If `log_(10)(sin(x+pi/4))=(log_(10)6-1)/2`, the value of `log_(10)(sinx)+log_(10)(cosx)` is

Options

A. -1
B. -2
C. 2
D. 1

Answer

To solve the equation \( \log_{10}(\sin(x + \frac{\pi}{4})) = \frac{\log_{10}(6) - 1}{2} \) and find the value of \( \log_{10}(\sin x) + \log_{10}(\cos x) \), we can follow these steps:

### Step 1: Rewrite the given equation
We start with the equation:
\[ \log_{10}(\sin(x + \frac{\pi}{4})) = \frac{\log_{10}(6) - 1}{2} \]

### Step 2: Multiply both sides by 2
Multiply both sides by 2 to eliminate the fraction:
\[ 2 \log_{10}(\sin(x + \frac{\pi}{4})) = \log_{10}(6) - 1 \]

### Step 3: Rewrite the right side
We can rewrite \( -1 \) as \( \log_{10}(10^{-1}) \):
\[ 2 \log_{10}(\sin(x + \frac{\pi}{4})) = \log_{10}(6) - \log_{10}(10) \]
Using the property of logarithms \( \log_{10}(a) - \log_{10}(b) = \log_{10}\left(\frac{a}{b}\right) \), we get:
\[ 2 \log_{10}(\sin(x + \frac{\pi}{4})) = \log_{10}\left(\frac{6}{10}\right) \]

### Step 4: Simplify the left side
Using the property \( n \log_{10}(a) = \log_{10}(a^n) \):
\[ \log_{10}(\sin^2(x + \frac{\pi}{4})) = \log_{10}\left(\frac{6}{10}\right) \]

### Step 5: Remove the logarithm
Since the logarithms are equal, we can equate the arguments:
\[ \sin^2(x + \frac{\pi}{4}) = \frac{6}{10} = \frac{3}{5} \]

### Step 6: Use the sine addition formula
Using the sine addition formula:
\[ \sin(x + \frac{\pi}{4}) = \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}(\sin x + \cos x) \]
Thus:
\[ \sin^2(x + \frac{\pi}{4}) = \left(\frac{1}{\sqrt{2}}(\sin x + \cos x)\right)^2 \]
This gives:
\[ \frac{1}{2}(\sin x + \cos x)^2 = \frac{3}{5} \]

### Step 7: Solve for \( (\sin x + \cos x)^2 \)
Multiply both sides by 2:
\[ (\sin x + \cos x)^2 = \frac{6}{5} \]

### Step 8: Use the identity
We know that:
\[ \sin^2 x + \cos^2 x = 1 \]
Thus:
\[ (\sin x + \cos x)^2 = 1 + 2\sin x \cos x \]
Setting this equal to our previous result:
\[ 1 + 2\sin x \cos x = \frac{6}{5} \]

### Step 9: Solve for \( \sin x \cos x \)
Subtract 1 from both sides:
\[ 2\sin x \cos x = \frac{6}{5} - 1 = \frac{1}{5} \]
Thus:
\[ \sin x \cos x = \frac{1}{10} \]

### Step 10: Find \( \log_{10}(\sin x) + \log_{10}(\cos x) \)
Using the logarithmic property:
\[ \log_{10}(\sin x) + \log_{10}(\cos x) = \log_{10}(\sin x \cos x) \]
Substituting the value we found:
\[ \log_{10}(\sin x \cos x) = \log_{10}\left(\frac{1}{10}\right) = -1 \]

### Final Answer
Thus, the value of \( \log_{10}(\sin x) + \log_{10}(\cos x) \) is:
\[ \boxed{-1} \]