If sin(theta)=X, then express sec(theta) in terms of x

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

If sin(theta)=X, then express sec(theta) in terms of x

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

The answer I got was \[\sqrt{1-(1/x)^2}\] Would this be correct?
|dw:1438813890812:dw| use pythagorean theorem to find the other side
and recall for secant use hyp/adj

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

I'm sorry, I'm a bit confused where to go from there
can you find the adjacent side by use of Pythagorean theorem ?
\[(\text{ adjacent })^2+x^2=1^2 \\ \text{ solve for adjacent }\]
let me know if you don't know how maybe I can give you a step
adjacent= -x+1?
\[( \text{ adjacent } )^2+x^2=1 \\ \text{ subtract} x^2 \text{ on both sides } \\ ( \text{ adjacent })^2=1-x^2 \\ \text{ take square root of both sides } \\ \text{ adjacent } = \pm \sqrt{1-x^2}\] you cannot simplify this ... if that is what you tried to do
|dw:1438814521770:dw| now find sec(theta) using the right triangle
The whole negative thing kinda threw me off. Thanks
What I tried to do initially was rewrite the identity sin^2+Cos^2=1 and go from there
you can use that too
\[\sin^2(\theta)+\cos^2(\theta)=1 \\ \text{ we are given } \sin(\theta)=x \text{ so } \sin^2(\theta)=x^2 \\ x^2+\cos^2(\theta)=1 \\ \\ \text{ subtract} x^2 \text{ on both sides } \\ \cos^2(\theta)=1-x^2 \\ \text{ take square root of both sides } \\ \cos(\theta)=\pm \sqrt{1-x^2}\] then just flip both sides to find sec(theta)
its the same thing really
Ooh, I tried to use the recirpocals first, hence why I had (1/x)^2 Thanks a bunch mate (:
np
well csc^2(x) would actually be (1/x)^2
Yeah, I figured if the identity works with sin and cos, it'd work with their reciprocals. I see the error
Do you think you might be able to help me with one more question?
i can try
Evaluate cos1+cos2+cos3+...+cos357+cos358+cos359 (all measurements are in degrees) The only thing I can imagine is that it repeats at a certain point or something
\[\cos(x)+\cos(180-x) \\=\cos(x)+\cos(180)\cos(x)-\sin(180)\sin(x) \\ =\cos(x)-1\cos(x)-0\sin(x)\\ =0\] So I think maybe we can use cos(x)+cos(180-x)
So I think maybe we can use cos(x)+cos(180-x)=0*
\[\cos(1)+\cos(179)=0 \\ \cos(2)+\cos(178)=0 \\ \cos(3)+\cos(177)=0 \\ \cos(4)+\cos(176) =0 \\ \cdots \\ \] and I have to come back to finish this i'm being called away unfortunately
Alright
notice above I put a minus but it doesn't matter that one part is still 0 \[\cos(x)+\cos(180-x) \\=\cos(x)+\cos(180)\cos(x)+\sin(180)\sin(x) \\ =\cos(x)-1\cos(x)+0\sin(x)\\ =0\] hmm... \[\cos(180)=-1 \\ \text{ hmmm... we need to think now for the rest of the finite series }\] like we already know cos(1)+cos(2)+...+cos(180)=-1 we need to figure out would to do with cos(181)+cos(182)+cos(183)+...+cos(359) we should be able to think of something similar
how about for that other sign try to see if something like cos(180+x)+cos(360-x)=0 works.. cos(180)cos(x)-sin(180)sin(x)+cos(360)cos(x)+sin(360)sin(x) =-cos(x)+cos(x)=0

Not the answer you are looking for?

Search for more explanations.

Ask your own question