Trig help please Find the exact value of each trigonometric equation

Mathematics

2 answers:

gavmur [86]3 years ago

7 0

The exact value for the equation is true but I don't really think that's the question so anyways...

- 15.) The exact form for this equation is -13pi/3 and the decimal form -13.613...

- 16.) The exact form for this equation is 23pi/4 and the decimal form 18/064...

- 17.) The exact form is -7pi/2 as the decimal is -10.995...

- 18.) The exact is -29pi/6 and the decimal is -15.184...

GaryK [48]3 years ago

6 0

We know any trig problem that asks for exact values probably has something to do with 30° or 45° and their multiples. That's $\pi/6$ and $\pi/4$ ; we're apparently doing radians in this one.

General rules off the top of my head: Coterminal angles (gotten by adding or subtracting multiples of 2π) have the same values for their trig functions , cosine is even, sine is odd, cosine negate supplementary angles, sine of supplementary angles is unchanged, and the cosine of an angle is the sine of the complementary angle.

$\cos (- \frac{13\pi}{3}) = \cos( 13\pi/3-6(2\pi)) =\cos(\pi/3) = \frac 1 2$

$\csc(\frac{23 \pi}{4}) = \dfrac{1}{\sin (23\pi/4 - 3(8\pi/4))} = \dfrac{1}{\sin(-\pi/4)}= \dfrac{1}{- 1 /\sqrt{2}} = - \sqrt{2}$

$\sec(-\frac {7 \pi}{2}) = \dfrac{1}{\cos(-7\pi/2+ (4/2)(2\pi) )}= \dfrac{1}{\cos(\pi/2)} = \dfrac 1 0$

That one is undefined

$\cot(-\frac{29\pi}{6}) = \cot(-29\pi/6 + (18/6) (2 \pi)) = \cot(7\pi/6) \\= \tan(\pi/2 - 7\pi/6) = \tan(-4\pi/6)= \tan(-2\pi/3 + \pi) = \tan(\pi/3)= \sqrt{3}$