Problem 74
Question
\(73-90\) Prove the identity. $$ \sin 8 x=2 \sin 4 x \cos 4 x $$
Step-by-Step Solution
Verified Answer
The identity holds true by applying the double angle formula for sine.
1Step 1: Understanding the Identity
We need to prove that \( \sin 8x = 2 \sin 4x \cos 4x \). This corresponds to understanding if the identity can be expressed using trigonometric identities already known.
2Step 2: Using Double Angle Identity
Recall the double angle identity for sine is \( \sin(2A) = 2 \sin A \cos A \). Here, we can apply this identity to express \( \sin 8x \) in terms of \( \sin 4x \) and \( \cos 4x \).
3Step 3: Expressing \( \sin 8x \)
Applying the double angle formula \( \sin(2A) = 2 \sin A \cos A \) to \( A = 4x \), we have: \( \sin 8x = \sin(2 \times 4x) = 2 \sin 4x \cos 4x \).
4Step 4: Conclusion
By applying the double angle identity to combine \( \sin 4x \) and \( \cos 4x \), we found that the given identity holds true: \( \sin 8x = 2 \sin 4x \cos 4x \).
Key Concepts
Double Angle IdentitySine FunctionCosine Function
Double Angle Identity
The double angle identity is a key concept in trigonometry that helps simplify expressions or solve equations involving trigonometric functions. It's particularly useful when dealing with angles that are multiples of a smaller angle, like in the identity \(\sin(2A) = 2 \sin A \cos A\). This identity shows the relationship between the sine of double an angle and the sine and cosine of the original angle.
Here are some important points about the double angle identity:
Here are some important points about the double angle identity:
- The identity \(\sin(2A) = 2 \sin A \cos A\) is derived from the sum of angles formula \(\sin(A + B) = \sin A \cos B + \cos A \sin B\) by setting \(A = B\).
- It can be applied for any angle \(A\), making it a versatile tool in solving trigonometric problems.
- The double angle identities also exist for cosine, such as \(\cos(2A) = \cos^2 A - \sin^2 A\) or \(\cos(2A) = 2 \cos^2 A - 1\).
Sine Function
The sine function is one of the primary functions in trigonometry, representing the ratio of the opposite side to the hypotenuse in a right-angled triangle. It is periodic, meaning it repeats its values in a regular pattern over intervals. The sine function is crucial for modeling wave-like phenomena and varies smoothly from -1 to 1.
Some key characteristics of the sine function include:
Some key characteristics of the sine function include:
- The sine of an angle \(x\), denoted as \( \sin x \), repeats every \(360^\circ\) or \(2\pi\) radians, known as its period.
- The function is odd, which indicates \( \sin(-x) = -\sin x \).
- It reaches its maximum value of 1 at \(x = \frac{\pi}{2} + 2k\pi\), where \(k\) is an integer.
- Zero values occur at multiples of \(\pi\), such as \(x = k\pi\), where \(k\) is an integer.
Cosine Function
The cosine function is another fundamental trigonometric function, expressing the ratio of the adjacent side to the hypotenuse in a right-angled triangle. Like the sine function, cosine is periodic and shares many properties with sine, but its graph is shifted horizontally by \(\frac{\pi}{2}\) radians.
Here are some essential features of the cosine function:
Here are some essential features of the cosine function:
- The cosine of an angle \(x\), represented as \( \cos x \), has a period of \(360^\circ\) or \(2\pi\) radians, like the sine function.
- It is an even function, meaning \( \cos(-x) = \cos x \).
- It reaches its maximum value of 1 at multipliers of \(2\pi\), such as \(x = 2k\pi\), where \(k\) is any integer.
- Zero values occur at odd multiples of \(\frac{\pi}{2}\), such as \(x = \frac{\pi}{2} + (2k+1)\pi\).
Other exercises in this chapter
Problem 73
\(73-90\) Prove the identity. $$ \cos ^{2} 5 x-\sin ^{2} 5 x=\cos 10 x $$
View solution Problem 73
Verify the identity. $$ \sec ^{4} x-\tan ^{4} x=\sec ^{2} x+\tan ^{2} x $$
View solution Problem 74
Verify the identity. $$ \frac{\cos \theta}{1-\sin \theta}=\sec \theta+\tan \theta $$
View solution Problem 75
\(73-90\) Prove the identity. $$ (\sin x+\cos x)^{2}=1+\sin 2 x $$
View solution