Use half angle identity to find cos (-3pi/8)
The Correct Answer and Explanation is:
To find the exact value of cos(−3π/8) using a half-angle identity, we follow these steps:
Step 1: Recognize the half-angle identity for cosine
The half-angle identity for cosine is:cos(θ2)=±1+cos(θ)2\cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos(\theta)}{2}}cos(2θ)=±21+cos(θ)
The sign depends on the quadrant in which the angle lies.
Step 2: Express −3π/8 as a half-angle
Letθ2=−3π8⇒θ=−3π4\frac{\theta}{2} = -\frac{3\pi}{8} \Rightarrow \theta = -\frac{3\pi}{4}2θ=−83π⇒θ=−43π
So we will use:cos(−3π8)=±1+cos(−3π4)2\cos\left(-\frac{3\pi}{8}\right) = \pm \sqrt{\frac{1 + \cos\left(-\frac{3\pi}{4}\right)}{2}}cos(−83π)=±21+cos(−43π)
Step 3: Evaluate cosine of the double angle
Use the even property of cosine:cos(−3π4)=cos(3π4)\cos\left(-\frac{3\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right)cos(−43π)=cos(43π)cos(3π4)=−22\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}cos(43π)=−22
Step 4: Plug in the value
cos(−3π8)=1+(−22)2=1−222\cos\left(-\frac{3\pi}{8}\right) = \sqrt{\frac{1 + \left(-\frac{\sqrt{2}}{2}\right)}{2}} = \sqrt{\frac{1 – \frac{\sqrt{2}}{2}}{2}}cos(−83π)=21+(−22)=21−22=2−24=2−22= \sqrt{\frac{2 – \sqrt{2}}{4}} = \frac{\sqrt{2 – \sqrt{2}}}{2}=42−2=22−2
Now determine the correct sign. Since −3π/8 is in the fourth quadrant, cosine is positive there.
Final Answer:
cos(−3π8)=2−22\cos\left(-\frac{3\pi}{8}\right) = \frac{\sqrt{2 – \sqrt{2}}}{2}cos(−83π)=22−2
Explanation:
To solve trigonometric expressions exactly, identities such as the half-angle formula are essential. The half-angle formula for cosine simplifies the cosine of an angle that is half of a known angle. In this case, −3π/8 is half of −3π/4. By applying the identity, we were able to rewrite the expression in terms of a more familiar angle: 3π/4. The cosine of 3π/4 is a standard value from the unit circle. Substituting this value into the formula leads to a square root expression. Because −3π/8 lies in the fourth quadrant where cosine is positive, the positive root was selected. This method ensures that we maintain the exact trigonometric value rather than estimating or using a calculator.
