If cos x = -7/25 in quadrant 2, what is tan and sin? Please help. I want to check if I have the right answers.

Question

2. tan x = (sq. root 3)/4 in quadrant 3
sin and cos?

3. sin x = -5/3 in quadrant 4
cos and tan?

It's really important that I make sure I have the right answers. I have to use sin, cos, and tan to plug them in trig equations. It I get mess up one thing, I can everything wrong b/c it screws up every single equation.

Answer 1

1. sin x = 24/25
tan x = -7/24
2. sin x = -(root 19*root 3)/19
cos x = -(4*root 19)/19
3. cos x = 4/5
tan x = 3/4

Answer 2

Please draw the following diagrams so you can see
what's happening. Going back to basics has always
been the the best way for me to understand a subject.

1) cos x = -7/25 in quadrant 2.

Draw X & Y axes and a triangle in quadrant 2.
Calculate x = arccos(-7/25) = 106.3º.
This is already in quadrant 2.
The angle of the triangle at the origin, that we are
interested in, is therefore, 180º - 106.3º = 73.7º.
Cos = adjacent / hypotenuse, so the
x-value side of the triangle is -7 units.
The hypotenuse is 25 units (always positive).
The third side is found using a^2 + b^2 = c^2,
so, a^2 + 7^2 = 25^2, therefore, a = ± 24 units,
but we take only the positive value, because this
vertical side of the triangle is above the X-axis.
Now we can find sin x from the diagram.
It will be 24/25. This tallies with the knowledge
that sin is positive in quadrant 2.
Tan x is just sin x / cos x, which is
(24/25) / (-7/25) = -24/7.
And we know that tan is negative in quadrant 2.
Summarising : sin x = 24/25 and tan x = -24/7.

2) tan x = sqrt(3)/4 in quadrant 3.

Draw another diagram again.
Calculate x = arctan[sqrt(3)/4] = 23.4º
This is in quadrant 1. To get the same angled triangle
in quadrant 3, it's necessary to add 180º, so the whole
angle is really 203.4º, but we only have to work on the
23.4º angled triangle in quadrant 3.
Tan is opposite / adjacent,
so opposite = -sqrt(3) and adjacent = -4.
Note from the diagram that these must be both negative.
The hypotenuse must therefore be :
sqrt{[sqrt(3)]^2 + 4^2} = sqrt(19) (always positive).
Sin will then be opposite / hypotenuse
= -sqrt(3) / sqrt(19) = -sqrt(57) / 19
Note: sin is negative in quadrant 3.
Cos is adjacent / hypotenuse
= -4 / sqrt(19) = -4 * sqrt(19) / 19,
noting that cos is negative in quadrant 3.
Summarising: sin x = -sqrt(57) / 19
and cos x = -4 * sqrt(19) / 19.

3) sin x = -5/3 in quadrant 4

This is not valid as sin x
must be >= -1 and <= 1.

If you really meant sin x = -3/5 instead,
then opposite = -3, hypotenuse = 5 and
by Pythagoras, adjacent = 4.
So, cos x = 4/5 and tan x = -3/4.

Answer 3

1. The triangle is a 7 24 25 right triangle. If the cosine has magnitude 7/25, then the sine must have magnitude 24/25, and the tangent is the quotient of these, i.e. 24/7. Both sine and tangent will be positive, as we are above the x-axis.

2. Find the hypotenuse of the tangent triangle by Pythagoras:
h = sqrt(1 + 3/16) = sqrt(19)/4. With all three sides known, division will then yield the sine and cosine.

3. Are imaginary. No real angle has a sine of magnitude greater than 1.

Answer 4

cosx=adjacent/hypotenuse
adjacent=7
hypotenuse=25
a^2+b^2=c^2
49+b^2=625
b^2=576
b=24=opposite

Quadrant 2 rules:
sinx is positive
cosx is negative
tanx is negative

sinx=opposite/hypotenuse=24/25
tanx=opposite/adjacent=-24/7

Quadrant 3 rules:
sinx is negative
cosx is negative
tanx is positive

Quadrant 4 rules:
sinx is negative
cosx is positive
tanx is negative

Hopefully this will help you do the rest.

Answer 5

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