5 Questions
Without using a calculator, find the value of $\cos(-765^\circ)$, $\sin(105^\circ)$, and $\sin(100^\circ)$.
The values are: $\cos(-765^\circ) = \cos(765^\circ) = \cos(360^\circ + 405^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2}$, $\sin(105^\circ) = \sin(90^\circ + 15^\circ) = \sin(15^\circ) = \frac{1}{2}(\sqrt{6} - \sqrt{2})$, and $\sin(100^\circ) = \sin(90^\circ + 10^\circ) = \sin(10^\circ) = \frac{1}{2}(\sqrt{5} - 1)$.
Without using a calculator, find the value of $\sin(150^\circ) + \cos(300^\circ) - \tan(315^\circ) + \sec(3660^\circ)$.
The value is: $\sin(150^\circ) + \cos(300^\circ) - \tan(315^\circ) + \sec(3660^\circ) = \frac{1}{2} + \frac{1}{2} - 1 + 1 = 1$.
Without using a calculator, find the value of $\sin(420^\circ) \cos(390^\circ) + \sin(-330^\circ) \cos(-300^\circ)$.
The value is: $\sin(420^\circ) \cos(390^\circ) + \sin(-330^\circ) \cos(-300^\circ) = \sin(60^\circ) \cos(30^\circ) + \sin(30^\circ) \cos(60^\circ) = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1$.
Prove the identity $\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{2}$.
Using the identity $\cos 2\theta = 2\cos^2\theta - 1$, we have $2\cos 20^\circ \cos 40^\circ \cos 80^\circ - 1 = 0$, so $\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{2}$.
Prove the identity $8\sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ) = \frac{3}{2}$.
Using the identity $\sin 2\theta = 2\sin\theta\cos\theta$, we have $4\sin(20^\circ) \sin(40^\circ) \sin(60^\circ) \sin(80^\circ) = \frac{3}{2}$, so $8\sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ) = \frac{3}{2}$.
Trigonometry Quiz - First Year BMS Course at TSSM’s Bhivarabai Sawant College of Engineering and Research, Narhe, Pune. Test your knowledge in trigonometry with this assignment for the academic year 2023-24. Assignment No. 5 with a maximum of 10 marks. Submit on or before the specified date and get assessed by faculty.
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