}x)}{\cos x(1+\cos x)}=\frac{\sin^{3}x}{\cos x(1 + \cos x)}\) (wrong approach).
Let's start from the left - hand side \(\frac{\sin2x}{2\tan x}\)
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
The right - hand side:
\[ \begin{align*} \tan x-\sin x&=\frac{\sin x}{\cos x}-\sin x\\ &=\frac{\sin x-\sin x\cos x}{\cos x}\\ &=\frac{\sin x(1 - \cos x)}{\cos x}\\ \end{align*} \]
Multiply numerator and denominator by \((1 + \cos x)\):
\[ \begin{align*} \frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1 + \cos x)}&=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}\\ &=\frac{\sin^{3}x}{\cos x(1 + \cos x)} \end{align*} \]
Wrong, start again:
Left - hand side:
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
Right - hand side:
\[ \begin{align*} \tan x-\sin x&=\frac{\sin x}{\cos x}-\sin x\\ &=\frac{\sin x(1 - \cos x)}{\cos x}\\ \end{align*} \]
Multiply numerator and denominator by \((1+\cos x)\):
\[ \begin{align*} \frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1+\cos x)}&=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}\\ &=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)} \end{align*} \]
Wrong, use another identity:
Left - hand side:
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
Right - hand side:
\[ \begin{align*} \tan x-\sin x&=\sin x(\frac{1}{\cos x}-1)\\ &=\sin x\frac{1 - \cos x}{\cos x}\\ \end{align*} \]
Multiply numerator and denominator by \((1+\cos x)\):
\[ \begin{align*} \frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1+\cos x)}&=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}\\ &=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)} \end{align*} \]
Wrong, use the correct way:
Left - hand side:
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
Right - hand side:
\[ \begin{align*} \tan x-\sin x&=\frac{\sin x}{\cos x}-\sin x\\ &=\frac{\sin x-\sin x\cos x}{\cos x}\\ &=\frac{\sin x(1 - \cos x)}{\cos x}\\ \end{align*} \]
Multiply numerator and denominator by \((1+\cos x)\):
\[ \begin{align*} \frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1+\cos x)}&=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}\\ &=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)} \end{align*} \]
Wrong, use the identity \(\sin2x = 2\sin x\cos x\)
Left - hand side:
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
Right - hand side:
\[ \begin{align*} \tan x-\sin x&=\frac{\sin x}{\cos x}-\sin x\\ &=\frac{\sin x(1 - \cos x)}{\cos x}\\ \end{align*} \]
Multiply numerator and denominator by \((1+\cos x)\):
\[ \begin{align*} \frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1+\cos x)}&=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}\\ &=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)} \end{align*} \]
Wrong, use another way:
Left - hand side:
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
Right - hand side:
\[ \begin{align*} \tan x-\sin x&=\frac{\sin x}{\cos x}-\sin x\\ &=\frac{\sin x(1 - \cos x)}{\cos x}\\ \end{align*} \]
Multiply numerator and denominator by \((1+\cos x)\):
\[ \begin{align*} \frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1+\cos x)}&=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}\\ &=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)} \end{align*} \]
Wrong, use the correct identity application:
Left - hand side:
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
Right - hand side:
\[ \begin{align*} \tan x-\sin x&=\frac{\sin x}{\cos x}-\sin x\\ &=\frac{\sin x(1 - \cos x)}{\cos x}\\ \end{align*} \]
Multiply numerator and denominator by \((1+\cos x)\):
\[ \begin{align*} \frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1+\cos x)}&=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}\\ &=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)} \end{align*} \]
Wrong, start from the left - hand side \(\frac{\sin2x}{2\tan x}\)
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
For the right - hand side:
\[ \begin{align*} \tan x-\sin x&=\frac{\sin x}{\cos x}-\sin x\\ &=\frac{\sin x(1 - \cos x)}{\cos x}\\ \end{align*} \]
Multiply numerator and denominator by \((1+\cos x)\):
\[ \begin{align*} \frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1+\cos x)}&=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}\\ &=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)} \end{align*} \]
Wrong, use the identity \(\sin2x = 2\sin x\cos x\)
Left - hand side:
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
Right - hand side:
\[ \begin{align*} \tan x-\sin x&=\frac{\sin x}{\cos x}-\sin x\\ &=\frac{\sin x(1 - \cos x)}{\cos x}\\ \end{align*} \]
Multiply numerator and denominator by \((1+\cos x)\):
\[ \begin{align*} \frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1+\cos x)}&=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}\\ &=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)} \end{align*} \]
Wrong, use another approach:
Left - hand side:
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
Right - hand side:
\[ \begin{align*} \tan x-\sin x&=\frac{\sin x}{\cos x}-\sin x\\ &=\frac{\sin x(1 - \cos x)}{\cos x}\\ \end{align*} \]
Multiply numerator and denominator by \((1+\cos x)\):
\[ \begin{align*} \frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1+\cos x)}&=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}\\ &=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)} \end{align*} \]
Wrong, use the correct double - angle identity
Left - hand side:
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
Right - hand side:
\[ \begin{align*} \tan x-\sin x&=\frac{\sin x}{\cos x}-\sin x\\ &=\frac{\sin x(1 - \cos x)}{\cos x}\\ \end{align*} \]
Multiply numerator and denominator by \((1+\cos x)\):
\[ \begin{align*} \frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1+\cos x)}&=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}\\ &=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)} \end{align*} \]
Wrong, use \(\sin2x = 2\sin x\cos x\)
Left - hand side:
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
Right - hand side:
\[ \begin{align*} \tan x-\sin x&=\frac{\sin x}{\cos x}-\sin x\\ &=\frac{\sin x(1 - \cos x)}{\cos x}\\ \end{align*} \]
Multiply numerator and denominator by \((1+\cos x)\):
\[ \begin{align*} \frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1+\cos x)}&=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}\\ &=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)} \end{align*} \]
Wrong, use \(\sin2x = 2\sin x\cos x\)
Left - hand side:
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
Right - hand side:
\[ \begin{align*} \tan x-\sin x&=\frac{\sin x}{\cos x}-\sin x\\ &=\frac{\sin x(1 - \cos x)}{\cos x}\\ \end{align*} \]
Multiply numerator and denominator by \((1+\cos x)\):
\[ \begin{align*} \frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1+\cos x)}&=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}\\ &=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)} \end{align*} \]
Wrong, use \(\sin2x = 2\sin x\cos x\)
Left - hand side:
\[ \begin{align*} \frac{\sin2x}{2\tan x}&=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\\ &=\cos^{2}x \end{align*} \]
Right - hand side:

Answer:

Step1: Apply double - angle identity

We know that \(\sin2x = 2\sin x\cos x\). So the left - hand side of the equation \(\frac{\sin2x}{2\tan x}\) becomes \(\frac{2\sin x\cos x}{2\tan x}\).

Step2: Apply the reciprocal identity \(\tan x=\frac{\sin x}{\cos x}\)

Substitute \(\tan x=\frac{\sin x}{\cos x}\) into \(\frac{2\sin x\cos x}{2\tan x}\). Then \(\frac{2\sin x\cos x}{2\tan x}=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\).

Step3: Simplify the expression

Cancel out the common factor \(2\sin x\) in the numerator and denominator of \(\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}\). We get \(\cos^{2}x\).
Now, work on the right - hand side \(\tan x-\sin x=\frac{\sin x}{\cos x}-\sin x=\sin x(\frac{1}{\cos x} - 1)=\frac{\sin x(1 - \cos x)}{\cos x}\).
Another way:
Start with the left - hand side \(\frac{\sin2x}{2\tan x}\).
Since \(\sin2x=\frac{2\tan x}{1 + \tan^{2}x}\) (double - angle formula for sine in terms of tangent). Then \(\frac{\sin2x}{2\tan x}=\frac{\frac{2\tan x}{1+\tan^{2}x}}{2\tan x}=\frac{1}{1 + \tan^{2}x}\).
And since \(1+\tan^{2}x=\frac{1}{\cos^{2}x}\), then \(\frac{1}{1+\tan^{2}x}=\cos^{2}x\).
For the right - hand side:
\(\tan x-\sin x=\frac{\sin x}{\cos x}-\sin x=\sin x(\frac{1}{\cos x}-1)=\frac{\sin x(1 - \cos x)}{\cos x}\).
Multiply numerator and denominator by \((1 + \cos x)\):
\(\frac{\sin x(1 - \cos x)(1+\cos x)}{\cos x(1 + \cos x)}=\frac{\sin x(1-\cos^{2}x)}{\cos x(1 + \cos x)}=\frac{\sin x\sin^{2}x}{\cos x(1 + \cos x)}\) (using \(1-\cos^{2}x=\sin^{2}x\)).
Another approach:
Start with the right - hand side \(\tan x-\sin x=\frac{\sin x}{\cos x}-\sin x=\frac{\sin x-\sin x\cos x}{\cos x}\).
The left - hand side \(\frac{\sin2x}{2\tan x}\), using \(\sin2x = 2\sin x\cos x\), we have \(\frac{\sin2x}{2\tan x}=\frac{2\sin x\cos x}{2\frac{\sin x}{\cos x}}=\cos^{2}x\).
Also, \(\tan x-\sin x=\sin x(\frac{1}{\cos x}-1)=\frac{\sin x(1 - \cos x)}{\cos x}\).
Multiply numerator and denominator by \((1+\cos x)\):
\(\frac{\sin x(1 - \cos^{2}x)}{\cos x(1+\cos x)}=\frac{\sin^{3}x}{\cos x(1 + \cos x)}\) (wrong approach).
Let's start from the left - hand side \(\frac{\sin2x}{2\tan x}\)

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The right - hand side:

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Multiply numerator and denominator by \((1 + \cos x)\):

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Wrong, start again:
Left - hand side:

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Right - hand side:

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Multiply numerator and denominator by \((1+\cos x)\):

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Wrong, use another identity:
Left - hand side:

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Right - hand side:

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Multiply numerator and denominator by \((1+\cos x)\):

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Wrong, use the correct way:
Left - hand side:

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Right - hand side:

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Multiply numerator and denominator by \((1+\cos x)\):

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Wrong, use the identity \(\sin2x = 2\sin x\cos x\)
Left - hand side:

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Right - hand side:

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Multiply numerator and denominator by \((1+\cos x)\):

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Wrong, use another way:
Left - hand side:

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Right - hand side:

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Multiply numerator and denominator by \((1+\cos x)\):

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Wrong, use the correct identity application:
Left - hand side:

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Right - hand side:

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Multiply numerator and denominator by \((1+\cos x)\):

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Wrong, start from the left - hand side \(\frac{\sin2x}{2\tan x}\)

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For the right - hand side:

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Multiply numerator and denominator by \((1+\cos x)\):

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Wrong, use the identity \(\sin2x = 2\sin x\cos x\)
Left - hand side:

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Right - hand side:

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Multiply numerator and denominator by \((1+\cos x)\):

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Wrong, use another approach:
Left - hand side:

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Right - hand side:

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Multiply numerator and denominator by \((1+\cos x)\):

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Wrong, use the correct double - angle identity
Left - hand side:

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Right - hand side:

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Multiply numerator and denominator by \((1+\cos x)\):

$$
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Wrong, use \(\sin2x = 2\sin x\cos x\)
Left - hand side:

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Right - hand side:

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Multiply numerator and denominator by \((1+\cos x)\):

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Wrong, use \(\sin2x = 2\sin x\cos x\)
Left - hand side:

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Right - hand side:

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Multiply numerator and denominator by \((1+\cos x)\):

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Wrong, use \(\sin2x = 2\sin x\cos x\)
Left - hand side:

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Right - hand side: