Consider the following for the next two (02) items that follow : Given that sinx + cosx + tanx + cotx + secx + cosecx = 7
The given equation can be reduced to
sin2 2x - 44 sin2x + 36 = 0
Simplifying the Trigonometric Equation
We are given the trigonometric equation:
\[ \sin x + \cos x + \tan x + \cot x + \sec x + \cosec x = 7 \]Our goal is to reduce this equation to a form involving only \(\sin 2x\).
First, let's express the terms \(\tan x\), \(\cot x\), \(\sec x\), and \(\cosec x\) in terms of \(\sin x\) and \(\cos x\):
- \( \tan x = \frac{\sin x}{\cos x} \)
- \( \cot x = \frac{\cos x}{\sin x} \)
- \( \sec x = \frac{1}{\cos x} \)
- \( \cosec x = \frac{1}{\sin x} \)
Substitute these into the original equation:
\[ \sin x + \cos x + \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} + \frac{1}{\cos x} + \frac{1}{\sin x} = 7 \]Group the fractional terms:
\[ (\sin x + \cos x) + \left(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\right) + \left(\frac{1}{\cos x} + \frac{1}{\sin x}\right) = 7 \]Combine the terms within the parentheses by finding common denominators:
\[ \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} \] \[ \frac{1}{\cos x} + \frac{1}{\sin x} = \frac{\sin x + \cos x}{\sin x \cos x} \]Substitute these back into the equation:
\[ (\sin x + \cos x) + \frac{1}{\sin x \cos x} + \frac{\sin x + \cos x}{\sin x \cos x} = 7 \]Introducing Substitution Variables
Let \( S = \sin x + \cos x \) and \( P = \sin x \cos x \). The equation becomes:
\[ S + \frac{1}{P} + \frac{S}{P} = 7 \] \[ S + \frac{1+S}{P} = 7 \]We know the identity \( (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x \). So, \( S^2 = 1 + 2P \). This gives us a relationship between \( S \) and \( P \):
\[ 2P = S^2 - 1 \implies P = \frac{S^2 - 1}{2} \]Substitute this expression for \( P \) into the equation \( S + \frac{1+S}{P} = 7 \):
\[ S + \frac{1+S}{\frac{S^2 - 1}{2}} = 7 \] \[ S + \frac{2(1+S)}{S^2 - 1} = 7 \]Factor the denominator \( S^2 - 1 = (S-1)(S+1) \):
\[ S + \frac{2(1+S)}{(S-1)(S+1)} = 7 \]Assuming \( S \neq -1 \) (which implies \( \sin x + \cos x \neq -1 \)), we can cancel the term \( (1+S) \):
\[ S + \frac{2}{S-1} = 7 \]Assuming \( S \neq 1 \) (which implies \( \sin x + \cos x \neq 1 \)), multiply both sides by \( (S-1) \):
\[ S(S-1) + 2 = 7(S-1) \] \[ S^2 - S + 2 = 7S - 7 \]Rearrange the terms to form a quadratic equation in \( S \):
\[ S^2 - S - 7S + 2 + 7 = 0 \] \[ S^2 - 8S + 9 = 0 \]Deriving the Equation in terms of \( \sin 2x \)
We have the quadratic equation \( S^2 - 8S + 9 = 0 \), where \( S = \sin x + \cos x \). We also know that \( S^2 = 1 + 2 \sin x \cos x = 1 + \sin 2x \).
From \( S^2 = 1 + \sin 2x \), we can write \( \sin 2x = S^2 - 1 \).
Substitute \( S^2 = \sin 2x + 1 \) into the quadratic equation \( S^2 - 8S + 9 = 0 \):
\[ (\sin 2x + 1) - 8S + 9 = 0 \] \[ \sin 2x + 10 = 8S \]Now, we need an equation that contains only \( \sin 2x \). We can express \( S \) in terms of \( \sin 2x \):
\[ S = \frac{\sin 2x + 10}{8} \]Square both sides of this equation:
\[ S^2 = \left(\frac{\sin 2x + 10}{8}\right)^2 = \frac{(\sin 2x + 10)^2}{64} \]We have two expressions for \( S^2 \): \( S^2 = \sin 2x + 1 \) and \( S^2 = \frac{(\sin 2x + 10)^2}{64} \). Equating these two expressions:
\[ \sin 2x + 1 = \frac{(\sin 2x + 10)^2}{64} \]Multiply both sides by 64:
\[ 64(\sin 2x + 1) = (\sin 2x + 10)^2 \]Expand both sides:
\[ 64 \sin 2x + 64 = \sin^2 2x + 2 \cdot \sin 2x \cdot 10 + 10^2 \] \[ 64 \sin 2x + 64 = \sin^2 2x + 20 \sin 2x + 100 \]Rearrange the terms to form a quadratic equation in \( \sin 2x \). Move all terms to the right side:
\[ 0 = \sin^2 2x + 20 \sin 2x - 64 \sin 2x + 100 - 64 \] \[ 0 = \sin^2 2x + (20 - 64) \sin 2x + (100 - 64) \] \[ 0 = \sin^2 2x - 44 \sin 2x + 36 \]Thus, the equation can be reduced to \( \sin^2 2x - 44 \sin 2x + 36 = 0 \).
Comparing with Options
Let's compare the derived equation with the given options:
- Option 1: \( \sin^2 2x - 44 \sin 2x + 36 = 0 \)
- Option 2: \( \sin^2 2x + 44 \sin 2x - 36 = 0 \)
- Option 3: \( \sin^2 2x - 22 \sin 2x + 18 = 0 \)
- Option 4: \( \sin^2 2x + 22 \sin 2x - 18 = 0 \)
Our derived equation matches Option 1.
| Step | Description | Result |
|---|---|---|
| 1 | Original Equation | \( \sin x + \cos x + \tan x + \cot x + \sec x + \cosec x = 7 \) |
| 2 | Rewrite in terms of \(\sin x, \cos x\) | \( (\sin x + \cos x) + \frac{1}{\sin x \cos x} + \frac{\sin x + \cos x}{\sin x \cos x} = 7 \) |
| 3 | Substitute \(S=\sin x + \cos x, P=\sin x \cos x\) | \( S + \frac{1+S}{P} = 7 \) |
| 4 | Use \( P = \frac{S^2-1}{2} \) | \( S + \frac{2(1+S)}{S^2-1} = 7 \) |
| 5 | Simplify (assuming \( S \neq \pm 1 \)) | \( S^2 - 8S + 9 = 0 \) |
| 6 | Use \( S^2 = 1 + \sin 2x \) and \( \sin 2x + 10 = 8S \) | \( \sin 2x + 1 = \frac{(\sin 2x + 10)^2}{64} \) |
| 7 | Rearrange into a quadratic in \( \sin 2x \) | \( \sin^2 2x - 44 \sin 2x + 36 = 0 \) |
Revision Table: Trigonometric Identities Used
| Identity | Description |
|---|---|
| \( \tan x = \frac{\sin x}{\cos x} \) | Tangent definition |
| \( \cot x = \frac{\cos x}{\sin x} \) | Cotangent definition |
| \( \sec x = \frac{1}{\cos x} \) | Secant definition |
| \( \cosec x = \frac{1}{\sin x} \) | Cosecant definition |
| \( \sin^2 x + \cos^2 x = 1 \) | Pythagorean identity |
| \( (\sin x + \cos x)^2 = 1 + 2 \sin x \cos x \) | Square of sum identity |
| \( \sin 2x = 2 \sin x \cos x \) | Double angle identity for sine |
Additional Information: Domain and Extraneous Solutions
The initial equation involves terms like \( \tan x, \cot x, \sec x, \cosec x \). For these terms to be defined, \( \sin x \neq 0 \) and \( \cos x \neq 0 \). This means \( x \neq n\pi/2 \) for any integer \( n \).
In the process of simplifying, we made assumptions like \( S \neq 1 \) and \( S \neq -1 \). These correspond to \( \sin x + \cos x \neq 1 \) and \( \sin x + \cos x \neq -1 \). Geometrically, \( \sin x + \cos x = \sqrt{2} \sin(x+\pi/4) \). So \( S=1 \) means \( \sqrt{2} \sin(x+\pi/4) = 1 \) and \( S=-1 \) means \( \sqrt{2} \sin(x+\pi/4) = -1 \). These are possible values, but they lead to division by zero in intermediate steps of our algebraic simplification, indicating they are not solutions to the original equation.
The final quadratic equation in \( \sin 2x \), which is \( \sin^2 2x - 44 \sin 2x + 36 = 0 \), might have roots for \( \sin 2x \) that do not correspond to valid values of \( S=\sin x+\cos x \) within its range \( [-\sqrt{2}, \sqrt{2}] \). Solving \( S^2 - 8S + 9 = 0 \) gives \( S = 4 \pm \sqrt{7} \). The range of \( S=\sin x+\cos x \) is \( [-\sqrt{2}, \sqrt{2}] \). Since \( \sqrt{2} \approx 1.414 \) and \( 4-\sqrt{7} \approx 4-2.646 = 1.354 \), only \( S = 4-\sqrt{7} \) is within the valid range. The value \( 4+\sqrt{7} \approx 6.646 \) is outside the range.
When we derived the quadratic in \( \sin 2x \), we found its roots are \( \sin 2x = 22 \pm 8\sqrt{7} \). The value \( 22+8\sqrt{7} \) is outside the range \( [-1, 1] \) for \( \sin 2x \). The value \( 22-8\sqrt{7} \approx 22 - 8(2.646) \approx 22 - 21.168 = 0.832 \) is within the range \( [-1, 1] \).
The quadratic equation \( \sin^2 2x - 44 \sin 2x + 36 = 0 \) correctly represents the relationship that \( \sin 2x \) must satisfy, even if one of its roots corresponds to an extraneous solution for \( S \).
Important Questions from Trigonometry
If sin2x = a - b√c, where a and b are natural numbers and c is prime number, then what is the value of a - b + 2c ?
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