15Find the exact value of the trigonometric expression given that sin(u) = -where 37/2

given: sin(u)=-35 where 3π2<u<2π cos(v)=1517 where 0<v<π2 we have to find the exact…

Answered: 15 Find the exact value of the…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter7: Analytic Trigonometry

Section7.3: The Addition And Subtraction Formulas

Problem 14E

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Related questions

Question

15
Find the exact value of the trigonometric expression given that sin(u) = -
where 37/2 <u< 27, and cos(v) =
17
where 0 < v < T/2.
csc(u - v)

Expert Solution

Step 1

given:

$\sin (u) = \frac{- 3}{5}$ where $\frac{3 π}{2} < u < 2 π$

$\cos (v) = \frac{15}{17}$ where $0 < v < \frac{π}{2}$

we have to find the exact value of the trigonometric expression csc(u-v)

Step 2

$\sin (u) = \frac{- 3}{5}$ , therefore

$|\sin (u)| = |\frac{- 3}{5}| = \frac{3}{5}$

as we know that:

$|\sin (u)| = \frac{p e r p e n d i c u l a r}{h y p o t e n u s e}$

therefore,

$|\sin (u)| = \frac{p e r p e n d i c u l a r}{h y p o t e n u s e} = \frac{3}{5}$

therefore,

perpendicular=3 and hypotenuse=5

Step 3

as we know that:

${(p e r p e n d i c u l a r)}^{2} + {(b a s e)}^{2} = {(h y p o t e n u s e)}^{2}$

therefore,

$\begin{array}{rcl} {(p e r p e n d i c u l a r)}^{2} + {(b a s e)}^{2} & = & {(h y p o t e n u s e)}^{2} \\ 3^{2} + {(b a s e)}^{2} & = & 5^{2} \\ 9 + {(b a s e)}^{2} & = & 25 \\ {(b a s e)}^{2} & = & 25 - 9 \\ {(b a s e)}^{2} & = & 16 \\ (b a s e) & = & \sqrt{16} \\ (b a s e) & = & 4 \end{array}$

now as we know that:

$|\cos (u)| = \frac{b a s e}{h y p o t e n u s e}$

Step 4

therefore,

$|\cos (u)| = \frac{b a s e}{h y p o t e n u s e} = \frac{4}{5}$

as $\frac{3 π}{2} < u < 2 π$ , that implies u lies in fourth quadrant.

as we know that in fourth quadrant cosine and secant trigonometric functions have positive values and rest of the trigonometric functions have negative values.

therefore,

as we have $|\cos (u)| = \frac{4}{5}$ therefore,

$\cos (u) = \frac{4}{5}$ as u lies in fourth quadrant.

Step 5

now,

$\cos (v) = \frac{15}{17}$ , therefore

$|\cos (v)| = |\frac{15}{17}| = \frac{15}{17}$

as we know that:

$|\cos (v)| = \frac{b a s e}{h y p o t e n u s e}$