# One More Trig Identity Problem

• DethRose
In summary: So, while it is definitely possible to do these problems without the formulas, they may become much more difficult later on."In summary, the person is having trouble with a trig assignment and asks for help. They explain that they need to find sinx, cosx, tanx, sin2x, cos2x, and tan2x. They also ask if anyone could help them with this problem. The person then provides the answers to the questions and the person asks for help again.
DethRose
There is one last problem i have on my trig assignment and i have no clue how to do it. The questions is:

Find sinx, cosx, tanx, sin2x, cos2x, and tan2x from the given information:

secx=5, sin is negative.

If anyone could show me how to do this problem it would be soooo appreciated.

Thanks

$$\sec x = 5 \Rightarrow \frac{1}{\sec x} = \frac{1}{5} = \cos x$$

$$\cos x$$ is positive and $$\sin x$$ is negative in the fourth quadrant.

$$\sin 2x = 2\sin x \cos x$$

$$\cos 2x = 1-2\sin^{2} x$$

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so how do i find sinx cause if you go from cos wouldn't you have to express sin in terms of tan since cos=sin/tan?

draw a right triangle. If you know the adjacent side is 1, and the hypotenuse is 5, then the opposite side is $$2\sqrt{6}$$.

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I don't understand how that will help?

not that I am saying your wrong lol I am just saying i don't understand how to do that

ah i get you

wouldnt the other side be square root of 24 if you use pythagorins theorem?

yes it would. sorry, a typo.

Ok just to make sure I am doing it correctly here are the answers i got, would be great if you could tell me if i totally messed it up lol.

I got:

sinx= -2 squareroot 6/5
tanx= -2 squareroot 6
sin2x= -4 squareroot 6/25
cos2x= 23/25
tan2x= -4 squareroot 6/5

thanks for all the help!

$$\cos 2x = -\frac{23}{25}$$

$$\tan 2x = \frac{\sin 2x}{\cos 2x}$$

All the rest look good.

why are those 2 like that?

how can cosx and cos2x both be 1/5?

Also here's how i did those 2 questions I am just wondering why they are wrong?

cos2x=1-2sin^2x
=1-2(-2 squareroot 6/5)squared
=1-2(24/25)
=23/25

tan2x=2tanx/1-tan^2x
=2(-2 squareroot 6)/1-6
=-4 squareroot 6/-5
=4 squareroot 6/5

cos2x=1-2sin^2x
=1-2(-2 squareroot 6/5)squared
=1-2(24/25)
=23/25
I said that it should be a -23/25.

And you are right.

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"$$\sin 2x = 2\sin x \cos x$$"

I think this might be a problem for some, because I was doing problems similar to this the other day, and only today did we begin learning the double-angle, half-angle, and power down formulas.

## 1) What is a trigonometric identity?

A trigonometric identity is a mathematical equation involving trigonometric functions that is always true for all values of the variables involved. Some common examples include the Pythagorean identity (sin²θ + cos²θ = 1) and the double angle identities (sin(2θ) = 2sinθcosθ).

## 2) What is the purpose of solving trigonometric identities?

Solving trigonometric identities helps to simplify and manipulate complex trigonometric expressions, making them easier to work with in various mathematical applications. It is also an important skill for understanding and proving other mathematical concepts.

## 3) How do I approach solving a trigonometric identity problem?

The most common approach is to use known identities and algebraic manipulations to transform one side of the equation into the other. It can also be helpful to look for patterns and use basic trigonometric identities as a starting point.

## 4) What are some common tips for solving trigonometric identities?

Some tips include using the unit circle to visualize and understand trigonometric functions, simplifying expressions using basic trigonometric identities, and being familiar with common trigonometric formulas and conversions. It can also be helpful to practice and become comfortable with algebraic manipulations.

## 5) Can a trigonometric identity problem have multiple solutions?

Yes, some trigonometric identities have multiple solutions or can be solved in different ways. It is important to check your work and make sure that your solution satisfies the original equation, as there may be extraneous solutions that do not work.

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