# Math Help - anotherr good old trig indentities

1. ## anotherr good old trig indentities

so i just had a math exam and i was wondering how to do this question... becausecause i did something crazy and it worked out. but i think it was wrong l

1/sin2x + 1/cos2x = (tanx + 1/tanx)^2

help would be appreciated. thanks

2. If we start off with a right angle triangle with a hypotenuse of length 1, we can use the Pythagorean theorem to get
x2 + y2 = 1
But, since
cos a = x/1 = x
sin a = y/1 = y
we can substitute to get
cos2a + sin2a = 1
The other things to remember are that tan = sin/cos, cot = cos/sin, sec = 1/cos and csc = 1/sin.
On top of that, it helps to know
cos(2a) = cos2a - sin2a
sin(2a) = 2(cos a) (sin a)
stemming from the facts that
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
Knowing these basic facts will help you solve any trig identity problems you might encounter. For example, if we take the basic squared equation above and divide both sides by cos2a we get
1 + tan2a = sec2a
Similarly, if we divide the basic squared equation by sin2a we get
cot2a + 1 = csc2a
In solving trig identity problems, I usually find that converting everything to sine and cosine keeps my life simpler. It also helps to know how to multiply and divide fractions.
Example 1: Prove sin2a + cos2a = (sina)(csca)
On the left hand side, we know that it equals 1. All we have to do is show that the right hand side is 1 as well. We know that csc = 1/sin. So,
(sina)(csca)
= (sina) x 1/(sina)
= (sina)/(sina) = 1
Sometimes, a little bit of intuition comes in handy, especially when you're missing something on one side of the equation that you have on the other. This next example is a case in point.
Example 2: Prove 2cos2a = cos(2a) + 1
There's not much we can do with the left hand side. So, let's expand the right hand side.
cos(2a) + 1
= cos2a - sin2a + 1
Isn't that wonderful? We've got a sin2 on the right hand side but not on the left. And what do we do with that 1 on the right hand side? This is where knowing your basic trig identities comes to the rescue. We can use the fact that sin2 + cos2 = 1.
cos2a - sin2a + 1
= cos2a - sin2a + sin2a + cos2a
= cos2a + cos2a
= 2cos2a
which is the same as the left hand side. Notice that in this problem, we didn't touch the left hand side at all.
This last example illustrates the point that converting everything to sin and cos and knowing how to multiply and divide fractions can make your life easier.
Example 3: Prove (csc a)/ (cot2a) = (tan a)(sec a).
They don't get much uglier than this do they? Well, lets start with the left hand side.
(csc a)/ (cot2a)
= (1/sin a)/ [(cos2a)/ (sin2a)]
= = (1/sin a) x [(sin2a)/ (cos2a)]
= (sin2a)/ [(sin a) (cos2a)]
= (sin a)/ (cos2a)
which is as far as we can go with the left hand side. Now the right hand side.
(tan a)(sec a)
= (sin a)/(cos a) x 1/(cos a)
= (sin a)/ (cos2a)
and the two sides are the same.

3. Originally Posted by iamdenis
so i just had a math exam and i was wondering how to do this question... becausecause i did something crazy and it worked out. but i think it was wrong l

1/sin2x + 1/cos2x = (tanx + 1/tanx)^2

help would be appreciated. thanks
I hope this is for you to solve for x, because it is NOT an identity. (For example, let $x = \pi/8$. Then the LHS is $2\sqrt{2}$ and the RHS is 8.)

I presume there's a typo in the question?

-Dan

4. Originally Posted by topsquark
I hope this is for you to solve for x, because it is NOT an identity. (For example, let $x = \pi/8$. Then the LHS is $2\sqrt{2}$ and the RHS is 8.)

I presume there's a typo in the question?

-Dan

this is what i actually meant, sorry

1/sin^2 x + 1/cos^2 x = (tanx + 1/tanx)^2

hopefully this changed the outcome?

5. Originally Posted by iamdenis
this is what i actually meant, sorry

1/sin^2 x + 1/cos^2 x = (tanx + 1/tanx)^2

hopefully this changed the outcome?
I will presume Mr. Patel's solution is good, but considering the spam nature of some of his other posts it might not be up for long so I'll do it over.

LHS:
$\frac{1}{sin^2(x)} + \frac{1}{cos^2(x)}$

$= \frac{cos^2(x) + sin^2(x)}{sin^2(x)cos^2(x)}$

$= \frac{1}{sin^2(x)cos^2(x)}$

RHS:
$\left ( tan(x) + \frac{1}{tan(x)} \right ) ^2$

$= \left ( \frac{sin(x)}{cos(x)} + \frac{cos(x)}{sin(x)} \right) ^2$

$= \left ( \frac{sin^2(x) + cos^2(x)}{sin(x)cos(x)} \right ) ^2$

$= \frac{1^2}{sin^2(x)cos^2(x)}$

$= \frac{1}{sin^2(x)cos^2(x)}$

which is the same as the LHS.

-Dan

6. thanks alot, thats the answer i got but i seriously made some retard step that made it work lol. i just squared all the terms as the first step.

- thanks for the help

7. Hello, iamdenis!

The identity is true if all those 2's are exponents . . .

$\frac{1}{\sin^2\!x} + \frac{1}{\cos^2\!x} \;= \;\left(\tan x + \frac{1}{\tan x}\right)^2$

The left side is: . $\frac{\cos^2\!x + \sin^2\!x}{\sin^2\!x\cos^2\!x} \;=\;\frac{1}{\sin^2\!x\cos^2\!x}$

Divide top and bottom by $\cos^4\!x\!:$ . $\frac{\frac{1}{\cos^4\!x}}{\frac{\sin^2\!x\cos^2\! x}{\cos^4\!x}}\;=\;\frac{\sec^4\!x}{\frac{\sin^2\! x}{\cos^2\!x}} \;=\;\frac{\sec^4\!x}{\tan^2\!x} \;=\;\left(\frac{\sec^2\!x}{\tan x}\right)^2$

Then: . $\left(\frac{\tan^2\!x +1}{\tan x}\right)^2 \;=\; \left(\frac{\tan^2\!x}{\tan x} + \frac{1}{\tan x}\right)^2 \;=\;\left(\tan x +\frac{1}{\tan x}\right)^2$