Problem 24 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Draw the graphs of $$ f(x)=\sin ^{2} x-1 $$

Step-by-Step Solution

Answer

The graph of \(f(x)=\sin ^{2} x-1\) oscillates between -1 and 0. It reaches its maximum value of 0 at \( \pi/2, 3 \pi/2 \) and so on and reaches its minimum value of -1 at \( 0, \pi, 2\pi \) and so on. The function has a period of \( 2\pi \).

Step 1: Identify the Basic Function

The base function here is \( \sin x \). The square of this function \( \sin^{2}x \) oscillates between 0 and 1 in its basic period \(0 \leq x \leq 2\pi \), and this will be our starting point.

Step 2: Apply the Transformation

Note that we have subtracted 1 from \( \sin^{2}x \). This will shift all values of the original function down by one. Therefore, the values of the function are shifted to range between -1 and 0.

Step 3: Sketch the Graph

To sketch the graph, plot key points such as \( x = 0, \pi/2, \pi, 3\pi/2, \) and \( 2\pi \). You will see that the graph starts at -1 at x = 0, goes up to 0 at \( \pi/2 \), returns to -1 at \( \pi \), goes up to 0 at \( 3\pi/2 \) and finally returns to -1 at \( 2\pi \). Do notice that the function is periodic with period 2\pi