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3 years ago

Given: cos θ=-4/5, sin x = -12/13, θ is in the third quadrant,

Mathematics

1 answer:

USPshnik [31]3 years ago

4 0

By definition of tangent,

tan(2θ) = sin(2θ) / cos(2θ)

Recall the double angle identities:

sin(2θ) = 2 sin(θ) cos(θ)

cos(2θ) = cos²(θ) - sin²(θ) = 2 cos²(θ) - 1

where the latter equality follows from the Pythagorean identity, cos²(θ) + sin²(θ) = 1. From this identity we can solve for the unknown value of sin(θ):

sin(θ) = ± √(1 - cos²(θ))

and the sign of sin(θ) is determined by the quadrant in which the angle terminates.

We're given that θ belongs to the third quadrant, for which both sin(θ) and cos(θ) are negative. So if cos(θ) = -4/5, we get

sin(θ) = - √(1 - (-4/5)²) = -3/5

Then

tan(2θ) = sin(2θ) / cos(2θ)

tan(2θ) = (2 sin(θ) cos(θ)) / (2 cos²(θ) - 1)

tan(2θ) = (2 (-3/5) (-4/5)) / (2 (-4/5)² - 1)

tan(2θ) = 24/7

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and for domain : 3 x + 6 ≠ 0
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3 years ago

Which of these is the graph of |x|=3

Luda [366]

Answer:

it is b

Step-by-step explanation:

Absolute value is how far a number is from 0, and since its not an inequality there shouldn't be arrows. It can only be -3 or 3 and nothing in between.

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3 years ago

Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first

BaLLatris [955]

Answer:

$\cos(x+y)$ goes with $-\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin(x+y)$ goes with $\frac{\sqrt{6}-\sqrt{2}}{4}$

$\tan(x+y)$ goes with $\sqrt{3}-2$

Step-by-step explanation:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

We are given:

$\sin(x)=\frac{\sqrt{2}}{2}$ which if we look at the unit circle we should see

$\cos(x)=\frac{\sqrt{2}}{2}$ .

We are also given:

$\cos(y)=\frac{-1}{2}$ which if we look the unit circle we should see

$\sin(y)=\frac{\sqrt{3}}{2}$ .

Apply both of these given to:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}$

$\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$

$\frac{-\sqrt{2}-\sqrt{6}}{4}$

$-\frac{\sqrt{6}+\sqrt{2}}{4}$

Apply both of the givens to:

$\sin(x+y)$

$\sin(x)\cos(y)+\sin(y)\cos(x)$ by addition identity for sine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}$

$\frac{-\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Now I'm going to apply what 2 things we got previously to:

$\tan(x+y)$

$\frac{\sin(x+y)}{\cos(x+y)}$ by quotient identity for tangent

$\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}$

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

$-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}$

$-\frac{8-2\sqrt{12}}{4}$

There is a perfect square in 12, 4.

$-\frac{8-2\sqrt{4}\sqrt{3}}{4}$

$-\frac{8-2(2)\sqrt{3}}{4}$

$-\frac{8-4\sqrt{3}}{4}$

Divide top and bottom by 4 to reduce fraction:

$-\frac{2-\sqrt{3}}{1}$

$-(2-\sqrt{3})$

Distribute:

$\sqrt{3}-2$

6 0

3 years ago

PLEASE HELP!!!!

Strike441 [17]

1 Dog weighs 50lbs because in the stem and leaf plot it says there is a 50lbs dog and a 55lbs dog.

5 0

3 years ago

yawa3891 [41]

Answer:

TBH I would go with either B or D.

Step-by-step explanation:

Well first off your going to 24 miles not 96 or 99 and it doesn't say how much they went on Monday or Tuesday

8 0

3 years ago

Read 2 more answers

Given: cos θ=-4/5, sin x = -12/13, θ is ﻿in ﻿the ﻿third ﻿quadrant, ﻿

Given: cos θ=-4/5, sin x = -12/13, θ is in the third quadrant,