Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 268: 65b

Answer

Either $t=0.142$ or $t=0.785$

Work Step by Step

$$s(t)=\sin t+2\cos t$$ For $s(t)=\frac{3\sqrt2}{2}$, we would have the equation: $$\sin t+2\cos t=\frac{3\sqrt2}{2}$$ - We do not have any identities that can relate $\sin t$ and $\cos t$, but we have this identity: $\sin^2 t+\cos^2t=1$, which might be useful. So the strategy now is to square both sides. Yet before doing so, we rearrange the equation: $$\sin t=\frac{3\sqrt2}{2}-2\cos t$$ Now square both sides: $$\sin^2t=\frac{(3\sqrt2)^2}{4}-2\times\frac{3\sqrt2}{2}\times2\cos t+4\cos^2t$$ $$\sin^2t=\frac{9}{2}-6\sqrt2\cos t+4\cos^2t$$ Now apply $\sin^2t=1-\cos^2t$: $$1-\cos^2t=\frac{9}{2}-6\sqrt2\cos t+4\cos^2t$$ $$5\cos^2t-6\sqrt2\cos t+\frac{7}{2}=0$$ We can treat the question as a quadratic formula, which $a=5, b=-6\sqrt2, c=\frac{7}{2}$ - Find $\Delta$: $$\Delta=b^2-4ac$$ $$\Delta=(-6\sqrt2)^2-4\times5\times\frac{7}{2}$$ $$\Delta=36\times2-2\times5\times7$$ $$\Delta=72-70$$ $$\Delta=2$$ - Find $\cos t$: $$\cos t=\frac{-b\pm\sqrt\Delta}{2a}=\frac{6\sqrt2\pm\sqrt{2}}{10}$$ 1) For $\cos t=\frac{6\sqrt2+\sqrt{2}}{10}=\frac{7\sqrt2}{10}\approx0.99$ $$t=\cos^{-1}0.99\approx0.142$$ - Try back to $s(t)$: $$\sin0.142+2\cos0.142\approx2.121=\frac{3\sqrt2}{2}$$ Therefore, $t=0.142$ would be accepted. 2) For $\cos t=\frac{6\sqrt2-\sqrt{2}}{10}=\frac{5\sqrt2}{10}=\frac{\sqrt2}{2}$ $$t=\cos^{-1}\frac{\sqrt2}{2}=\frac{\pi}{4}$$ - Try back to $s(t)$: $$\sin\frac{\pi}{4}+2\cos\frac{\pi}{4}=\frac{\sqrt2}{2}+2\times\frac{\sqrt2}{2}=\frac{3\sqrt2}{2}$$ Therefore, $t=\frac{\pi}{4}\approx0.785$ would also be accepted. Yet the question asks for only one value of $t$, so you can either choose $t=0.142$ or $t=0.785$
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