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2 years ago

Find the exact value of the indicated trigonometric function of θ.

Mathematics

1 answer:

Mariana [72]2 years ago

8 0

Answer:

cosθ = -√(35/36)

tanθ = -1/√35

Step-by-step explanation:

We know that

sin²θ + cos²θ = 1

(1/6)² + cos²θ = 1

1 - (1/6)² = cos²θ

35/36 = cos²θ

sec θ = 1/ cosθ

1/ secθ = cos θ

Because secθ < 0, 1/ secθ would also be < 0, so cos θ < 0

cosθ = -√(35/36)

tanθ = sinθ/cosθ = (1/6)/-√(35/36)

= (1/6) / -(√35/6)

= -1/√35

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2 years ago

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Given sin(u)= -7/25 and cos(v) = -4/5, what is the exact value of cos(u-v) if both angles are in quadrant 3

solmaris [256]

Given:

$\sin (u)=-\dfrac{7}{25}$

$\cos (v)=-\dfrac{4}{5}$

To find:

The exact value of cos(u-v) if both angles are in quadrant 3.

Solution:

In 3rd quadrant, cos and sin both trigonometric ratios are negative.

We have,

$\sin (u)=-\dfrac{7}{25}$

$\cos (v)=-\dfrac{4}{5}$

Now,

$\cos (u)=-\sqrt{1-\sin^2 (u)}$

$\cos (u)=-\sqrt{1-(-\dfrac{7}{25})^2}$

$\cos (u)=-\sqrt{1-\dfrac{49}{625}}$

$\cos (u)=-\sqrt{\dfrac{625-49}{625}}$

On further simplification, we get

$\cos (u)=-\sqrt{\dfrac{576}{625}}$

$\cos (u)=-\dfrac{24}{25}$

Similarly,

$\sin (v)=-\sqrt{1-\cos^2 (v)}$

$\sin (v)=-\sqrt{1-(-\dfrac{4}{5})^2}$

$\sin (v)=-\sqrt{1-\dfrac{16}{25}}$

$\sin (v)=-\sqrt{\dfrac{25-16}{25}}$

$\sin (v)=-\sqrt{\dfrac{9}{25}}$

$\sin (v)=-\dfrac{3}{5}$

Now,

$\cos (u-v)=\cos u\cos v+\sin u\sin v$

$\cos (u-v)=\left(-\dfrac{24}{25}\right)\left(-\dfrac{4}{5}\right)+\left(-\dfrac{7}{25}\right)\left(-\dfrac{3}{25}\right)$

$\cos (u-v)=\dfrac{96}{625}+\dfrac{21}{625}$

$\cos (u-v)=\dfrac{1 17}{625}$

Therefore, the value of cos (u-v) is 0.1872.

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2 years ago