Trigonometry – Unit Circle, Identities, Compound Angles, Solving Equations Quiz - ICSE Class 11 Mathematics
Free practice quiz: Trigonometry – Unit Circle, Identities, Compound Angles, Solving Equations (ICSE Class 11 Mathematics). 40 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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If $\sin(\theta) = \frac{3}{5}$ and $\theta$ is in the second quadrant, find the value of $\cos(\theta)$.
- $\frac{4}{5}$
- $-\frac{4}{5}$
- $-\frac{3}{5}$
- $\frac{3}{5}$
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Answer: B. $-\frac{4}{5}$
Given $\sin(\theta) = \frac{3}{5}$. Since $\theta$ is in the second quadrant, $\cos(\theta)$ will be negative. Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we have $(\frac{3}{5})^2 + \cos^2(\theta) = 1$. This gives $\frac{9}{25} + \cos^2(\theta) = 1$, so $\cos^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25}$. Therefore, $\cos(\theta) = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}$. Since $\theta$ is in the second quadrant, $\cos(\theta) = -\frac{4}{5}$.
If $\sin(\theta) = \frac{1}{2}$ and $\theta$ is in the second quadrant, find the value of $\cos(\theta)$.
- $\frac{\sqrt{3}}{2}$
- $-\frac{\sqrt{3}}{2}$
- $\frac{1}{2}$
- $-\frac{1}{2}$
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Answer: B. $-\frac{\sqrt{3}}{2}$
Given $\sin(\theta) = \frac{1}{2}$. Since $\theta$ is in the second quadrant, $\cos(\theta)$ will be negative. Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we have $(\frac{1}{2})^2 + \cos^2(\theta) = 1$. This gives $\frac{1}{4} + \cos^2(\theta) = 1$, so $\cos^2(\theta) = 1 - \frac{1}{4} = \frac{3}{4}$. Therefore, $\cos(\theta) = \pm\sqrt{\frac{3}{4}} = \pm\frac{\sqrt{3}}{2}$. Since $\theta$ is in the second quadrant, $\cos(\theta) = -\frac{\sqrt{3}}{2}$.
If $\tan(\theta) = \frac{3}{4}$ and $\theta$ is in the third quadrant, find the value of $\sin(\theta)$.
- $-\frac{3}{5}$
- $\frac{3}{5}$
- $-\frac{4}{5}$
- $\frac{4}{5}$
Show answer
Answer: A. $-\frac{3}{5}$
Given $\tan(\theta) = \frac{3}{4}$. Since $\theta$ is in the third quadrant, both $\sin(\theta)$ and $\cos(\theta)$ are negative. We can construct a right-angled triangle where the opposite side is 3 and the adjacent side is 4. The hypotenuse is $\sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$. Therefore, $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$. As $\theta$ is in the third quadrant, $\sin(\theta)$ is negative. So, $\sin(\theta) = -\frac{3}{5}$.
If $\cos(\theta) = \frac{4}{5}$ and $\theta$ is in the fourth quadrant, find the value of $\sin(\theta)$.
- $-\frac{3}{5}$
- $\frac{3}{5}$
- $-\frac{4}{5}$
- $\frac{4}{5}$
Show answer
Answer: A. $-\frac{3}{5}$
Given $\cos(\theta) = \frac{4}{5}$. Since $\theta$ is in the fourth quadrant, $\sin(\theta)$ will be negative. Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we have $\sin^2(\theta) + (\frac{4}{5})^2 = 1$. This gives $\sin^2(\theta) + \frac{16}{25} = 1$, so $\sin^2(\theta) = 1 - \frac{16}{25} = \frac{9}{25}$. Therefore, $\sin(\theta) = \pm\sqrt{\frac{9}{25}} = \pm\frac{3}{5}$. Since $\theta$ is in the fourth quadrant, $\sin(\theta) = -\frac{3}{5}$.
If $\sin(\theta) = -\frac{3}{5}$ and $\theta$ is in the third quadrant, find the value of $\cos(\theta)$.
- $\frac{4}{5}$
- $-\frac{4}{5}$
- $\frac{3}{5}$
- $-\frac{3}{5}$
Show answer
Answer: B. $-\frac{4}{5}$
Given $\sin(\theta) = -\frac{3}{5}$. Since $\theta$ is in the third quadrant, $\cos(\theta)$ will be negative. Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we have $(-\frac{3}{5})^2 + \cos^2(\theta) = 1$. This gives $\frac{9}{25} + \cos^2(\theta) = 1$, so $\cos^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25}$. Therefore, $\cos(\theta) = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}$. Since $\theta$ is in the third quadrant, $\cos(\theta) = -\frac{4}{5}$.