Chapter 3.12 - More Trigonometric Identities

In the previous section, we derived nine trigonometric identities. In this section, we will derive a few more identities.

Let us derive the tenth identity. It can be derived in 5 steps:
1. Consider tan (x+y).
It can be written as: $\tan (x+y)=\frac{\sin (x+y)}{\cos (x+y)}$
2. The numerator and denominator can be expanded using identities 7 and 3 respectively. We get:

$$\tan (x+y)=\frac{\sin (x+y)}{\cos (x+y)}=\frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y - \sin x \sin y}$$
3. Dividing both numerator and denominator by cos x cos y, we get:
$$\tan (x+y)=\frac{\frac{\sin x \cos y}{\cos x \cos y} + \frac{\cos x \sin y}{\cos x \cos y}}{\frac{\cos x \cos y}{\cos x \cos y} - \frac{\sin x \sin y}{\cos x \cos y}}$$
$$\Rightarrow \tan (x+y)=\frac{\frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}}{\frac{\cos y}{ \cos y} - \frac{\sin x \sin y}{\cos x \cos y}}$$
4. Thus we get the tenth identity:
Identity 10: $\tan (x+y)=\frac{\tan x + \tan y}{1 - \tan x \tan y}$
5. Note:
    ♦ If (x+y) is an odd multiple of $\frac{\pi}{2}$, tan (x+y) will not be defined.
    ♦ If x is an odd multiple of $\frac{\pi}{2}$, tan x will not be defined.
    ♦ If y is an odd multiple of $\frac{\pi}{2}$, tan y will not be defined.
• So before applying this identity, we must ensure that none of (x+y), x and y is an odd multiple of $\frac{\pi}{2}$

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Let us derive the eleventh identity. It can be derived in 4 steps:
1. Consider identity 10: $\tan (x+y)=\frac{\tan x + \tan y}{1 - \tan x \tan y}$
• Put '-y' in place of y. We get:

$$\tan (x+(-y))=\frac{\tan x + \tan (-y)}{1 - \tan x \tan (-y)}$$
$$\Rightarrow \tan (x-y)=\frac{\tan x + \frac{\sin (-y)}{\cos (-y)}}{1 - \tan x \times \frac{\sin (-y)}{\cos (-y)}}$$
2. Now we use identities 1 and 2. We get:
$$\tan (x-y)=\frac{\tan x + \frac{-\sin y}{\cos y}}{1 - \tan x \times \frac{-\sin y}{\cos y}}$$
3. Thus we get the eleventh identity:
Identity 11: $\tan (x-y)=\frac{\tan x - \tan y}{1 + \tan x \tan y}$
4. Here also, before applying this identity, we must ensure that none of (x-y), x and y is an odd multiple of $\frac{\pi}{2}$

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Let us derive the twelfth identity. It can be derived in 5 steps:
1. Consider cot (x+y).
It can be written as: $\cot (x+y)=\frac{\cos (x+y)}{\sin (x+y)}$
2. The numerator and denominator can be expanded using identities 3 and 7 respectively. We get:

$$\cot (x+y)=\frac{\cos (x+y)}{\sin (x+y)}=\frac{\cos x \cos y - \sin x \sin y}{\sin x \cos y + \cos x \sin y}$$
3. Dividing both numerator and denominator by sin x sin y, we get:
$$\cot (x+y)=\frac{\frac{\cos x \cos y}{\sin x \sin y} - \frac{\sin x \sin y}{\sin x \sin y}}{\frac{\sin x \cos y}{\sin x \sin y} + \frac{\cos x \sin y}{\sin x \sin y}}$$
$$\Rightarrow \cot (x+y)=\frac{\cot x \cot y - 1}{\cot y + \cot x}$$
4. Thus we get the twelfth identity:
Identity 12: $\cot (x+y)=\frac{\cot x \cot y - 1}{\cot y + \cot x}$
5. Note:
    ♦ If (x+y) is a multiple of 𝞹, cot (x+y) will not be defined.
    ♦ If x is a multiple of 𝞹, cot x will not be defined.
    ♦ If y is a multiple of 𝞹, cot y will not be defined.
• So before applying this identity, we must ensure that none of (x+y), x and y is a multiple of 𝞹.

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Let us derive the thirteenth identity. It can be derived in 4 steps:
1. Consider identity 12: $\cot (x+y)=\frac{\cot x \cot y - 1}{\cot y + \cot x}$
• Put '-y' in place of y. We get:

$$\cot (x+(-y))=\frac{\cot x \cot (-y) - 1}{\cot (-y) + \cot x}$$
$$\Rightarrow \cot (x-y)=\frac{\cot x \times \frac{\cos (-y)}{\sin (-y)} - 1}{\frac{\cos (-y)}{\sin (-y)} + \cot x}$$
2. Now we use identities 1 and 2. We get:
$$\cot (x-y)=\frac{\cot x \times \frac{\cos y}{-\sin y} - 1}{\frac{\cos y}{-\sin y} + \cot x}$$
$\Rightarrow \cot (x-y)=\frac{-\cot x \cot y - 1}{-\cot y + \cot x}$
$\Rightarrow \cot (x-y)=\frac{-1 \times (\cot x \cot y + 1)}{-1 \times (\cot y - \cot x)}$
3. Thus we get the thirteenth identity:
Identity 13: $\cot (x-y)=\frac{\cot x \cot y + 1}{\cot y - \cot x}$
4. Here also, before applying this identity, we must ensure that none of (x-y), x and y is a multiple of 𝞹

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Let us derive the fourteenth identity. It can be derived in 6 steps:
1.Consider identity 3: cos (x+y) = cos x cos y - sin x sin y
• Put 'x' in place of y. We get:
cos (x+x) = cos x cos x - sin x sin x = cos²x - sin²x
⇒ cos 2x = cos²x - sin²x
2. Consider the result in (1).
• We can replace cos²x by: 1 - sin²x
We get: cos 2x = 1 - sin²x - sin²x
⇒ cos 2x = 1 - 2 sin²x
3. Consider the result in (1).
• We can replace sin²x by: 1 - cos²x
We get: cos 2x = cos²x - 1 - cos²x
⇒ cos 2x = 2 cos²x - 1
4. Consider the result in (1).
• It can be written as: $\cos 2x = \frac{\cos^2x-\sin^2x}{\cos^2x+\sin^2x}$
(This is because, cos²x + sin²x = 1)
• Dividing numerator and denominator by cos²x, we get:

$$\cos 2x = \frac{\frac{\cos^2x}{\cos^2x}-\frac{\sin^2x}{\cos^2x}}{\frac{\cos^2x}{\cos^2x}+\frac{\sin^2x}{\cos^2x}}$$
$\Rightarrow \cos 2x = \frac{1-\tan^2x}{1+\tan^2x}$
5. From the results in steps (1), (2), (3) and (4), we get the fourteenth identity:
Identity 14:
$\cos 2x \;\;= \cos^2x-\sin^2x\;\;=1-2\sin^2x\;\;=2 \cos^2x-1\;\;=\frac{1-\tan^2x}{1+\tan^2x}$
6. If we intend to use the identity involving tan, we must ensure that x is not an odd multiple of $\frac{\pi}{2}$, Where n is any integer.

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Let us derive the fifteenth identity. It can be derived in 4 steps:
1.Consider identity 7: sin (x+y) = sin x cos y + cos x sin y
• Put 'x' in place of y. We get:
sin (x+x) = sin x cos x + cos x sin x = cos²x - sin²x
⇒ sin 2x = 2 sin x cos x
2. Consider the result in (1).
• It can be written as: $\sin 2x = \frac{2\sin x \cos x}{\cos^2x+\sin^2x}$
(This is because, cos²x + sin²x = 1)
• Dividing numerator and denominator by cos²x, we get:

$$\sin 2x = \frac{\frac{2\sin x \cos x}{\cos^2x}}{\frac{\cos^2x}{\cos^2x}+\frac{\sin^2x}{\cos^2x}}$$
3. From the results in (1) and (2), we get the fifteenth identity:
Identity 15: $\sin 2x \;\;= 2\sin x \cos x \;\;= \frac{2\tan x}{1+\tan^2x}$
4. Before applying this identity, we must ensure that x is not an odd multiple of $\frac{\pi}{2}$, Where n is any integer.

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Let us derive the sixteenth identity. It can be derived in 2 steps:
1.Consider identity 10: $\tan (x+y)=\frac{\tan x + \tan y}{1 - \tan x \tan y}$
• Put 'x' in place of y. We get:
$\tan (x+x)=\frac{\tan x + \tan x}{1 - \tan x \tan x}$
• Thus we get the sixteenth identity:
Identity 16: $\tan 2x=\frac{2\tan x}{1 - \tan^2 x}$
2. Before applying this identity, we must ensure that x is not an odd multiple of $\frac{\pi}{2}$, Where n is any integer.

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Let us derive the seventeenth identity. It can be derived in 2 steps:
1.Consider identity 7: sin (x+y) = sin x cos y + cos x sin y
• Put '2x' in place of x and 'x' in place of y. We get:
sin (2x+x) = sin 2x cos x + cos 2x sin x
2. Applying identities 14 and 15, we get:
sin (2x+x) = 2 sin x cos x cos x + (1 - 2 sin²x) sin x
⇒ sin 3x = 2 sin x cos²x + (sin x - 2 sin³x)
⇒ sin 3x = 2 sin x (1 - sin²x) + (sin x - 2 sin³x)
⇒ sin 3x = 2 sin x  - 2sin³x + sin x - 2 sin³x
• Thus we get the seventeenth identity:
Identity 17: sin 3x = 3 sin x  - 4 sin³x

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Let us derive the eighteenth identity. It can be derived in 2 steps:
1.Consider identity 3: cos (x+y) = cos x cos y - sin x sin y
• Put '2x' in place of x and 'x' in place of y. We get:
cos (2x+x) = cos 2x cos x - sin 2x sin x
2. Applying identities 14 and 15, we get:
cos (2x+x) = (2 cos²x - 1) cos x - 2 sin x cos x sin x
⇒ cos 3x = (2 cos²x - 1) cos x - 2 sin²x cos x
⇒ cos 3x = 2 cos³x - cos x - 2 (1 - cos²x) cos x
⇒ cos 3x = 2 cos³x - cos x - 2 cos x + cos³x
• Thus we get the eighteenth identity:
Identity 18: cos 3x = 4 cos³x - 3 cos x

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Let us derive the nineteenth identity. It can be derived in 3 steps:
1. Consider identity 10: $\tan (x+y)=\frac{\tan x + \tan y}{1 - \tan x \tan y}$
• Put '2x' in place of x and 'x' in place of y. We get:
$\tan (2x+x)=\frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}$
2. Applying identities 16, we get:

$$\tan 3x=\frac{\frac{2\tan x}{1 - \tan^2x} + \tan x}{1 - \frac{2\tan x \times \tan x}{1 - \tan^2}}$$
$$\Rightarrow \tan 3x=\frac{\frac{2\tan x + \tan x - \tan^3x}{1 - \tan^2x}}{\frac{1 - \tan^2-2\tan x \times \tan x}{1 - \tan^2}}=\frac{2\tan x + \tan x - \tan^3x}{1 - \tan^2-2\tan x \times \tan x}$$
• Thus we get the nineteenth identity:
Identity 19: $\tan 3x=\frac{3\tan x - \tan^3x}{1 - 3\tan^2}$
3. While using this identity, we must ensure that 3x is not an odd multiple of $\frac{\pi}{2}$, Where n is any integer.
• But what about x? tan x is present on the right side. So x also should not be an odd multiple of $\frac{\pi}{2}$, Where n is any integer.
• However, we need not mention this restriction on 'x'. Because, the restriction on 3x will automatically imply the restriction on x. This can be explained in 4 steps:
(i) Let x be an odd multiple of $\frac{\pi}{2}$
Then x = $(2n+1)\frac{\pi}{2}$
(ii) Whatever be the value of integer n, the value of (2n+1) will be odd. That is why, we say that, to avoid division by zero, x should not be $(2n+1)\frac{\pi}{2}$.
(iii) This is same as saying: x should not be $n\pi + \frac{\pi}{2}$ because,
$(2n+1)\frac{\pi}{2}= n\pi + \frac{\pi}{2}$
(iv) Now, if x is $(2n+1)\frac{\pi}{2}$, 3x is $3(2n+1)\frac{\pi}{2}$
3(2n+1) will always be an odd number.
• So it is enough to mention that, 3x should not be equal to $n\pi + \frac{\pi}{2}$.
In other words, it is enough to mention that, 3x should not be an odd multiple of $\frac{\pi}{2}$.

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Let us derive the twentieth identity. It can be derived in 8 steps:
1. Consider the cosine identities:
    ♦ Identity 3: cos (x+y) = cos x cos y - sin x sin y
    ♦ Identity 4: cos (x-y) = cos x cos y + sin x sin y
• Adding the identities 3 and 4, we get:
cos (x+y) + cos (x-y) = 2 cos x cos y
• Subtracting the identity 4 from the identity 3, we get:
cos (x+y) - cos (x-y) = -2 sin x sin y
2. Consider the sine identities:
    ♦ Identity 7: sin (x+y) = sin x cos y + cos x sin y
    ♦ Identity 8: sin (x-y) = sin x cos y - cos x sin y
• Adding the identities 7 and 8, we get:
sin (x+y) + sin (x-y) = 2 sin x cos y
• Subtracting the identity 8 from the identity 7, we get:
sin (x+y) - sin (x-y) = 2 cos x sin y
3. Let (x+y) = 𝜃 and (x-y) = ɸ
    ♦ Adding, we get: x+y+x-y = 2x = (𝜃 + ɸ) ⇒ $x = \frac{\theta + \phi}{2}$
    ♦ Subtracting, we get: x+y-x+y = 2y = (𝜃 - ɸ) ⇒ $y = \frac{\theta - \phi}{2}$
4. Then the results in (1) become:
$\cos \theta + \cos \phi = 2 \cos \frac{\theta + \phi}{2} \cos \frac{\theta - \phi}{2}$
$\cos \theta - \cos \phi = -2 \sin \frac{\theta + \phi}{2} \sin \frac{\theta - \phi}{2}$
5. Also the result in (2) become:
$\sin \theta + \sin \phi = 2 \sin \frac{\theta + \phi}{2} \cos \frac{\theta - \phi}{2}$
$\sin \theta - \sin \phi = 2 \cos \frac{\theta + \phi}{2} \sin \frac{\theta - \phi}{2}$
6. The results in 4 and 5 uses 𝜃 and ɸ
• We know that, we can write the sine and cosine of any real number.
• So it does not matter if it is 𝜃, ɸ, x or y.
7. So from (4), we get two identities:
Identity 20(a): $\cos x + \cos y = 2 \cos \frac{x + y}{2} \cos \frac{x - y}{2}$
Identity 20(b): $\cos x - \cos y = -2 \sin \frac{x + y}{2} \sin \frac{x - y}{2}$
8. Also from (5), we get two identities:
Identity 20(c): $\sin x + \sin y = 2 \sin \frac{x + y}{2} \cos \frac{x - y}{2}$
Identity 20(d): $\sin x - \sin y = 2 \cos \frac{x + y}{2} \sin \frac{x - y}{2}$

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Let us derive the twenty first identity. It can be derived in steps:
1. Consider the results in step 4 of the previous identity 20. We will write them again:
$\cos \theta + \cos \phi = 2 \cos \frac{\theta + \phi}{2} \cos \frac{\theta - \phi}{2}$
$\cos \theta - \cos \phi = -2 \sin \frac{\theta + \phi}{2} \sin \frac{\theta - \phi}{2}$
2. Also consider the results in step 5 of the previous identity 20. We will write them again:
$\sin \theta + \sin \phi = 2 \sin \frac{\theta + \phi}{2} \cos \frac{\theta - \phi}{2}$
$\sin \theta - \sin \phi = 2 \cos \frac{\theta + \phi}{2} \sin \frac{\theta - \phi}{2}$
3. But 𝜃 = (x+y) and ɸ =(x-y)
Also $\frac{\theta + \phi}{2}=x$ and $\frac{\theta - \phi}{2}=y$
4. So we get two identities from step (1):
Identity 21(a): $2 \cos x \cos y=\cos (x+y) + \cos (x-y)$
Identity 21(b): $-2 \sin x \sin y=\cos (x+y) + \cos (x-y)$
5. Similarly we get two identities from step (2):
Identity 21(c): $2 \sin x \cos y=\sin (x+y) + \sin (x-y)$
Identity 21(d): $2 \cos x \sin y=\sin (x+y) - \sin (x-y)$

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Let us derive the twenty second identity.


◼ Remarks:
1 (magenta color): Here we use identity 14.

• Thus we get two identities:
Identity 22(a): $\sin x ~=~ \pm \sqrt{\frac{1 - \cos 2x}{2}}$
Identity 22(b): $\sin \frac{x}{2} ~=~ \pm \sqrt{\frac{1 - \cos x}{2}}$

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Let us derive the twenty third identity.

◼ Remarks:
1 (magenta color): Here we use identity 14.

• Thus we get two identities:
Identity 23(a): $\cos x ~=~ \pm \sqrt{\frac{1 + \cos 2x}{2}}$
Identity 23(b): $\cos \frac{x}{2} ~=~ \pm \sqrt{\frac{1 + \cos x}{2}}$ 

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Let us derive the twenty fourth identity.

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\tan x}    & {~=~}    &{\pm \sqrt{\frac{1 - \cos 2x}{2}}~\div~\pm \sqrt{\frac{1 + \cos 2x}{2}}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\pm \sqrt{\frac{1 - \cos 2x}{2}}~\times~\pm \sqrt{\frac{2}{1 + \cos 2x}}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\pm \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\pm \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}~\times~\frac{\sqrt{1 - \cos 2x}}{\sqrt{1 - \cos 2x}}}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{1 - \cos 2x}{\sqrt{1 - \cos^2 2x}}}    \\ {~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{1 - \cos 2x}{\sin 2x}}    \\ \end{array}$

◼ Remarks:
1 (magenta color): Here we use identities 23 and 24.

• Similarly, we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\tan x}    & {~=~}    &{\pm \sqrt{\frac{1 - \cos 2x}{2}}~\div~\pm \sqrt{\frac{1 + \cos 2x}{2}}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\pm \sqrt{\frac{1 - \cos 2x}{2}}~\times~\pm \sqrt{\frac{2}{1 + \cos 2x}}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\pm \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\pm \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}~\times~\frac{\sqrt{1 + \cos 2x}}{\sqrt{1 + \cos 2x}}}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sqrt{1 - \cos^2 2x}}{1 + \cos 2x}}    \\ {~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sin 2x}{1 + \cos 2x}}    \\ \end{array}$

◼ Remarks:
1 (magenta color): Here we use identities 23 and 24.                           


• Thus we get four identities:
Identity 24(a): $\tan x ~=~\frac{1 - \cos 2x}{\sin 2x}$

Identity 24(b): $\tan x ~=~ \frac{\sin 2x}{1 + \cos 2x}$

Identity 24(c): $\tan \frac{x}{2} ~=~ \frac{1 - \cos x}{\sin x}$

Identity 24(d): $\tan \frac{x}{2} ~=~ \frac{\sin x}{1 + \cos x}$

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We have seen all the 24 identities. Let us see some solved examples:
Solved example 3.29
Prove that $3\sin \frac{\pi}{6} \sec \frac{\pi}{3}- 4 \sin \frac{5\pi}{6} \cot \frac{\pi}{4}=1$
Solution:
1. We already know the following values:
    ♦ $\sin \frac{\pi}{6}=\frac{1}{2}$
    ♦ $\sec \frac{\pi}{3}=\frac{2}{1}=2$
    ♦ $\cot \frac{\pi}{4}=\frac{1}{1}=1$
2. We need the value of $\sin \frac{5\pi}{6}$
• We can write it in terms of values that we know:
$\sin \frac{5\pi}{6}=\sin(\frac{6\pi}{6}-\frac{\pi}{6})$
⇒ $\sin \frac{5\pi}{6}=\sin(\pi-\frac{\pi}{6})$
3. We can expand the right side using identity 8:
$\begin{eqnarray} \sin(\pi-\frac{\pi}{6}) &=& \sin \pi \cos \frac{\pi}{6}- cos \pi \sin \frac{\pi}{6} \nonumber \\ &=& 0 \times \cos \frac{\pi}{6}- (-1) \times \frac{1}{2} \nonumber \\ &=& 0+ \frac{1}{2} \nonumber \\ &=& \frac{1}{2} \nonumber \ \end{eqnarray}$
4. Now we can substitute the values in the LHS of the given expression:
$3\times \frac{1}{2} \times 2 - 4 \times \frac{1}{2} \times 1=(3-2)=1=\text{RHS}$

Easier method:
1. Same as (1) above.
2. Same as (2) above.
3. $\sin(\pi-\frac{\pi}{6})=\sin \frac{\pi}{6}=\frac{1}{2}$ (using identity 9.d)
4. Same as (4) above.

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In the next section, we will see a few more solved examples.

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