Chapter 6: Q. 57 (page 396)

Find the exact value of each of the remaining trigonometric functions of $θ .$
$tanθ = - \frac{1}{3}, sinθ > 0$

Short Answer

Expert verified

The values of trigonometric functions are,

$secθ = - \frac{\sqrt{10}}{3} cosθ = - \frac{3 \sqrt{10}}{10} sinθ = \frac{\sqrt{10}}{10} cscθ = \sqrt{10}$

Step by step solution

Step 1. Given Information

Given that the trigonometric function is $tanθ = - \frac{1}{3}, sinθ > 0 .$ To find the exact values of trigonometric function using equation substitution.

Step 2. Solving a equation

It implies that $cosθ < 0,$ it appears in II quadrant which is $cotθ = - 3 .$ On using the equation to solve,

$\tan^{2} θ + 1 = \sec^{2} θ secθ = \pm \sqrt{\tan^{2} θ + 1}$

On substituting the equation,

$secθ = - \sqrt{1 + {(- \frac{1}{3})}^{2}} = - \sqrt{1 + \frac{1}{9}} secθ = - \frac{\sqrt{10}}{3}$

Step 3. Solving the trigonometric function

On solving the rest of trigonometric function,

$cosθ = \frac{1}{s e c θ} = \frac{1}{- \frac{\sqrt{10}}{3}} cosθ = - \frac{3 \sqrt{10}}{10}$

On solving the equation of $sinθ = \pm \sqrt{1 - \cos^{2} θ}$

It appears in II quadrant with $sinθ > 0 .$

$sinθ = \sqrt{1 - \cos^{2} θ} = \sqrt{1 - {(- \frac{3 \sqrt{10}}{10})}^{2}} sinθ = \frac{\sqrt{10}}{10} cscθ = \frac{1}{sinθ} = \frac{1}{\frac{\sqrt{10}}{10}} cscθ = \sqrt{10}$

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Find the exact value of each of the remaining trigonometric functions of $θ .$
$tanθ = - \frac{1}{3}, sinθ > 0$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solving a equation

Step 3. Solving the trigonometric function

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Find the exact value of each of the remaining trigonometric functions of θ.tanθ=-13,sinθ>0

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solving a equation

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One App. One Place for Learning.

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Find the exact value of each of the remaining trigonometric functions of $θ .$
$tanθ = - \frac{1}{3}, sinθ > 0$