Question

Without using tables,find the value of sin260+cos 260

A 2
B 5
C 1
D 2

Answer 1

Answer on Question #51813 – Math – Trigonometry

Without using tables, find the value of $\sin 260 + \cos 260$

A 2

B 5

C 1

D 2

Solution:

The statement of question does not seem clear.

Case 1.

$$\sin 26{}^\circ + \cos 26{}^\circ \quad (1)$$ $$\cos(90{}^\circ - \alpha) = \sin \alpha \Rightarrow \cos(26{}^\circ) = \sin(64{}^\circ) \quad (2)$$

(2) in(1):

$$\sin 26{}^\circ + \cos 26{}^\circ = \sin 26{}^\circ + \sin 64{}^\circ$$ $$\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \Rightarrow$$ $$\sin 26{}^\circ + \sin 64{}^\circ = -2 \sin \frac{26{}^\circ + 64{}^\circ}{2} \cos \frac{64{}^\circ - 26{}^\circ}{2} = -2 \sin 45{}^\circ \cos 38{}^\circ$$ $$\sin 45{}^\circ = \frac{1}{\sqrt{2}}$$

We can't find value of $\cos 38{}^\circ$ without using tables:

$$\cos 38{}^\circ = 0.788$$ $$\sin 26{}^\circ + \sin 64{}^\circ = -2 \cdot \frac{1}{\sqrt{2}} \cdot 0.788 = -0.8111 = -1.1144$$

**Answer**: $\sin 26{}^\circ + \cos 26{}^\circ = -1.1144$

Case 2.

$$\sin 260{}^\circ + \cos 260{}^\circ = \sin(260{}^\circ - 360{}^\circ) + \cos(260{}^\circ - 360{}^\circ) = \sin(-100{}^\circ) + \cos(-100{}^\circ) = \cos(100{}^\circ) - \sin(100{}^\circ) \quad (1)$$ $$\cos(90{}^\circ - \alpha) = \sin \alpha \Rightarrow \cos(100{}^\circ) = \sin(-10{}^\circ) = -\sin(10{}^\circ) \quad (2)$$

(2) in(1):

$$\Rightarrow \cos(100{}^\circ) - \sin(100{}^\circ) = -\sin 10{}^\circ - \sin 100{}^\circ = -(\sin 10{}^\circ + \sin 100{}^\circ)$$ $$\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \Rightarrow$$ $$- (\sin 10{}^{\circ} + \sin 100{}^{\circ}) = -2 \sin \frac{10{}^{\circ} + 100{}^{\circ}}{2} \cos \frac{100{}^{\circ} - 10{}^{\circ}}{2} = -2 \sin 55{}^{\circ} \cos 45{}^{\circ}.$$

We can't find value of $\sin 55{}^{\circ}$ without using tables:

$$\cos 45{}^{\circ} = \frac{1}{\sqrt{2}}; \quad \sin 55{}^{\circ} = \cos (90{}^{\circ} - 55{}^{\circ}) = \cos 35{}^{\circ} \approx 0.5735764$$ $$- (\sin 10{}^{\circ} + \sin 100{}^{\circ}) = -2 \cdot \frac{1}{\sqrt{2}} \cdot 0.5735764 = -0.8111$$

Answer: $\sin 260{}^{\circ} + \cos 260{}^{\circ} = -0.8111$

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