If \(\cos^2 x=\frac{9}{16}\) and (x) is in the fourth quadrant, what is the value of (\sin x)?
Advertisement
Answer and explanation
Correct answer: \(-\frac{\sqrt{7}}{4}\)
\(\sin^2 x=1-\frac{9}{16}=\frac{7}{16}\). In the fourth quadrant, (\sin x) is negative.
Frequently asked questions
What is the correct answer to this question?
\(-\frac{\sqrt{7}}{4}\)
Why is this the correct answer?
\(\sin^2 x=1-\frac{9}{16}=\frac{7}{16}\). In the fourth quadrant, (\sin x) is negative.
Which subject and chapter does this question cover?
This is a Class 12 Mathematics question. Chapter: Trigonometric Functions. Topic: Trigonometric functions and their properties.
Advertisement