The period of sin² θ is (A) π² (B) π (C) 2π (D) π/2.

Q: Wait, why do we reject the negative root here? Or is it just about the period?

Great question! We aren't rejecting a root; we are analyzing the function's behavior. The period is about how often the function repeats its values. Squaring the function changes the values, but the period is determined by how quickly the function completes one full cycle of its pattern. The negative sign in the identity just shifts or reflects the graph, but the frequency remains determined by the $2\theta$ term.

Q: Can I just use the graph to find the period?

Absolutely! Visualizing the graph is a powerful tool. If you draw $\sin \theta$ and then square it, you will see the negative parts flip up, creating a wave that repeats twice as often. This confirms that the period is halved. However, in JEE Advanced, an analytical proof using identities is essential for complex functions where graphing might be difficult.

Q: What if the function was $\sin^3 \theta$? Would the period be $\pi/3$?

That's a fantastic intuition! For $\sin^n \theta$, the period depends on whether $n$ is even or odd. If $n$ is even, the period is $\pi$. If $n$ is odd, the period remains $2\pi$. For $\sin^3 \theta$, you would use the triple-angle identity $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$, which shows the period is still $2\pi$. Keep exploring these patterns!

Question

The period of $\sin^2 	heta$ is

Options:

(A) $\pi^2$

(B) $\pi$

(C) $2\pi$

(D) $\pi/2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to dive into the elegant world of trigonometric functions. We often treat functions like $\sin 	heta$ as simple, static waves, but when we start manipulating them—squaring them, scaling them, shifting them—they reveal a deeper, rhythmic beauty.

Our mission today is to find the fundamental period of $f(	heta) = \sin^2 	heta$. Let's embark on this journey together.

### The Visual Dance of the Sine Wave

Imagine standing on the shore, watching the waves roll in. The function $y = \sin 	heta$ is the perfect mathematical representation of that motion. It starts at zero, climbs to one, dips to negative one, and returns to zero, completing one full cycle in $2\pi$ radians.

Now, what happens when we square this function? When you square a positive number, it stays positive; when you square a negative number, it also becomes positive.

This is the key. All those negative valleys of the sine wave, which previously dipped below the $x$-axis, are now forced to flip upwards. Visually, we can already guess that the period has been halved.

### The Analytical Bridge: Power Reduction

To prove our visual intuition, we need to transform our squared function into a linear one. We turn to our trusty toolkit of trigonometric identities, specifically the double-angle identity for cosine:

$$\cos 2	heta = 1 - 2\sin^2 	heta$$

This identity is a bridge between the squared world and the linear world. Let's rearrange it to isolate our target, $\sin^2 	heta$:

$$2\sin^2 	heta = 1 - \cos 2	heta$$

$$\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$$

By splitting this into two terms, we get the following expression:

$$f(	heta) = \frac{1}{2} - \frac{1}{2}\cos 2	heta$$

Now, look at this expression. It is no longer a squared function; it is a simple, linear cosine function with a vertical shift and a scaled amplitude.

### The Final Revelation: The Frequency Factor

The constant term $\frac{1}{2}$ simply shifts the entire graph vertically and has no impact on the period. The coefficient $-\frac{1}{2}$ scales the amplitude, which also leaves the period untouched.

The entire "rhythm" of the function is dictated by the term $\cos 2	heta$. We know that for any function of the form $\cos(k	heta)$, the period $T$ is given by the formula:

$$T = \frac{2\pi}{|k|}$$

In our case, $k = 2$. Substituting this into our formula, we get:

$$T = \frac{2\pi}{2} = \pi$$

The math confirms our visual intuition perfectly! The **fundamental period of $\sin^2 	heta$ is $\pi$**.

Remember, whenever you encounter higher powers of sine or cosine, use your power-reduction identities to linearize the function. The period will reveal itself through the frequency factor.

Animated Solution for Mathematics - Trigonometry: The period of $sin^{2} θ$ is

Select Answer:

Visualized Solution

Introduction to the Function

The Base Function $sin θ$

Visualizing $sin^{2} θ$

Observing the New Period

The Power-Reduction Identity

Applying the Identity

Analyzing the Components

Period of $cos (k θ)$

Calculating the Final Period

Conclusion

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Visual Dance of the Sine Wave

The Analytical Bridge

The Final Revelation

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The period of sin2θ is

Select Answer:

Visualized Solution

Introduction to the Function

The Base Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinθ

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2θ