I just started studying my trig for my summer class which will start in a few weeks but I'am going crazy over a a certain indenity problem. It seems simple enough: "Express sin t in terms of sec t".I just can't get it right though I come close. I have the answer in the book but I can't figure out how to get it! I hope that one of you would be able to show me the steps so I can stop fretting over it and resume my studies.
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Having trouble with the fundamental identities of trig.
- Thread starter Al72579
- Start date
Hello, Al72579!
You're expected to know the basic identities and their variations.
The reciprocal identities:
. . \(\displaystyle \sin\theta\:=\:\frac{1}{\csc\theta}\;\;\;\;\csc\theta \:=\:\frac{1}{\sin\theta}\)
. . \(\displaystyle \cos\theta\:=\:\frac{1}{\sec\theta}\;\;\;\;\sec\theta\:=\:\frac{1}{\cos\theta}\)
. . \(\displaystyle \tan\theta\:=\:\frac{1}{\cot\theta}\;\;\;\;\cot\theta\:=\:\frac{1}{\tan\theta}\)
The "Pythagorean Identities":
. . \(\displaystyle \sin^2\theta\,+\,\cos^2\theta\:=\:1\;\) [1]
. . \(\displaystyle \sec^2\theta\:=\:\tan^2\theta\,+\,1\;\) [2]
These two have a variety of variations.
For example, [1] can be rearranged like this:
. . . .\(\displaystyle \sin^2\theta\,+\,\cos^2\theta\:=\:1\;\;\Rightarrow\;\;\sin^2\theta \:=\:1\,-\,\cos^2\theta\;\;\Rightarrow\;\;\sin\theta\:=\:\sqrt{1\,-\,\cos^2\theta}\)
or: \(\displaystyle \:\sin^2\theta\,+\,\cos^2\theta\:=\:1\;\;\Rightarrow\;\;\cos^2\theta\:=\:1\,-\,\sin^2\theta\;\;\Rightarrow\;\;\cos\theta\:=\:\sqrt{1\,-\,\sin^2\theta}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
We are given: \(\displaystyle \,\sin\theta\) . . . and we want it in terms of \(\displaystyle sec\theta\)
We can write: \(\displaystyle \L\:\sin\theta\;=\;\sqrt{1\,-\,\cos^2\theta}\) . . . right?
. . and we know that: \(\displaystyle \,\cos\theta\:=\:\frac{1}{\sec\theta}\)
Substitute: \(\displaystyle \L\:\sin\theta\;=\;\sqrt{1\,-\,\left(\frac{1}{\sec\theta}\right)^2}\;\) . . . There!
Express \(\displaystyle \sin\theta\) in terms of \(\displaystyle \sec\theta\)
You're expected to know the basic identities and their variations.
The reciprocal identities:
. . \(\displaystyle \sin\theta\:=\:\frac{1}{\csc\theta}\;\;\;\;\csc\theta \:=\:\frac{1}{\sin\theta}\)
. . \(\displaystyle \cos\theta\:=\:\frac{1}{\sec\theta}\;\;\;\;\sec\theta\:=\:\frac{1}{\cos\theta}\)
. . \(\displaystyle \tan\theta\:=\:\frac{1}{\cot\theta}\;\;\;\;\cot\theta\:=\:\frac{1}{\tan\theta}\)
The "Pythagorean Identities":
. . \(\displaystyle \sin^2\theta\,+\,\cos^2\theta\:=\:1\;\) [1]
. . \(\displaystyle \sec^2\theta\:=\:\tan^2\theta\,+\,1\;\) [2]
These two have a variety of variations.
For example, [1] can be rearranged like this:
. . . .\(\displaystyle \sin^2\theta\,+\,\cos^2\theta\:=\:1\;\;\Rightarrow\;\;\sin^2\theta \:=\:1\,-\,\cos^2\theta\;\;\Rightarrow\;\;\sin\theta\:=\:\sqrt{1\,-\,\cos^2\theta}\)
or: \(\displaystyle \:\sin^2\theta\,+\,\cos^2\theta\:=\:1\;\;\Rightarrow\;\;\cos^2\theta\:=\:1\,-\,\sin^2\theta\;\;\Rightarrow\;\;\cos\theta\:=\:\sqrt{1\,-\,\sin^2\theta}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
We are given: \(\displaystyle \,\sin\theta\) . . . and we want it in terms of \(\displaystyle sec\theta\)
We can write: \(\displaystyle \L\:\sin\theta\;=\;\sqrt{1\,-\,\cos^2\theta}\) . . . right?
. . and we know that: \(\displaystyle \,\cos\theta\:=\:\frac{1}{\sec\theta}\)
Substitute: \(\displaystyle \L\:\sin\theta\;=\;\sqrt{1\,-\,\left(\frac{1}{\sec\theta}\right)^2}\;\) . . . There!
Al72579 said:I just started studying my trig for my summer class which will start in a few weeks but I'am going crazy over a a certain indenity problem. It seems simple enough: "Express sin t in terms of sec t".I just can't get it right though I come close. I have the answer in the book but I can't figure out how to get it! I hope that one of you would be able to show me the steps so I can stop fretting over it and resume my studies.
You are expected to know the basic Pythagorean identity
sin<SUP>2</SUP>t + cos<SUP>2</SUP>t = 1
You're also expected to know that sec t = 1/cos t, so cos t = 1 / sec t
sin<SUP>2</SUP>t + (1 / sec t)<SUP>2</SUP> = 1
Now....solve that for sin t.....
(you're too fast for me, Soroban)