Question

1.Express cos4tita in terms of powere of
cos tita
2.G iven that sin tita=-3/5 and cos alpha=-2/5. Find sin(A+B) if 1. tita and alpha are acute 2. tita is acute and alpha is obtuse

Answer 1

Answer to Question #83933 – Math – Trigonometry

Question

1. Express cos4tita in terms of powers of $\cos$ tita

2. Given that $\sin \mathrm{tita} = -3/5$ and $\cos \mathrm{alpha} = -2/5$. Find $\sin (A + B)$ if 1. tita and alpha are acute 2. tita is acute and alpha is obtuse.

Solution

1. We know that $\cos 2\theta = 2\cos^2\theta - 1$

$$\begin{array}{l} \cos 4\theta = \cos 2(2\theta) = 2\cos^2 2\theta - 1 \\ = 2(2\cos^2 \theta - 1)^2 - 1 \\ = 2(4\cos^4 \theta - 4\cos^2 \theta + 1) - 1 \\ = 8\cos^4 \theta - 8\cos^2 \theta + 1 \\ \end{array}$$

2. Given $\sin \theta = -\frac{3}{5} \Rightarrow \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(-\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$

$$\cos \alpha = -\frac{2}{5} \Rightarrow \sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \left(-\frac{2}{5}\right)^2} = \sqrt{1 - \frac{4}{25}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}$$

CASE I: $\theta$ and $\alpha$ are acute then $\sin \theta = \frac{3}{5}$, $\cos \theta = \frac{4}{5}$ and $\cos \alpha = \frac{2}{5}$, $\sin \alpha = \frac{\sqrt{21}}{5}$

$$\sin(\theta + \alpha) = \sin \theta \cos \alpha + \cos \theta \sin \alpha = \frac{3}{5} \cdot \frac{2}{5} + \frac{4}{5} \cdot \frac{\sqrt{21}}{5} = \frac{6 + 4\sqrt{21}}{25}$$

CASE II: $\theta$ is acute and $\alpha$ is obtuse then $\sin \theta = \frac{3}{5}$, $\cos \theta = \frac{4}{5}$ and $\cos \alpha = -\frac{2}{5}$, $\sin \alpha = \frac{\sqrt{21}}{5}$

$$\sin(\theta + \alpha) = \sin \theta \cos \alpha + \cos \theta \sin \alpha = \frac{3}{5} \cdot \left(-\frac{2}{5}\right) + \frac{4}{5} \cdot \frac{\sqrt{21}}{5} = \frac{-6 + 4\sqrt{21}}{25}.$$

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