You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Trig identities - help quick
- Thread starter TubaKid22
- Start date
These identities can be proved by strategic factoring, which I figure is the point of them. Feel free to be creative and use different approaches.
1. LHS: factor out \(\displaystyle \mbox{\cos{x}}\).
2. LHS: factor out \(\displaystyle \mbox{\frac{1}{\sin{x}}}\).
3. LHS: what might we factor out of the left-hand parentheses?
Enjoy.
1. LHS: factor out \(\displaystyle \mbox{\cos{x}}\).
2. LHS: factor out \(\displaystyle \mbox{\frac{1}{\sin{x}}}\).
3. LHS: what might we factor out of the left-hand parentheses?
Enjoy.
We consider each side of the equation separately (because, well, we need to show it is indeed an equation).
The LHS of the first one factorises:
\(\displaystyle \mbox{LHS = \cos{x}\cdot\tan^2{x} + \cos{x} = \cos{x}\left(tan^2{x} + 1\right)}\)
The LHS of the first one factorises:
\(\displaystyle \mbox{LHS = \cos{x}\cdot\tan^2{x} + \cos{x} = \cos{x}\left(tan^2{x} + 1\right)}\)
OK, so to verify you use the other trig identities, but that is where i get mixed up. Some how they dont come up even. here is my work so far
cos x tan ^2 x +cos x = sec x
1/sin x (times) sec ^2 x = sec x
sec x = sec x
is that right, or what is going on
cos x tan ^2 x +cos x = sec x
1/sin x (times) sec ^2 x = sec x
sec x = sec x
is that right, or what is going on
You're skipping steps. Write it down as though someone else was reading who could not do the intermediate steps mentally.TubaKid22 said:OK, so to verify you use the other trig identities, but that is where i get mixed up. Some how they dont come up even. here is my work so far
cos x tan ^2 x +cos x = sec x
1/sin x (times) sec ^2 x = sec x
sec x = sec x
is that right, or what is going on
And how about you say
LHS = . . .
LHS = . . .
.
.
LHS = RHS
Saying sec(x) = sec(x) doesn't, on appearance (albeit to someone being myopic), prove anything; the "LHS/RHS" style shows the original LHS = cos(x) = RHS more apparently.
TubaKid22 said:on the third problem my work is
(sec x - tan x)(1+sin x)=cos x
(1/cos x - sin x/ cos x)(1+sin x) = cos x
Now factor out 1/cos(x) from the left-hand parentheses..
The one in my post at 4.05pm.TubaKid22 said:what steps do i need to add in and how would i go about writing that
3.
\(\displaystyle \mbox{ LHS = \left(\sec{x} - \tan{x}\right)\left(1 + \sin{x}\right) \\
\\
\\
\\
= \left(\frac{1}{\cos{x}} - \frac{\sin{x}}{\cos{x}}\right)\left(1 + \sin{x}\right) \\
\\
\\
\\
= \frac{1}{\cos{x}}\left(1 - \sin{x}\right)\left(1 + \sin{x}\right)} \\\)
\(\displaystyle \mbox{ LHS = \left(\sec{x} - \tan{x}\right)\left(1 + \sin{x}\right) \\
\\
\\
\\
= \left(\frac{1}{\cos{x}} - \frac{\sin{x}}{\cos{x}}\right)\left(1 + \sin{x}\right) \\
\\
\\
\\
= \frac{1}{\cos{x}}\left(1 - \sin{x}\right)\left(1 + \sin{x}\right)} \\\)
There is never one approach you >should< take. Let the creative juices flow . . .
You might go with my original suggestion.
\(\displaystyle \mbox{LHS = \csc{x} - \cot{x}\cos{x}}\)
\(\displaystyle \mbox{ = \frac{1}{\sin{x}} - \frac{\cos{x}}{\sin{x}} \cos{x}}\)
\(\displaystyle \mbox{ = \frac{1}{\sin{x}} - \frac{\cos^2{x}}{\sin{x}}}\)
I count at least three steps for you to write out.
You might go with my original suggestion.
\(\displaystyle \mbox{LHS = \csc{x} - \cot{x}\cos{x}}\)
\(\displaystyle \mbox{ = \frac{1}{\sin{x}} - \frac{\cos{x}}{\sin{x}} \cos{x}}\)
\(\displaystyle \mbox{ = \frac{1}{\sin{x}} - \frac{\cos^2{x}}{\sin{x}}}\)
I count at least three steps for you to write out.