Question
Which is the exact value of cos 75°?
(1 point)
Responses
Start Fraction Start Root 6 End Root minus Start Root 2 End Root over 4 End Fraction
Image with alt text: Start Fraction Start Root 6 End Root minus Start Root 2 End Root over 4 End Fraction
Start Fraction Start Root 6 End Root plus Start Root 2 End Root over 4 End Fraction
Image with alt text: Start Fraction Start Root 6 End Root plus Start Root 2 End Root over 4 End Fraction
Start Fraction Start Root 2 End Root minus Start Root 6 End Root over 4 End Fraction
Image with alt text: Start Fraction Start Root 2 End Root minus Start Root 6 End Root over 4 End Fraction
Start Fraction Start Root 6 End Root minus Start Root 2 End Root over 2 End Fraction
(1 point)
Responses
Start Fraction Start Root 6 End Root minus Start Root 2 End Root over 4 End Fraction
Image with alt text: Start Fraction Start Root 6 End Root minus Start Root 2 End Root over 4 End Fraction
Start Fraction Start Root 6 End Root plus Start Root 2 End Root over 4 End Fraction
Image with alt text: Start Fraction Start Root 6 End Root plus Start Root 2 End Root over 4 End Fraction
Start Fraction Start Root 2 End Root minus Start Root 6 End Root over 4 End Fraction
Image with alt text: Start Fraction Start Root 2 End Root minus Start Root 6 End Root over 4 End Fraction
Start Fraction Start Root 6 End Root minus Start Root 2 End Root over 2 End Fraction
Answers
Answered by
GPT-4o mini
To find the exact value of \(\cos 75^\circ\), we can use the angle sum identity for cosine:
\[
\cos(a + b) = \cos a \cos b - \sin a \sin b
\]
Choosing \(a = 45^\circ\) and \(b = 30^\circ\), we have:
\[
\cos 75^\circ = \cos(45^\circ + 30^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ
\]
Using the known values of the trigonometric functions:
\[
\cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 30^\circ = \frac{1}{2}
\]
Now we substitute these values into the formula:
\[
\cos 75^\circ = \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right)
\]
This simplifies to:
\[
\cos 75^\circ = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}
\]
Combining the terms gives:
\[
\cos 75^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}
\]
Thus, the exact value of \(\cos 75^\circ\) is:
\[
\frac{\sqrt{6} - \sqrt{2}}{4}
\]
The correct answer from the options provided is:
**Start Fraction Start Root 6 End Root minus Start Root 2 End Root over 4 End Fraction**
\[
\cos(a + b) = \cos a \cos b - \sin a \sin b
\]
Choosing \(a = 45^\circ\) and \(b = 30^\circ\), we have:
\[
\cos 75^\circ = \cos(45^\circ + 30^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ
\]
Using the known values of the trigonometric functions:
\[
\cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 30^\circ = \frac{1}{2}
\]
Now we substitute these values into the formula:
\[
\cos 75^\circ = \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right)
\]
This simplifies to:
\[
\cos 75^\circ = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}
\]
Combining the terms gives:
\[
\cos 75^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}
\]
Thus, the exact value of \(\cos 75^\circ\) is:
\[
\frac{\sqrt{6} - \sqrt{2}}{4}
\]
The correct answer from the options provided is:
**Start Fraction Start Root 6 End Root minus Start Root 2 End Root over 4 End Fraction**