Question

Найти sin a, tg a, ctg a, если cos a = -(8/17), п/2 < a < п

Answer 1

Ответ и Объяснение:

Информация: Верны следующие тригонометрические равенства:

$\tt \displaystyle sina=\pm \sqrt{1-cos^2a}; \; tga=\frac{sina}{cosa} ;\; ctga=\frac{cosa}{sina} .$

Решение. Известно $\dfrac{\pi }{2} < a < \pi$ и $\tt \displaystyle cosa=-\frac{8}{17}.$

При $\dfrac{\pi }{2} < a < \pi$, то есть во втором квадранте sina > 0. Поэтому

$\tt \displaystyle sina=\sqrt{1-cos^2a}=\sqrt{1- \left (-\frac{8}{17} \right )^2}=\sqrt{1- \frac{64}{289} }=\sqrt{\frac{225}{289} }=\frac{15}{17}.$

Теперь можем найти tga и ctga:

$\tt \displaystyle tga=\frac{\dfrac{15}{17} }{-\dfrac{8}{17} } =-\frac{15}{8} ; \; ctga=\frac{-\dfrac{8}{17} }{\dfrac{15}{17} } =-\frac{8}{15}.$

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Answer 2

To find the values of sine (sin a), tangent (tan a), and cotangent (cot a) given that cos a = -(8/17) and π/2 < a < π, we can use the trigonometric identity:

sin^2 a + cos^2 a = 1

First, let's find sin a:

We know cos a = -(8/17)

Using the trigonometric identity, sin^2 a + cos^2 a = 1

(sin a)^2 + (-(8/17))^2 = 1

(sin a)^2 + 64/289 = 1

(sin a)^2 = 1 - 64/289

(sin a)^2 = (289 - 64) / 289

(sin a)^2 = 225 / 289

sin a = ±√(225 / 289)

sin a = ±(15 / 17)

Since π/2 < a < π, a lies in the second quadrant where the sine is positive. Therefore,

sin a = 15 / 17

Next, let's find tan a:

tan a = sin a / cos a

tan a = (15 / 17) / (-(8 / 17))

tan a = -(15 / 8)

tan a = -1.875

Finally, let's find cot a:

cot a = 1 / tan a

cot a = 1 / -(15 / 8)

cot a = -8 / 15

So, the values are:

sin a = 15 / 17 tan a = -15 / 8 cot a = -8 / 15