Question

If cos Θ = negative 4 over 7, what are the values of sin Θ and tan Θ?

Answer 1

what do you mean by that are you trying to find and angle?
or we are using -4/7 to find the angle of cos , sin, and tan

Answer 2

It is Pythagorean Identity

Answer 3

I can do most of it, but when the cos gets squared off, I don't understand how the fraction changes.

Answer 4

\[\cos \theta = \frac{ adj }{ hyp} \rightarrow \cos \theta=\frac{ -4 }{ 7 }\]
\[\sin \theta=\frac{ opp }{hyp } and \tan \theta=\frac{ opp }{ adj}\]

Answer 5

Are you certain that there was not more to the problem such as limitations on the angle?

Answer 6

Not that I know of.

Answer 7

If the terminal side of the angle is in the second quadrant, then the sine will be positve and the tangent negative; if the terminal side is in the third quadrant than the signs of sine and tangent would be reversed.

Answer 8

Cosine is negative in the second and third quadrants. There wasnt' something like\[90<\theta<180\]or\[180<\theta<270\]as part of the problem?

Answer 9

No, not for this one, it's done like this sample one, Substitute 1 over 4 into the Pythagorean Identity sin2 θ + cos2 θ = 1 and solve for cos θ.

sin2 θ + cos2 θ = 1

the fraction 1 over 4 squared+ cos2 θ = 1

1 over 15 + cos2 θ = 1

cos2 θ = 15 over 16

the square root of cosine squared theta equals plus or minus the square root of the fraction 15 over 16

cos θ = plus or minus the square root of 15 over 4

Once you know sin θ = 1 over 4 and cos θ = plus or minus the square root of 15 over 4 , you can find the value of tan θ.

tan θ = sine of theta over cosine of theta

tan θ = the fraction 1 over 4 over the fraction plus or minus the square root of 15 over 4

tan θ = 1 over 4 times 4 over plus or minus square root of 15

tan θ = 1 over plus or minus square root of 15

tan θ = plus or minus square root of 15 over 15

Answer 10

OK, they seem to be accepting the plus or minus as part of the answer indicating the ambiguity I have been asking about. I'll make a drawing of the situation on the next post.

Answer 11

There is also a third quadrant version of this also