Principal value in trigonometry refers to the unique angle within a specific range for which a trigonometric function takes a given value. For inverse trigonometric functions, the principal value is the angle that lies within a defined interval, ensuring a single and consistent output for each input.
Class 12 Maths Chapter 2 Inverse Trigonometric Functions deals with the inverse of trigonometric functions like sine and cosine and tangent or cosecant and secant and cotangent. It explains how to find angles from given function values. The chapter includes domains ranges graphs and solving equations involving inverse trigonometric functions.
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To solve sin[π/3 + sin⁻¹(1/2)], let’s break it into steps:
Step 1: Simplify sin⁻¹(1/2)
The angle whose sine is 1/2 in the range of sin⁻¹ (i.e., [-π/2, π/2]) is:
sin⁻¹(1/2) = π/6
Step 2: Substitute into the given expression
Now substitute sin⁻¹(1/2) = π/6 into the expression:
sin[π/3 + sin⁻¹(1/2)] = sin(π/3 + π/6)
Step 3: Simplify π/3 + π/6
Find a common denominator:
π/3 + π/6 = 2π/6 + π/6 = 3π/6 = π/2
Step 4: Simplify sin(π/2)
From the unit circle, sin(π/2) = 1.
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