Problem 25
Question
Calculate the quantity without using the the trigonometric functions on your calculator. You are given that \(\sin (\pi / 12)=0.259\) and \(\cos (\pi / 5)=0.809 .\) You may want to draw a picture showing the angles involved and check your answer on a calculator. $$\sin \frac{\pi}{5}$$
Step-by-Step Solution
Verified Answer
\( \sin \frac{\pi}{5} \approx 0.588 \)
1Step 1: Understanding the Formula
To find \( \sin \frac{\pi}{5} \), we'll use the angle addition formula: \( \sin(a + b) = \sin a \cos b + \cos a \sin b \). We need to express \( \frac{\pi}{5} \) in terms of known angles \( \frac{\pi}{12} \) and \( \frac{\pi}{5} \).
2Step 2: Decompose Angles if possible
We notice that \( \frac{\pi}{5} \approx \frac{3\pi}{15} \), and \( \frac{\pi}{12} \approx \frac{5\pi}{60} \). We need to find close values or combinations, but since \( \sin \frac{\pi}{5} \) is already given to us through cosine complement, we might directly relate to known identities.
3Step 3: Recall Double Angle Identities
We leverage the identity \( \cos(2x) = 1 - 2\sin^2(x) \). Knowing \( \cos\frac{\pi}{5} \approx 0.809 \), solve for \( \sin \frac{\pi}{5} \) using \( \sin \frac{\pi}{5} = \sqrt{1 - \cos^2 \frac{\pi}{5}} \).
4Step 4: Calculate \( \sin \frac{\pi}{5} \)
Compute \( \sin \frac{\pi}{5} = \sqrt{1 - (0.809)^2} \). Calculate this to find \( \sin \frac{\pi}{5} = \sqrt{1 - 0.654481} = \sqrt{0.345519} \approx 0.588 \).
5Step 5: Verify Solutions
Finally, verify your calculation using a calculator to ensure \( \sin \frac{\pi}{5} \approx 0.588 \) is accurate and matches closely with a sine value for \( \frac{\pi}{5} \).
Key Concepts
Angle Addition FormulaDouble Angle IdentitiesSine and Cosine Calculation
Angle Addition Formula
The Angle Addition Formula is a powerful tool in trigonometry that helps calculate the sine, cosine, or tangent of an angle that can be expressed as the sum or difference of two other angles. This formula is particularly useful when the angles involved are not standard values that can be easily found on a calculator.
To illustrate, the formula for the sine of a sum of two angles is given by:
In the context of the exercise, this formula was initially considered to tackle the problem, as students were tasked to compute \( \sin \frac{\pi}{5} \), but given its approximation, exploring other identities proved to be more efficient.
To illustrate, the formula for the sine of a sum of two angles is given by:
- \( \sin(a + b) = \sin a \cos b + \cos a \sin b \)
In the context of the exercise, this formula was initially considered to tackle the problem, as students were tasked to compute \( \sin \frac{\pi}{5} \), but given its approximation, exploring other identities proved to be more efficient.
Double Angle Identities
Double Angle Identities are used to express trigonometric functions of double angles, which are angles that are two times a given angle. These identities simplify complex trigonometric problems and assist in manipulating expressions for calculations.
For cosine, one common double angle identity is:
For cosine, one common double angle identity is:
- \( \cos(2x) = 1 - 2\sin^2(x) \)
Sine and Cosine Calculation
Accurate calculation of sine and cosine without a calculator can often feel tricky, but with identities, it's quite manageable. Both sine and cosine are fundamental trigonometric ratios that are always worth understanding deeply.
When calculating \( \sin \frac{\pi}{5} \) using given approximations, recalling the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) is immensely helpful. With the cosine provided as \( 0.809 \), the sine can be found as follows:
When calculating \( \sin \frac{\pi}{5} \) using given approximations, recalling the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) is immensely helpful. With the cosine provided as \( 0.809 \), the sine can be found as follows:
- First calculate: \( \cos^2 \frac{\pi}{5} = (0.809)^2 = 0.654481 \)
- Then, use the identity to find sine: \( \sin \frac{\pi}{5} = \sqrt{1 - 0.654481} \)
- Finally, \( \sin \frac{\pi}{5} = \sqrt{0.345519} \approx 0.588 \)
Other exercises in this chapter
Problem 24
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