Problem 27 Evaluate \(\sin \left(\cos ^{-1}... [FREE SOLUTION]

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Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

1 Edition

Chapter 6: Problem 27

Evaluate $\sin \left(\cos ^{-1} \frac{1}{4}+\tan ^{-1} 2\right)$.

Short Answer

Expert verified

$ \sin \left(\cos ^{-1} \frac{1}{4}+\tan ^{-1} 2\right) = \frac{\sqrt{75} + 2\sqrt{5}}{20} $

Step by step solution

Calculate sin(a) and cos(a)

To find the sine and cosine of $a = \cos^{-1}\frac{1}{4}$, we can consider a right-angled triangle, with an adjacent side of length 1 and a hypotenuse of length 4. With this information, we can find the length of the opposite side using Pythagoras' theorem, and then calculate sin(a) and cos(a): Opposite side = $\sqrt{Hypotenuse^2 - Adjacent^2} = \sqrt{4^2 - 1^2} = \sqrt{15}$ Now, we can find sin(a) and cos(a): sin(a) = $\frac{Opposite}{Hypotenuse} = \frac{\sqrt{15}}{4}$ cos(a) = $\frac{Adjacent}{Hypotenuse} = \frac{1}{4}$

Calculate sin(b) and cos(b)

To find the sine and cosine of $b = \tan^{-1}2$, we can consider a right-angled triangle, with an opposite side of length 2 and an adjacent side of length 1. With this information, we can find the length of the hypotenuse using Pythagoras' theorem, and then calculate sin(b) and cos(b): Hypotenuse = $\sqrt{Opposite^2 + Adjacent^2} = \sqrt{2^2 + 1^2} = \sqrt{5}$ Now, we can find sin(b) and cos(b): sin(b) = $\frac{Opposite}{Hypotenuse} = \frac{2}{\sqrt{5}}$ cos(b) = $\frac{Adjacent}{Hypotenuse} = \frac{1}{\sqrt{5}}$

Apply the sum-of-angles formula for sine

Next, we will use the sum-of-angles formula for sine to evaluate $\sin(a+b)$: $$\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$$ Substitute the values of sin(a), cos(a), sin(b), and cos(b) that we found in Steps 1 and 2: $\sin(a+b) = \sin(\cos^{-1} \frac{1}{4} + \tan^{-1}2) = \frac{\sqrt{15}}{4} \cdot \frac{1}{\sqrt{5}} + \frac{1}{4} \cdot \frac{2}{\sqrt{5}}$

Simplify the expression

Finally, let's simplify the expression: $\sin(a+b) = \frac{\sqrt{15}}{4\sqrt{5}} + \frac{2}{4\sqrt{5}} = \frac{\sqrt{15} + 2}{4\sqrt{5}}$

Rationalize the denominator

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{5}$: $\sin(a+b) = \frac{(\sqrt{15} + 2)\sqrt{5}}{4\cdot 5} = \frac{\sqrt{75} + 2\sqrt{5}}{20}$ The final answer is: $$\sin \left(\cos ^{-1} \frac{1}{4}+\tan ^{-1} 2\right) = \frac{\sqrt{75} + 2\sqrt{5}}{20}$$

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91影视

Evaluate \(\sin \left(\cos ^{-1} \frac{1}{4}+\tan ^{-1} 2\right)\).

Short Answer

Step by step solution

Calculate sin(a) and cos(a)

Calculate sin(b) and cos(b)

Apply the sum-of-angles formula for sine

Simplify the expression

Rationalize the denominator

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